MyOpenMath/Solutions/Maxwell's integral equations

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Gauss's law
Gauss's law equates any integral involving the electric field over a closed surface to an integral of the electric charge density enclosed by that closed surface.

$$\varepsilon_0\oint^{\color{red}{GS}} \vec E\cdot d\vec A=Q_{enc} = \int^{\color{red}{EV}} \rho(\vec \tilde r) d\tilde\tau$$

In contrast to the situation with one-dimensional integrals encountered in a first-year calculus course, it is not possible to specify the limit of integration using two indexes at the top and bottom of the integral sign. Henceforth we shall use a superscript on the integral sign, with the understanding that the reader will understand the details from the context of the discussion. In this case, GS refers to a Gaussian surface, and EV denotes the volume enclosed by the same Gaussian surface. The Greek lower-case tau is sometimes used to denote a volume element, which can also be expressed in Cartesian, cylindrical, and spherical coordinates, respectively as:

$$d\tilde\tau = d\tilde x d\tilde y d\tilde z =\tilde r^2 \sin\tilde\theta d\tilde\theta d\tilde\phi =\tilde r d\tilde r d\tilde z d\tilde\theta $$

The use of a tilde (or prime) on variables of integration is not mandatory, but sometimes useful to remind ourselves that these variables are arbitrary and do not appear when the final result is obtained. In our case, we will use a bit of intuition to deduce the electric field at some point $$(x,y,z)$$ in space, $$\vec E =\vec E(x,y,z)$$ using surface and volume integrals where the variables of integration might be $$(\tilde x, \tilde y, \tilde z).$$

Fortunately, the charge distributions discussed here will be so symmetrically simple that the integration may be performed virtually without calculus.

Gauss's law with spherical symmetry
guess law with symmetry

Shown to the right is a uniformly charged sphere of radius $$R$$ and a concentric Gaussian sphere of radius $$r$$. The surface of this Gaussian sphere identifies the location where we want to calculate the electric field: We seek to find $$E(r)$$, where $$r$$ is called the field point. To be useful in calculating the field, field point must lie on the Gaussian surface. The four red arrows on the Gaussian sphere indicate the direction of the outward unit normal.

r>R: Field point outside charged sphere
Since the Gaussian sphere has a radius of $$r,$$ we shall replace $$GS$$ by the symbol $$*r,$$ and use the fact that the electric field is uniform and always parallel to the outward unit normal to write $$\vec E\cdot d\vec A=$$$$\vec E\cdot \hat n dA=$$$$EdA,$$ where $$E$$ is constant and $$dA$$ is an element of area. Since the surface area of a sphere is $$4\pi r^2,$$ we have:


 * $$\oint^{\color{red}{GS}} \vec E\cdot d\vec A= $$ $$\oint^{\color{red}{*r}} \vec E\cdot d\vec A= $$$$E\oint^{\color{red}{*r}} dA= 4\pi r^2 E.$$

$$Q_{enc}$$, or the charge enclosed by Gaussian surface, is obtained by integrating charge density over larger sphere of radius $$r$$. But the charge density is non-zero only within the smaller sphere of radius $$R$$. In other words, the function $$\rho(\tilde r)$$ is discontinuous for a uniformly charged sphere of radius $$R$$:


 * $$\begin{align}

\rho(\tilde r)& =\rho_0 & \text{if } rR \end{align}$$

Therefore,
 * $$Q_{enc}=\int^{\color{red}{EV}}\rho(\tilde\vec r) d\tilde\tau = \int^{\color{red}{*r}}\rho(\tilde\vec r) d\tilde\tau $$ $$=\int^{\color{red}{*R}}\rho_0 d\tau$$ $$=\rho_0 \int^{\color{red}{*R}}d\tau$$$$=\frac{4 \pi R^3}{3}\rho_0 ,$$

where $$\frac 4 3 \pi R^3$$ is the volume of a sphere. Once these steps have been digested by the reader, they can be combined into a simple two step calculation based on Gauss's law for the electric field outside a charged sphere:


 * $$\varepsilon_0\underbrace{\oint^{\color{red}{*r}} \vec E\cdot d\vec A}_{E \cdot 4\pi r^2}=

