Gauss's Law for Electric Fields

1. Mathematical Background

The Divergence Theorem, which states that for any compact space $V \subset \mathbb{R}^3$ having piecewise smooth boundary $\partial V = S$ , if the vector field $\mathbf{F}$ is continuously differentiable over $V$ , then

$\iiint_{V} \nabla \cdot \mathbf{F}\ dV = \iint_{S} \mathbf{F} \cdot d\mathbf{S}$

2. What is Gauss's Law?

Maxwell's equations are fundamental results that govern the behavior of - and interactions betewen- electric and magnetic fields. We see how Maxwell's equations arise from a few simple physical principles coupled with the vector analysis.

If $\mathbf{E}$ is an electric field, then the flux of $\mathbf{E}$ across a closed surface $S$ is given by

$\Phi_{E}= \iint_{S} \mathbf{E} \cdot d \mathbf{S}$

where the integral is held over the surface $S$ (surface integral). Applying Divergence Theorem, we find that

$\Phi_{E}= \iiint_{D} \nabla \cdot \mathbf{E}\ dV$

where $D$ is the region enclosed by $S$ . If the electric field $\mathbf{E}$ is determined by a single point charge of $q$ coulombs located at the origin, then $\mathbf{E}$ is given by

$\mathbf{E}(\mathbf{x}) = \frac{q}{4\pi \epsilon_0} \frac{\mathbf{x}}{\lVert \mathbf{x} \rVert^3}$

where $\mathbf{x} = x \mathbf{i} + y\mathbf{j} + z \mathbf{k} \in \mathbb{R}^3 \setminus \left\{ 0 \right\}$ . In SI units, $\mathbf{E}$ is measured in Volts/meter. The constant $\epsilon_0$ is known as the permittivity of free space; its value is about $8.854 \times 10^{-12}\ \text{C}^2/\text{N}\cdot \text{m}^2$ .

2-1. Surface that does not enclose origin

For the electric field defined above, we can easily verify that

$\nabla \cdot \mathbf{E} = \sum_{cyc} (x^2 + y^2 + z^2)^{-3/2} - 3x^2 (x^2 + y^2 + z^2)^{-5/2} = 0$
wherever the field is defined. So if $D$ does not contain origin, then automatically flux is zero.

2-2. Surface that does enclose origin

Now the question is, how do we calculate the flux of the electric field across surfaces that do enclose the origin? The trick is to find an appropriate way to exclude the origin from consideration. To that end, first suppose that the surface $S_r$ denotes a sphere of radius $r>0$ centered at origin.

$S_r: x^2 + y^2 + z^2 = r^2$

Then the outward unit normal to $S_r$ is

$\mathbf{n} = \frac{x \mathbf{i} + y \mathbf{j} + z \mathbf{k}}{r} = \frac{\mathbf{x}}{r}$

Now using original equation,

$\iint_{S_r} \mathbf{E} \cdot d \mathbf{S} = \frac{q}{4\pi \epsilon_0} \iint_{S_r} \frac{\mathbf{x}}{\lVert \mathbf{x} \rVert^3} \cdot \frac{\mathbf{x}}{b} dS$

Since $\lVert \mathbf{x} \rVert = b$ on the sphere,

$\begin{align*} \iint_{S_r} \mathbf{E} \cdot d\mathbf{S} &= \frac{q}{4\pi\epsilon_0}\iint_{S_r} \frac{\mathbf{x}}{b^3} \cdot \frac{\mathbf{x}}{b} dS = \frac{q}{4\pi\epsilon_0} \iint_{S_r} \frac{\lVert \mathbf{x}\rVert^2}{b^4} dS \\ &= \frac{q}{4\pi \epsilon_0 b^2} \iint_{S_r} dS \\ &= \frac{q}{4\pi \epsilon_0 b^2} (4 \pi b^2) = \frac{q}{\epsilon_0} \end{align*}$

Now, suppose $S$ is any surface enclosing the origin. Let $S_r$ be a small sphere centered at the origin and contained inside $S$ . Let $D$ be the solid region in $\mathbb{R}^3$ between $S_r$ and $S$ . First, $\nabla \cdot \mathbf{E} = 0$ on $D$ , since it does not contain the origin. Orienting $S$ and $S_r$ with normals that point away from $D$ , we obtain the following.

