Quizbank/Electricity and Magnetism: Gauss' Law/T2

calcPhyEM_2GaussQuizzes/T2 ID153728160820

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Exams:  A0  A1  A2   B0  B1  B2   C0  C1  C2   D0  D1  D2   E0  E1  E2   F0  F1  F2   G0  G1  G2   H0  H1  H2   I0  I1  I2   J0  J1  J2   K0  K1  K2   L0  L1  L2   M0  M1  M2   N0  N1  N2   O0  O1  O2   P0  P1  P2   Q0  Q1  Q2   R0  R1  R2   S0  S1  S2   T0  T1  T2

Answers:  A0  A1  A2   B0  B1  B2   C0  C1  C2   D0  D1  D2   E0  E1  E2   F0  F1  F2   G0  G1  G2   H0  H1  H2   I0  I1  I2   J0  J1  J2   K0  K1  K2   L0  L1  L2   M0  M1  M2   N0  N1  N2   O0  O1  O2   P0  P1  P2   Q0  Q1  Q2   R0  R1  R2   S0  S1  S2   T0  T1  T2

60 Tests = 3 versions x 20 variations: Each of the 20 variations (A, B, ...) represents a different random selection of questions taken from the |study guide.The 3 versions (0,1,..) all have the same questions but in different order and with different numerical inputs. Unless all students take  version "0" it is best to reserve it for the instructor because the questions are grouped according to the order in which they appear on the study guide.

Links:  Quizbank/Instructions   |Study guide    file:QB-calcPhyEM_2GaussQuizzes-T2.pdf

Contact me at User talk:Guy vandegrift if you need any help.

T2 A0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant magnitude over a portion of the Gaussian surface
 * b) constant direction over a portion of the Gaussian surface
 * c) constant direction and magnitude over the entire Gaussian surface
 * d) constant in direction over the entire Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 A1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant magnitude over a portion of the Gaussian surface
 * b) constant in direction over the entire Gaussian surface
 * c) constant direction over a portion of the Gaussian surface
 * d) constant direction and magnitude over the entire Gaussian surface

T2 A2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction over a portion of the Gaussian surface
 * b) constant magnitude over a portion of the Gaussian surface
 * c) constant in direction over the entire Gaussian surface
 * d) constant direction and magnitude over the entire Gaussian surface

T2 B0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant magnitude over a portion of the Gaussian surface
 * b) constant in direction over the entire Gaussian surface
 * c) constant direction over a portion of the Gaussian surface
 * d) constant direction and magnitude over the entire Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 B1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant in direction over the entire Gaussian surface
 * b) constant magnitude over a portion of the Gaussian surface
 * c) constant direction and magnitude over the entire Gaussian surface
 * d) constant direction over a portion of the Gaussian surface

T2 B2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction and magnitude over the entire Gaussian surface
 * b) constant in direction over the entire Gaussian surface
 * c) constant direction over a portion of the Gaussian surface
 * d) constant magnitude over a portion of the Gaussian surface

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

T2 C0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction over a portion of the Gaussian surface
 * b) constant in direction over the entire Gaussian surface
 * c) constant direction and magnitude over the entire Gaussian surface
 * d) constant magnitude over a portion of the Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 C1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant in direction over the entire Gaussian surface
 * b) constant direction and magnitude over the entire Gaussian surface
 * c) constant direction over a portion of the Gaussian surface
 * d) constant magnitude over a portion of the Gaussian surface

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

T2 C2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant magnitude over a portion of the Gaussian surface
 * b) constant direction over a portion of the Gaussian surface
 * c) constant direction and magnitude over the entire Gaussian surface
 * d) constant in direction over the entire Gaussian surface

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 D0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction and magnitude over the entire Gaussian surface
 * b) constant in direction over the entire Gaussian surface
 * c) constant direction over a portion of the Gaussian surface
 * d) constant magnitude over a portion of the Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

T2 D1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction over a portion of the Gaussian surface
 * b) constant magnitude over a portion of the Gaussian surface
 * c) constant direction and magnitude over the entire Gaussian surface
 * d) constant in direction over the entire Gaussian surface