\underbrace{\int^{\color{red}{*R}}\rho d\tau}_{\frac{4\pi R^3}{3}\rho_0}$$    $$\Rightarrow \quad \vec E(r) = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \hat r.$$

r<R: Field point inside charged sphere
If the field point is inside the charged surface, the enclosed charge $$Q_{enc}$$ includes only the portion of the charge inside the Gaussian sphere. First we calculate the uniform charge density inside the sphere:

$$\rho_0 = \frac{\text{total charge}}{\text{volume of charged object}} = \frac{Q_{tot}}{\frac 4 3 \pi R^3}$$

Now the enclosed charge $$Q_{enc}$$ includes only the volume inside the Gaussian sphere of radius $$r<R$$. Since the sphere is uniformly charged, $$\rho_0$$ is a constant that can be taken out integral sign:

$$Q_{enc} = \int^{\color{red}{*r}} \rho d\tau = \rho_0\int^{\color{red}{*r}} d\tau = \rho_0\cdot\frac 4 3 \pi r^3$$

$$\varepsilon_0 \underbrace{\oint\vec E\cdot d\vec A}_{4\pi r^2 E(r)}\quad =\underbrace{ \int \rho d\tau}_{\rho_0 \frac 4 3 \pi r^3}$$    $$\Rightarrow \quad \vec E(r)= \frac{\rho_0}{3\varepsilon_0} \vec r$$

Gauss's law with cylindrical symmetry
In cylindrical coordinates $$(r,\theta,z)$$ a symmetry exists if the field variables do not depend on $$\theta$$ or $$z$$. This requires that the system extend to infinity in both directions on the z-axis. In the case of a charged cylinder, we may use a Gaussian Surface that is a cylinder of finite length, $$L$$. Such a cylinder has three distinct surfaces: A hollow cylinder with a curved surface of length $$L$$, and two flat circular discs at each end of the same radius. But if the field is in the radial $$(\hat r)$$ direction only, there is no flux through the two ends.

The charge enclosed by a Gaussian surface of length $$L$$ depends on whether the surface is inside or outside the charged sphere. The area of the cylinder's curved surface is the product of circumference and length:

$$A=2\pi r L$$

r>R: Field point outside charged cylinder
Outside the radius $$R$$ of charged cylinder, the enclosed charge is:

$$Q_{enc}=\pi R^2 \rho_0 L = \lambda L,$$

where the line charge density $$\lambda = \pi R^2 \rho_0$$ (Coulombs per meter.) Gauss's law now yields:

$$\varepsilon_0\underbrace{\oint^{\color{red}*r}\vec E\cdot d\vec A}_{E\cdot2\pi r L}= \int^{\color{red}{*r}}\rho d\tau= \underbrace{\rho_0\int^{\color{red}{*R}} d\tau}_{\rho_0\pi R^2 L}$$     $$\Rightarrow \quad \vec E(r)= \frac{\rho_0\pi R^2}{2\pi\varepsilon_0 r} \hat r= \frac{1}{2\pi\varepsilon_0} \frac \lambda r \hat r$$

r<R: Field point inside charged cylinder
Here, as in the case where the field point was inside the sphere, the amount of enclosed does does depend on $$r:$$

$$\varepsilon_0\underbrace{\oint^{\color{red}{*r}}\vec E\cdot d\vec A}_{E\cdot2\pi r L}= \int^{\color{red}{*r}}\rho d\tau= \underbrace{\rho_0\int^{\color{red}{*r}} d\tau}_{\rho_0\pi r^2 L}$$     $$\Rightarrow \quad \vec E(r)= \frac{\rho_0}{2\varepsilon_0} r \, \hat r =\frac{\rho_0}{2\varepsilon_0} \vec r.$$

Gauss's law in one dimension
This so-called "slab geometry" is extremely important for two reasons. First, it models both the parallel plate capacitor, as well as the field near a flat metal surface. Perhaps more significant is that it leads directly to a superior form of Maxwell's equations that uses partial derivatives instead of integrals.

