Talk:QB/d cp2.gaussC

First to allow and display discussion of each question, and second, to store the quiz in raw-script for.

1
- True + False
 * 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

True false questions are easy to write, but a collection of them forces the student to remember the answer (undestanding it is the student's responsibility) 

2
- True + False
 * 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

Part of a triplet of similar TF questions, designed to force those who memorize to work harder than those who understand. 

3
+ True - False
 * 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

Think of the Gaussian surface as a "probe" that you place where you want to know the field 

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

students need to be coached on the meaning of "necessary" and "sufficient". 

5
- True + False
 * In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$ [[file:GAUSS2.svg|thumb|175px]]

Sort of a trick question, perhaps, but this is an important topic. Also, we pair this on up with one where it is true. 

6
+ True - False
 * In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$ [[file:GAUSS2.svg|thumb|175px]]

This partners with the similar one that is false. 

7
- True + False
 * In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$ [[file:GAUSS2.svg|thumb|175px]]

False because they are equal in absolute magnitude but of the same sign. 

8
+ True - False
 * In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$ [[file:GAUSS2.svg|thumb|175px]]

True because they are of opposite sign. 

Raw Text

 * t QB/e_cp2.6


 * ! Public Domain CC0 user:Guy vandegrift
 * ? 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
 * - True
 * + False
 * $ True false questions are easy to write, but a collection of them forces the student to remember the answer (undestanding it is the student's responsibility)


 * ! Public Domain CC0 user:Guy vandegrift
 * ? 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
 * - True
 * + False
 * $ Part of a triplet of similar TF questions, designed to force those who memorize to work harder than those who understand.


 * ! Public Domain CC0 user:Guy vandegrift
 * ? 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
 * + True
 * - False
 * $ Think of the Gaussian surface as a "probe" that you place where you want to know the field


 * ! Public Domain CC0 user:Guy vandegrift
 * ? If Gauss' law can be reduced to an algebraic expression that easily calculates the electric field $$(\varepsilon_0EA^*= \rho V^*)$$, $$\vec E$$ had
 * - constant direction and magnitude over the entire Gaussian surface
 * + constant magnitude over a portion of the Gaussian surface
 * - constant direction over a portion of the Gaussian surface
 * - constant in direction over the entire Gaussian surface
 * $ students need to be coached on the meaning of "necessary" and "sufficient".


 * ! Public Domain CC0 user:Guy vandegrift
 * ? In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$dA_1=dA_3$$ [[file:GAUSS2.svg|thumb|175px]]
 * - True
 * + False
 * $ Sort of a trick question, perhaps, but this is an important topic. Also, we pair this on up with one where it is true.


 * ! Public Domain CC0 user:Guy vandegrift
 * ? In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1=\vec E_3\cdot dA_3$$ [[file:GAUSS2.svg|thumb|175px]]
 * + True
 * - False
 * $ This partners with the similar one that is false.


 * ! Public Domain CC0 user:Guy vandegrift
 * ? In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure, $$\vec E_1\cdot dA_1+\vec E_3\cdot dA_3 =0$$ [[file:GAUSS2.svg|thumb|175px]]
 * - True
 * + False
 * $ False because they are equal in absolute magnitude but of the same sign.


 * ! Public Domain CC0 user:Guy vandegrift
 * ? In this description of the flux element, $$d\vec S = \hat n dA_j$$ (j=1,2,3) where $$\hat n$$ is the outward unit normal, and a positive charge is assumed at point O, inside the Gaussian surface shown. The field lines exit at $$S_1$$ and $$S_3$$ but enter at $$S_2$$. In this figure,  $$\vec E_1\cdot dA_1+\vec E_2\cdot dA_3 =0$$ [[file:GAUSS2.svg|thumb|175px]]
 * + True
 * - False
 * $ True because they are of opposite sign.