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

T2 D2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant in direction over the entire Gaussian surface
 * b) constant direction over a portion of the Gaussian surface
 * c) constant magnitude over a portion of the Gaussian surface
 * d) constant direction and magnitude over the entire Gaussian surface

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

T2 E0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction and magnitude over the entire Gaussian surface
 * b) constant in direction over the entire Gaussian surface
 * c) constant magnitude over a portion of the Gaussian surface
 * d) constant direction over a portion of the Gaussian surface

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 E1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction over a portion of the Gaussian surface
 * b) constant direction and magnitude over the entire Gaussian surface
 * c) constant in direction over the entire Gaussian surface
 * d) constant magnitude over a portion of the Gaussian surface

T2 E2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant magnitude over a portion of the Gaussian surface
 * b) constant in direction over the entire Gaussian surface
 * c) constant direction over a portion of the Gaussian surface
 * d) constant direction and magnitude over the entire Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 F0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 F1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

T2 F2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

T2 G0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 G1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

T2 G2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

T2 H0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction and magnitude over the entire Gaussian surface
 * b) constant magnitude over a portion of the Gaussian surface
 * c) constant direction over a portion of the Gaussian surface
 * d) constant in direction over the entire Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 H1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant magnitude over a portion of the Gaussian surface
 * b) constant in direction over the entire Gaussian surface
 * c) constant direction over a portion of the Gaussian surface
 * d) constant direction and magnitude over the entire Gaussian surface

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

T2 H2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction and magnitude over the entire Gaussian surface
 * b) constant direction over a portion of the Gaussian surface
 * c) constant in direction over the entire Gaussian surface
 * d) constant magnitude over a portion of the Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

T2 I0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction and magnitude over the entire Gaussian surface
 * b) constant magnitude over a portion of the Gaussian surface
 * c) constant in direction over the entire Gaussian surface
 * d) constant direction over a portion of the Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 I1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant magnitude over a portion of the Gaussian surface
 * b) constant direction over a portion of the Gaussian surface
 * c) constant direction and magnitude over the entire Gaussian surface
 * d) constant in direction over the entire Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

T2 I2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant in direction over the entire Gaussian surface
 * b) constant direction and magnitude over the entire Gaussian surface
 * c) constant magnitude over a portion of the Gaussian surface
 * d) constant direction over a portion of the Gaussian surface

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

T2 J0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant magnitude over a portion of the Gaussian surface
 * b) constant direction over a portion of the Gaussian surface
 * c) constant in direction over the entire Gaussian surface
 * d) constant direction and magnitude over the entire Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 J1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction and magnitude over the entire Gaussian surface
 * b) constant in direction over the entire Gaussian surface
 * c) constant magnitude over a portion of the Gaussian surface
 * d) constant direction over a portion of the Gaussian surface

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

T2 J2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction and magnitude over the entire Gaussian surface
 * b) constant magnitude over a portion of the Gaussian surface
 * c) constant direction over a portion of the Gaussian surface
 * d) constant in direction over the entire Gaussian surface

T2 K0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant magnitude over a portion of the Gaussian surface
 * b) constant in direction over the entire Gaussian surface
 * c) constant direction and magnitude over the entire Gaussian surface
 * d) constant direction over a portion of the Gaussian surface

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 K1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant magnitude over a portion of the Gaussian surface
 * b) constant in direction over the entire Gaussian surface
 * c) constant direction and magnitude over the entire Gaussian surface
 * d) constant direction over a portion of the Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

T2 K2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant in direction over the entire Gaussian surface
 * b) constant magnitude over a portion of the Gaussian surface
 * c) constant direction and magnitude over the entire Gaussian surface
 * d) constant direction over a portion of the Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

T2 L0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 L1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

T2 L2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

T2 M0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 M1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

T2 M2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

T2 N0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 N1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 N2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

T2 O0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 O1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

T2 O2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

T2 P0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 P1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