Here we make three assumptions:
 * 1) The charge density depends only on the x variable: $$\rho=\rho(x)$$
 * 2) The electric field depends also depends only on x.
 * And, the electric field points only in the x-direction: $$\vec E = E_x(x)\hat i$$

The charged object is shown as a thin sheet that extends between $$-b/2 <x<b/2$$. First we apply Gauss's law for a field point outside the charged sheet. The GS is the six-sided box with area $$A$$ and length $$L$$ as shown in the figure. But since the electric field is in the x-direction, our surface integral involves only the two faces of area $$A.$$ Since these faces have opposite outward unit normals, the sum of the fluxes $$\Sigma \vec E\cdot \vec A$$ becomes a difference in electric fields:




 * $$\varepsilon_0 (E_2 - E_1) = \sigma,$$

where $$\sigma = Q_{enc}/A$$ is the surface charge density associated with the sheet, and $$E_2$$ represents the field to the right of the sheet. The ambiguity of this equation needs to be understood. In the previous examples we could resolve such ambiguities with a symmetry argument. It can be easily argued that an isolated charge with spherical symmetry would generate a field that points towards or away from the center. But an isolated infinite plane is difficult to imagine, as discussed in the figure to the right. This ambiguous formula has two important applications. The most obvious is the symmetric case where equal and opposite field exist on both sides of an isolated plane of surface charge:


 * $$E_2=-E_1=\frac{1}{2\varepsilon_0}\sigma.$$

A more common application calls for the electric field to vanish on one side of the plate. This can occur on the smooth surface of a solid conducting object, and also between the plates of a parallel plate capacitor. In that case:


 * $$E= \frac 1 \varepsilon_0 \sigma.$$

Gauss's law in differential form
The previous examples illustrate how a student can quickly calculate the electric field for simple geometries. A hint as to the profound impact these methods have had on physics can be illustrated by considering how the x-component of the electric field is related to a charge density that only depends on x: $$\rho(x,y,z)=\rho(x).$$ Let us select an arbitrary value of x and consider a box shaped Gaussian surface that occupies a large area $$A$$ in the $$xy$$ plane, but has a small differential length of $$\Delta x$$ along the x axis. Gauss's law is:


 * $$ E_x(x+\Delta x)-E_x(x)=\frac 1 \varepsilon_0\sigma(x)=\frac 1 \varepsilon_0\rho(x)\Delta x$$  $$\Rightarrow \varepsilon_0\frac{dE_x(x)}{dx}=\rho$$

Though the result is beyond the scope of most first-year physics courses, the vector methods introduced in this course permit one to derive one of the four Maxwell's equations in differential form:




 * $$\frac{\partial E_x}{\partial x} +\frac{\partial E_y}{\partial y} +\frac{\partial E_z}{\partial z} \equiv \vec\nabla\cdot\vec E = \frac 1 \varepsilon_0 \rho.$$

Ampere's law and beyond
Ampere's law can be written as:

$$\oint^{\color{red}{AL}}\vec B\cdot d\vec l = \mu_0I_{eff} = \mu_0\left[ \int \vec J\cdot d\vec A + \left(\begin{matrix} \text{Maxwell's}\\ \text{displacement}\\ \text{current}\\ \end{matrix}\right) \right]$$

Here, the effective current $$I_{eff}$$ includes not only the conventional current carried by moving charge, $$\int\vec J\cdot d\vec A$$, but also a term introduced by Maxwell to ensure that these four integral equations are self-consistent:

$$ \left(\begin{matrix} \text{Maxwell's}\\ \text{displacement}\\ \text{current}\\ \end{matrix}\right) = \varepsilon_0\frac{d}{dt} \int \vec E\cdot d\vec A, $$

where $$c=1/\sqrt{\varepsilon_0\mu_0}$$ is the speed of light in vacuum. The line integral associated with Ampere's law is marked in red and denoted by $$\color{red}{AI}$$ in the figures to the right. The right hand rule identifies the direction of the differential area element $$d\vec A.$$ In contrast with the situation in Gauss's law, where the Gaussian surface $$\color{red}{GS}$$ uniquely defines the volume, the surface integral is valid for any smooth surface that has the Ampere's loop $$\color{red}{AL}$$ as its boundary (for example the top and bottom surfaces in the column to the right have the same boundary, which is a circle.)  Since we are mostly interested in using these integral laws to quickly calculate electromagnetic fields for simple geometries, we will usually integrate $$\vec B\cdot d\vec A$$ over flat surfaces.