$\iint_{S} \mathbf{E} \cdot d \mathbf{S} - \iint_{S_r} \mathbf{E} \cdot d\mathbf{S} = \iint_{\partial D} \mathbf{E} \cdot d\mathbf{S} = \iiint_{D} \nabla \cdot \mathbf{E} = 0$

so that

$\iint_{S} \mathbf{E} \cdot d\mathbf{S} = \frac{q}{\epsilon_0}$

3. What about multiple charges?

Wel,, suppose a point charge of $q_i$ coulombs is located at position $r_i$ . The electric field is then (by superposition principle),

$\mathbf{E}(\mathbf{x}) = \frac{1}{4\pi \epsilon_0} \sum_{i=1}^{n} q_i \frac{\mathbf{x} - \mathbf{r}_i}{\lVert \mathbf{x} - \mathbf{r}_i \rVert^3}$

Since the electric field is just linear sum of each electric fields generated by point charges, $\nabla \cdot \mathbf{E} = 0$ for $\mathbf{x} = \mathbf{r}_i\ (i=1,2,...,n)$ . If $S$ is any piecewise smooth, oriented surface containing the charges, then we may use Divergence theorem to find the flux of $\mathbf{E}$ across $S$ by taking $n$ small spheres $S_1, S_2, ..., S_n$ each enclosing a single point charge. If $D$ is the region inside $S$ , but outside all the spheres, we have, by choosing appropriate orientations,

$\iint_{S} \mathbf{E} \cdot d\mathbf{S} - \sum_{i=1}^{n} \iint_{S_i}\mathbf{E} \cdot d\mathbf{S} = \iint_{\partial D} \mathbf{E} \cdot d\mathbf{S} = \iiint_{D} \nabla \cdot \mathbf{E}\ dV = 0$

And using the Gauss's Law for single charge,

$\iint_{S} \mathbf{E} \cdot d\mathbf{S} = \sum_{i=1}^{n} \iint_{S_i}\mathbf{E} \cdot d\mathbf{S} = \sum_{i=1}^{n} \frac{q_i}{\epsilon_0} = \frac{\text{total charge enclosed by }S}{\epsilon_0}$

4. What about continuous charge Distribution?

Finally, consider the case not of an electric field determined by discrete point charges, but rather of one determined by a continuous charge distribution given by a charge density $\rho(\mathbf{x})$ . The total charge over a region $D$ in space is given by

$\iiint_{D} \rho(\mathbf{x}) dV$

so that, we have

$\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi \epsilon_0} \iiint_{D} \rho(\mathbf{x}) \frac{\mathbf{r} - \mathbf{x}}{\lVert \mathbf{r} - \mathbf{x} \rVert^3}\ dV$

Now the question arises. Does this integral converges at points $\mathbf{r} \in D$ where $\rho(\mathbf{r}) \neq 0$ ? Well, in result, it DOES but it is NON-Trivial.

4-1. Proof of Convergence.

First write $\mathbf{E}(\mathbf{r})$ using individual components.

$\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi \epsilon_0} \left( \iiint_{D} \rho \frac{r_1 - x}{\lVert \mathbf{r} - \mathbf{x} \rVert^3}\ dV, \iiint_{D} \rho \frac{r_2 - y}{\lVert \mathbf{r} - \mathbf{x} \rVert^3}\ dV, \iiint_{D} \rho \frac{r_3 - z}{\lVert \mathbf{r} - \mathbf{x} \rVert^3}\ dV\right )$

where $\mathbf{r} = (r_1, r_2, r_3),\ \mathbf{x} = (x,y,z)$ . Looking at the first component of $\mathbf{E}$ ,

$\left| \frac{1}{4\pi \epsilon_0} \left( \rho \frac{r_1 - x}{\lVert \mathbf{r} - \mathbf{x} \rVert^3}\ \right) \right| \leq \frac{|\rho|}{4\pi\epsilon_0} \cdot \frac{1}{\lVert \mathbf{r} - \mathbf{x} \rVert^2} \leq \frac{K}{\lVert \mathbf{r} - \mathbf{x} \rVert^2}$

for some constant $K$ . (Because $\rho$ is continuous at bounded region $D$ , such constant exists!). So

$|E_1| \leq \iiint_{D} \frac{K}{\lVert \mathbf{r} - \mathbf{x} \rVert^2}\ dV$

Now we transform cartesian coordinate to spherical coordinate. Then

$\iiint_{D} \frac{K}{\lVert \mathbf{r} - \mathbf{x} \rVert^2}\ dV = \iiint_{D + \mathbf{r}} K \sin \varphi d\rho d\varphi d\theta < \infty$

because the integrand is continuous function over bounded region. Other coordinates $E_2, E_3$ are exactly same, so $\mathbf{E}$ is a well defined field vector.

4-2. Back to the formulation.

Now the intergral form of Gauss's Law, analogous to that of formula for discrete charge distribution is then

$\iint_{S} \mathbf{E} \cdot d\mathbf{S} = \frac{1}{\epsilon} \iiint_{D} \rho\ dV$

by Divergence theorem,

$\iint_{S} \mathbf{E} \cdot d\mathbf{S} = \frac{1}{\epsilon} \iiint_{D} \rho\ dV = \iiint_{D} \nabla \cdot \mathbf{E}\ dV$

Now we get the differential form of Gauss's Law,

$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$

5. So why is it useful?

It can be applied to

Any closed surface
Furthermore, we only have to consider the total charge enclosed by the surface (without calculating the charges or charge distributions in the total region!)

[Physics & EE #2] Gauss's Law in Electromagnetism