T2 P2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

T2 Q0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction over a portion of the Gaussian surface
 * b) constant direction and magnitude over the entire Gaussian surface
 * c) constant in direction over the entire Gaussian surface
 * d) constant magnitude over a portion of the Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 Q1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant magnitude over a portion of the Gaussian surface
 * b) constant in direction over the entire Gaussian surface
 * c) constant direction and magnitude over the entire Gaussian surface
 * d) constant direction over a portion of the Gaussian surface

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

T2 Q2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant magnitude over a portion of the Gaussian surface
 * b) constant direction and magnitude over the entire Gaussian surface
 * c) constant in direction over the entire Gaussian surface
 * d) constant direction over a portion of the Gaussian surface

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

T2 R0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 R1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

T2 R2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * a) True
 * b) False

T2 S0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant in direction over the entire Gaussian surface
 * b) constant direction and magnitude over the entire Gaussian surface
 * c) constant direction over a portion of the Gaussian surface
 * d) constant magnitude over a portion of the Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 S1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction over a portion of the Gaussian surface
 * b) constant direction and magnitude over the entire Gaussian surface
 * c) constant in direction over the entire Gaussian surface
 * d) constant magnitude over a portion of the Gaussian surface

T2 S2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction over a portion of the Gaussian surface
 * b) constant direction and magnitude over the entire Gaussian surface
 * c) constant in direction over the entire Gaussian surface
 * d) constant magnitude over a portion of the Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * a) True
 * b) False

T2 T0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction and magnitude over the entire Gaussian surface
 * b) constant direction over a portion of the Gaussian surface
 * c) constant magnitude over a portion of the Gaussian surface
 * d) constant in direction over the entire Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

T2 T1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction over a portion of the Gaussian surface
 * b) constant magnitude over a portion of the Gaussian surface
 * c) constant direction and magnitude over the entire Gaussian surface
 * d) constant in direction over the entire Gaussian surface

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

T2 T2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * a) True
 * b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * a) True
 * b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * a) constant direction and magnitude over the entire Gaussian surface
 * b) constant magnitude over a portion of the Gaussian surface
 * c) constant direction over a portion of the Gaussian surface
 * d) constant in direction over the entire Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * a) True
 * b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * a) True
 * b) False


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 * 1) of 10 blank lines to separate exams from keys
 * 2) of 10 blank lines to separate exams from keys
 * 3) of 10 blank lines to separate exams from keys
 * 4) of 10 blank lines to separate exams from keys
 * 5) of 10 blank lines to separate exams from keys
 * 6) of 10 blank lines to separate exams from keys
 * 7) of 10 blank lines to separate exams from keys
 * 8) of 10 blank lines to separate exams from keys
 * 9) of 10 blank lines to separate exams from keys
 * 10) of 10 blank lines to separate exams from keys

Key: A0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * +a) constant magnitude over a portion of the Gaussian surface
 * -b) constant direction over a portion of the Gaussian surface
 * -c) constant direction and magnitude over the entire Gaussian surface
 * -d) constant in direction over the entire Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: A1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * +a) constant magnitude over a portion of the Gaussian surface
 * -b) constant in direction over the entire Gaussian surface
 * -c) constant direction over a portion of the Gaussian surface
 * -d) constant direction and magnitude over the entire Gaussian surface

Key: A2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction over a portion of the Gaussian surface
 * +b) constant magnitude over a portion of the Gaussian surface
 * -c) constant in direction over the entire Gaussian surface
 * -d) constant direction and magnitude over the entire Gaussian surface

Key: B0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * +a) constant magnitude over a portion of the Gaussian surface
 * -b) constant in direction over the entire Gaussian surface
 * -c) constant direction over a portion of the Gaussian surface
 * -d) constant direction and magnitude over the entire Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: B1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant in direction over the entire Gaussian surface
 * +b) constant magnitude over a portion of the Gaussian surface
 * -c) constant direction and magnitude over the entire Gaussian surface
 * -d) constant direction over a portion of the Gaussian surface

