📂Electrodynamics

Electric Flux and Gauss's Law

Definition¹

The flux of an electric field $\mathbf{E}$ passing through a surface $\mathcal S$ is defined as follows.

$$ \Phi_{E} \equiv \int_{\mathcal S} \mathbf{E} \cdot d\mathbf{a} $$

Let’s consider $\mathcal{S}$ as some closed surface. Let the total charge inside the closed surface be $Q_{\text{in}}$. Then, the following equation holds.

$$ \oint_{\mathcal{S}} \mathbf{E} \cdot d\mathbf{a} = \frac{1}{\epsilon_{0}}Q_{\mathrm{in}} $$

This is known as Gauss’s law.

Flux

Flux, or flux density, refers to the amount of a physical quantity passing perpendicularly through a given surface. For example, water or gas flowing through a pipe does so parallel to the perpendicular surface of the pipe, thus, the amount flowing is the same as the flux. However, electric fields do not flow along a conduit. Therefore, the flux of an electric field is determined by taking the dot product. When taking the dot product, components that are not parallel result in zero.

Assuming the electric field lines pass through a surface as shown in the figure above. What we are interested in here is the degree to which it passes perpendicular to the surface. As we all know, vectors can be decomposed. Let’s break down the electric field lines into directions perpendicular and parallel to the surface. Then it would look like the figure below. Our goal is to find the $\mathbf{E}_\parallel$ indicated in the figure. Since the magnitude of the surface vector $d\mathrm{a}$ is $1$, it can be calculated with the following equation using the dot product.

$$ \mathbf{E}_\parallel=\mathbf{E} \cdot d\mathbf{a} $$

Following the method above, the flux of the electric field with respect to the given surface is defined as follows.

$$ \Phi_{E} \equiv \int_{\mathcal S} \mathbf{E} \cdot d\mathbf{a} $$

Gauss’s Law (Integral Form)

The essence of Gauss’s Law is that charges outside a closed surface have no effect on the flux.

As illustrated, the flux caused by electric field lines piercing through a closed surface from the outside is calculated twice, at both ends of the surface. The direction of the surface vector on a closed surface is always defined outward. Since the direction of the surface vectors at both ends is opposite, the magnitude of the flux passing through is the same but the direction is different. When these two are added together, it results in $0$, indicating that charges outside the closed surface do not affect the flux.

Derivation

Now, let’s consider a point charge $Q$ located at the center of a sphere with radius $r$. Let’s calculate the flux of electric field $\mathbf{E}$ passing through the spherical surface. Representing the electric field by Coulomb’s law yields the following result.

$$ \begin{align*} \Phi_{E} &= \oint \mathbf{E} \cdot d\mathbf{a} \\ &= \int_{0}^{2\pi} \int_{0}^\pi \left( \frac{1}{4\pi\epsilon_{0} q r^{2}} \hat{\mathbf{r}} \right) \cdot \left( r^{2}\sin\theta d\theta d \phi \hat{\mathbf{r}} \right) \\ &= \frac{1}{4\pi\epsilon_{0}}Q \int_{0}^{2\pi}d\phi \int_{0}^\pi \sin\theta d\theta \\ &= \frac{1}{4\pi\epsilon_{0}}Q (2\pi)(2) \\ &= \frac{1}{\epsilon_{0}}Q \end{align*} $$

It can be seen that the result is independent of the radius $r$. This is because the surface area of the sphere is proportional to $r^{2}$, and the electric field is inversely proportional to $r^{2}$. They cancel each other out, not affecting the result. In case there are multiple point charges inside the surface, they are simply added together. This is because the electric field of multiple point charges is simply the sum according to the superposition principle. For example, if there are point charges $Q_{1}$ and $Q_{2}$, and if it is $Q=Q_{1}+Q_{2}$, the result is as follows.

$$ \begin{align*} \oint \mathbf{E} \cdot d\mathbf{a} &= \oint \left( \sum \limits_{i=1}^{2} \mathbf{E}_{i} \right) \cdot d\mathbf{a} \\ &= \oint \left( \mathbf{E}_{1} + \mathbf{E}_{2} \right) \cdot d\mathbf{a} \\ &= \oint \mathbf{E}_{1} \cdot d\mathbf{a} + \oint \mathbf{E}_{2} \cdot d\mathbf{a} \\ &= \frac{1}{\epsilon_{0}}Q_{1}+\frac{1}{\epsilon_{0}}Q_{2} \\ &= \frac{1}{\epsilon_{0}}(Q_{1}+Q_{2})=\frac{1}{\epsilon_{0}}Q \end{align*} $$

Naturally, the result would be the same for more than three point charges. Thus, when the total charge inside a closed surface is $Q_{\text{in}}$ and the total electric field created by the charges is $\mathbf{E}$, the following equation holds.

$$ \begin{equation} \oint \mathbf{E} \cdot d\mathbf{a} = \dfrac{1}{\epsilon_{0}}Q_{\text{in}} \label{1} \end{equation} $$

Differential Form

Divergence Theorem

$$ \int_\mathcal{V} \nabla \cdot \mathbf{v} d\tau = \oint _{S} \mathbf{v} \cdot d \mathbf{a} $$

Applying the divergence theorem to $\eqref{1}$ gives us the following equation.

$$ \int_{\mathcal{V}} \nabla \cdot \mathbf{E} d\tau = \oint _{\mathcal{S}} \mathbf{E} \cdot d \mathbf{a} = \frac{1}{\epsilon_{0}}Q_{\text{in}} $$

Let’s denote the charge per unit volume as the volume charge density $\rho$. Then, the relationship between the total charge within a volume $Q_\mathrm{in}$ and $\rho$ is as follows.

$$ Q_\mathrm{in}=\int_\mathcal{V} \rho d\tau $$

Combining the results of the two equations above yields the following.

$$ \begin{align*} && \int_\mathcal{V} \nabla \cdot \mathbf{E} d\tau &= \int_\mathcal{V} \frac{1}{\epsilon_{0}}\rho d\tau \\ \implies && \nabla \cdot \mathbf{E} &= \frac{1}{\epsilon_{0}}\rho \end{align*} $$

This is referred to as Gauss’s Law in differential form and is one of Maxwell’s equations.