Magnetic field due to a straight current-carrying wire
The integral integral of $$\vec B\cdot d\vec l$$ around the circle of radius $$r$$ in the figure shown to the right is related to the current, $$I,$$ carried by the wire of radius $$RR  \\ \end{align}$$

There are a number of equivalent ways to express $$J_0$$:


 * $$\int^{\color{red}{*r}} J(\tilde r) dA = J_0\int^{\color{red}{*R}}dA=J_0 \cdot\pi R^2 = I$$  $$\rightarrow J_0=\frac{I}{\pi R^2}$$

When the field point is inside the wire $$(r<R)$$, we can express Ampere's law using a circle of radius $$r:$$

$$\underbrace{\oint^{\color{red}{*r}}\vec B\cdot d\vec l}_{2\pi r B_\theta}=\mu_0 \underbrace{\int^{\color{red}{*r}}J_0 dA}_{J_0\pi r^2} $$  $$\rightarrow \vec B = \frac{\mu_0J_0}{2}r \hat\theta $$ (inside wire.)

Long, thin solenoid
We can avoid the fallacy of circular reasoning by beginning with the somewhat correct representation of magnetic field due to a solenoid, shown to the left. The circles with dots indicate wires with current coming towards the reader. We see that near the center of this solenoid the magnetic field is relatively (1) strong, (2) horizontal, and (3) uniform with respect to position.

At this point, we postulate that the magnetic field away from the edges of a long, thin solenoid is significant only inside the solenoid, and that the magnetic field is parallel to the axis. Our Ampere's loop $$(\color{red}{AL})$$ is a rectangle with four sides, but contributes to the integral only for a length $$L$$ inside the solenoid. Defining $$n=N/L$$ to be the number of turns per unit length, we have:


 * $$\underbrace{\oint^{\color{red}{AL}}\vec B\cdot d\vec l}_{B_zL}=\mu_0 NI$$  $$\rightarrow \vec B = \mu_0nI\hat z.

$$

Toroid
The figure to the right shows a rectangular toroid, but regardless of the shape, the ampere loop shown establishes a magnetic field that depends on the distance from a vertical axis running though the center:


 * $$\underbrace{\oint^{\color{red}{AL}}\vec B\cdot d\vec l}_{B_\theta 2\pi r}=\mu_0 N I$$  $$\rightarrow \vec B = \mu_0\frac{NI}{2\pi r}\hat\theta.$$

Faraday's law with solenoid and loop
Faraday's law relates the emf, or voltage drop around a closed loop, to the change in the magnetic flux through that loop:


 * $$\underbrace{\oint\vec E\cdot d\vec l}_{E_\theta 2\pi r}=-\frac{d}{dt}\underbrace{\int\vec B\cdot d\vec A}_{BA}$$  $$\rightarrow E_

\theta= -A \frac{dB_0}{dt}.$$



Maxwell's induction current
The Ampere loop if labeled $$\partial S$$ circles a wire carrying a current I. The $$\partial$$ indicates that it is the boundary of an open surface that get pierced by this current. But this time, we consider two different surfaces with the same boundary. The surface $$S_1$$ is the area of the circle defined by $$\partial S$$, and it is pierced by an actual current $$I$$ shown in the figure. But if the surface is deformed into $$S_2$$ it instead goes through the capacitor, where a time-varying electric exists, but no current is present. Maxwell concluded that another way to generate a magnetic field was to have a time-varying electric field be present in the surface integral. Hence, he postulated that:
 * $$\oint_{\partial S} \vec B\cdot d\vec l = \mu_0 \left(I_\text{conventional} + I_\text{displacement} \right),$$

where,

$$I_\text{conventional}=\int_S \vec J\cdot d\vec A,$$ and $$I_\text{displacement}=\frac{dQ}{dt}.$$

Maxwell reasoned that since charge $$Q$$ creates an electric field, and current $$I=dQ/dt$$, it was the electric field between the plates that created the displacement current. Hence, for some unknown constant $$K$$:


 * $$I_\text{displacement}=K\frac{d}{dt}\int_S \vec E\cdot d\vec A.$$

To to show that this constant $$K=\varepsilon_0$$ we use the fact that Gauss's law yields a constant electric field for a parallel plate capacitor:


 * $$ \varepsilon_0 E A = Q$$ $$\Rightarrow I_\text{displacement}=\varepsilon_0 \frac{dE}{dt} A.$$

Hence, "Maxwell's missing term" is:


 * $$I_\text{displacement}=\varepsilon_0\frac{d}{dt}\int_S\vec E\cdot d\vec A.$$

Having rigorously derived this expression for the ideal parallel plate capacitor, it seems reasonable to speculate that his law holds for non-uniform electric fields.