Key: B2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction and magnitude over the entire Gaussian surface
 * -b) constant in direction over the entire Gaussian surface
 * -c) constant direction over a portion of the Gaussian surface
 * +d) constant magnitude over a portion of the Gaussian surface

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

Key: C0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction over a portion of the Gaussian surface
 * -b) constant in direction over the entire Gaussian surface
 * -c) constant direction and magnitude over the entire Gaussian surface
 * +d) constant magnitude over a portion of the Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: C1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant in direction over the entire Gaussian surface
 * -b) constant direction and magnitude over the entire Gaussian surface
 * -c) constant direction over a portion of the Gaussian surface
 * +d) constant magnitude over a portion of the Gaussian surface

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

Key: C2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * +a) constant magnitude over a portion of the Gaussian surface
 * -b) constant direction over a portion of the Gaussian surface
 * -c) constant direction and magnitude over the entire Gaussian surface
 * -d) constant in direction over the entire Gaussian surface

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: D0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction and magnitude over the entire Gaussian surface
 * -b) constant in direction over the entire Gaussian surface
 * -c) constant direction over a portion of the Gaussian surface
 * +d) constant magnitude over a portion of the Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

Key: D1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction over a portion of the Gaussian surface
 * +b) constant magnitude over a portion of the Gaussian surface
 * -c) constant direction and magnitude over the entire Gaussian surface
 * -d) constant in direction over the entire Gaussian surface

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

Key: D2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant in direction over the entire Gaussian surface
 * -b) constant direction over a portion of the Gaussian surface
 * +c) constant magnitude over a portion of the Gaussian surface
 * -d) constant direction and magnitude over the entire Gaussian surface

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

Key: E0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction and magnitude over the entire Gaussian surface
 * -b) constant in direction over the entire Gaussian surface
 * +c) constant magnitude over a portion of the Gaussian surface
 * -d) constant direction over a portion of the Gaussian surface

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: E1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction over a portion of the Gaussian surface
 * -b) constant direction and magnitude over the entire Gaussian surface
 * -c) constant in direction over the entire Gaussian surface
 * +d) constant magnitude over a portion of the Gaussian surface

Key: E2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * +a) constant magnitude over a portion of the Gaussian surface
 * -b) constant in direction over the entire Gaussian surface
 * -c) constant direction over a portion of the Gaussian surface
 * -d) constant direction and magnitude over the entire Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: F0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: F1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

Key: F2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

Key: G0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: G1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

Key: G2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

Key: H0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction and magnitude over the entire Gaussian surface
 * +b) constant magnitude over a portion of the Gaussian surface
 * -c) constant direction over a portion of the Gaussian surface
 * -d) constant in direction over the entire Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: H1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * +a) constant magnitude over a portion of the Gaussian surface
 * -b) constant in direction over the entire Gaussian surface
 * -c) constant direction over a portion of the Gaussian surface
 * -d) constant direction and magnitude over the entire Gaussian surface

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

Key: H2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction and magnitude over the entire Gaussian surface
 * -b) constant direction over a portion of the Gaussian surface
 * -c) constant in direction over the entire Gaussian surface
 * +d) constant magnitude over a portion of the Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

Key: I0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction and magnitude over the entire Gaussian surface
 * +b) constant magnitude over a portion of the Gaussian surface
 * -c) constant in direction over the entire Gaussian surface
 * -d) constant direction over a portion of the Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: I1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * +a) constant magnitude over a portion of the Gaussian surface
 * -b) constant direction over a portion of the Gaussian surface
 * -c) constant direction and magnitude over the entire Gaussian surface
 * -d) constant in direction over the entire Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

Key: I2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant in direction over the entire Gaussian surface
 * -b) constant direction and magnitude over the entire Gaussian surface
 * +c) constant magnitude over a portion of the Gaussian surface
 * -d) constant direction over a portion of the Gaussian surface

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

Key: J0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * +a) constant magnitude over a portion of the Gaussian surface
 * -b) constant direction over a portion of the Gaussian surface
 * -c) constant in direction over the entire Gaussian surface
 * -d) constant direction and magnitude over the entire Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: J1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction and magnitude over the entire Gaussian surface
 * -b) constant in direction over the entire Gaussian surface
 * +c) constant magnitude over a portion of the Gaussian surface
 * -d) constant direction over a portion of the Gaussian surface

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

Key: J2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction and magnitude over the entire Gaussian surface
 * +b) constant magnitude over a portion of the Gaussian surface
 * -c) constant direction over a portion of the Gaussian surface
 * -d) constant in direction over the entire Gaussian surface

Key: K0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * +a) constant magnitude over a portion of the Gaussian surface
 * -b) constant in direction over the entire Gaussian surface
 * -c) constant direction and magnitude over the entire Gaussian surface
 * -d) constant direction over a portion of the Gaussian surface

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: K1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * +a) constant magnitude over a portion of the Gaussian surface
 * -b) constant in direction over the entire Gaussian surface
 * -c) constant direction and magnitude over the entire Gaussian surface
 * -d) constant direction over a portion of the Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

Key: K2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant in direction over the entire Gaussian surface
 * +b) constant magnitude over a portion of the Gaussian surface
 * -c) constant direction and magnitude over the entire Gaussian surface
 * -d) constant direction over a portion of the Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

Key: L0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: L1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

Key: L2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

Key: M0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: M1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

Key: M2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

Key: N0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: N1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: N2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

Key: O0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: O1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

Key: O2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

Key: P0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: P1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

Key: P2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

Key: Q0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction over a portion of the Gaussian surface
 * -b) constant direction and magnitude over the entire Gaussian surface
 * -c) constant in direction over the entire Gaussian surface
 * +d) constant magnitude over a portion of the Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: Q1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * +a) constant magnitude over a portion of the Gaussian surface
 * -b) constant in direction over the entire Gaussian surface
 * -c) constant direction and magnitude over the entire Gaussian surface
 * -d) constant direction over a portion of the Gaussian surface

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

Key: Q2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * +a) constant magnitude over a portion of the Gaussian surface
 * -b) constant direction and magnitude over the entire Gaussian surface
 * -c) constant in direction over the entire Gaussian surface
 * -d) constant direction over a portion of the Gaussian surface

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

Key: R0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: R1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

Key: R2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated inside the Gaussian surface
 * -a) True
 * +b) False

Key: S0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant in direction over the entire Gaussian surface
 * -b) constant direction and magnitude over the entire Gaussian surface
 * -c) constant direction over a portion of the Gaussian surface
 * +d) constant magnitude over a portion of the Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: S1
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction over a portion of the Gaussian surface
 * -b) constant direction and magnitude over the entire Gaussian surface
 * -c) constant in direction over the entire Gaussian surface
 * +d) constant magnitude over a portion of the Gaussian surface

Key: S2
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated on the Gaussian surface
 * +a) True
 * -b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction over a portion of the Gaussian surface
 * -b) constant direction and magnitude over the entire Gaussian surface
 * -c) constant in direction over the entire Gaussian surface
 * +d) constant magnitude over a portion of the Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$
 * -a) True
 * +b) False

Key: T0
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

2) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction and magnitude over the entire Gaussian surface
 * -b) constant direction over a portion of the Gaussian surface
 * +c) constant magnitude over a portion of the Gaussian surface
 * -d) constant in direction over the entire Gaussian surface

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

Key: T1
1) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

3) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

4) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction over a portion of the Gaussian surface
 * +b) constant magnitude over a portion of the Gaussian surface
 * -c) constant direction and magnitude over the entire Gaussian surface
 * -d) constant in direction over the entire Gaussian surface

5) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

Key: T2
1) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$
 * +a) True
 * -b) False

2) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$
 * -a) True
 * +b) False

3) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * -a) constant direction and magnitude over the entire Gaussian surface
 * +b) constant magnitude over a portion of the Gaussian surface
 * -c) constant direction over a portion of the Gaussian surface
 * -d) constant in direction over the entire Gaussian surface

4) In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$
 * +a) True
 * -b) False

5) If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ was calculated outside the Gaussian surface
 * -a) True
 * +b) False