📂Electrodynamics

Electric Potential and Electromagnetic Fields

Overview¹

When the charge and current distribution change over time, the electric field and magnetic field are as follows.

$$\mathbf{E}= -\nabla V-\frac{\partial \mathbf{A}}{\partial t}$$

$$\mathbf{B} = \nabla \times \mathbf{A}$$

$V$ is the scalar potential, and $\mathbf{A}$ is the vector potential.

Description

When the charge density $\rho (\mathbf{r}, t)$ and current density $\mathbf{J}(\mathbf{r},t)$ are constant¹, knowing the Coulomb’s law and the Biot-Savart law allows us to find the electric field $\mathbf{E}(\mathbf{r},t)$ and the magnetic field $\mathbf{B}(\mathbf{r},t)$. It is somewhat more difficult to do this when the charge and current change over time.

In electrostatics, $\nabla\times \ \mathbf{E}=0$ and the curl of a gradient $0$, so it could be represented by $\mathbf{E}=-\nabla V$. However, in electromagnetism, since $\nabla \times \mathbf{E} = -\dfrac{\partial \mathbf{B}}{\partial t}$, it cannot be represented as the gradient of a scalar potential like in electrostatics. However, the divergence of a magnetic field still is $0$ and the divergence of a curl $0$, so like in magnetostatics, the magnetic field can still be represented as the curl of a vector potential.

$$\begin{equation} \mathbf{B} = \nabla \times \mathbf{A} \end{equation}$$

When applied to Faraday’s law,

$$\begin{align*} && \nabla \times \mathbf{E} &= -\dfrac{\partial \mathbf{B}}{\partial t} \\ \implies && \nabla \times \mathbf{E} &= -\frac{\partial }{\partial t} (\nabla \times \mathbf{A}) \\ \implies && \nabla \times \mathbf{E} &= -(\nabla \times \frac{\partial \mathbf{A}}{\partial t}) \\ \implies && \nabla \times \left( \mathbf{E} +\frac{\partial \mathbf{A}}{\partial t}\right) &= 0 \end{align*}$$

Therefore, since the curl of $\mathbf{E} +\frac{\partial \mathbf{A}}{\partial t}$ is $0$, it can be represented by the gradient of a scalar potential.

$$\begin{align} && \mathbf{E} +\frac{\partial \mathbf{A}}{\partial t} &= -\nabla V \nonumber \\ \implies && \mathbf{E} &= -\nabla V-\frac{\partial \mathbf{A}}{\partial t} \end{align}$$

Maxwell’s Equations

$$\begin{align} \nabla \cdot \mathbf{E} &= 0 \tag{a} \\[1em] \nabla \cdot \mathbf{B} &= 0 \tag{b} \\[1em] \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t} \tag{c} \\[1em] \nabla \times \mathbf{B} &= \mu_{0}\epsilon_{0}\frac{\partial \mathbf{E}}{\partial t} \tag{d} \end{align}$$

If $\mathbf{A}$ is constant, then $\mathbf{B}=\nabla \times \mathbf{A}=0$, the same as in electrostatics. Inserting $(2)$ into Gauss’s law $(a)$ gives:

$$\begin{align} && \nabla \cdot ( \nabla V) +\nabla \cdot \left( \frac{\partial \mathbf{A}}{\partial t} \right) &= -\frac{1}{\epsilon_{0}}\rho \nonumber \\ \implies && \nabla ^2 V +\dfrac{\partial }{\partial t}(\nabla \cdot \mathbf{A}) &= -\frac{1}{\epsilon_{0}}\rho \end{align}$$

Also, inserting $(1)$, $(3)$ into Ampère’s law $(d)$ gives:

$$\nabla \times (\nabla \times \mathbf{A})=\mu_{0} \mathbf{J}-\mu_{0}\epsilon_{0} \nabla\left( \dfrac{\partial V}{\partial t}\right)-\mu_{0}\epsilon_{0} \dfrac{\partial ^2 \mathbf{A} }{\partial t^2}$$

In this case, the curl of a curl $\nabla \times (\nabla \times \mathbf{A})=\nabla ( \nabla \cdot \mathbf{A}) - \nabla ^2 \mathbf{A}$, so the above equation becomes:

$$\begin{align} && \nabla ( \nabla \cdot \mathbf{A}) - \nabla ^2 \mathbf{A} = \mu_{0} \mathbf{J}-\mu_{0}\epsilon_{0} \nabla\left( \dfrac{\partial V}{\partial t}\right)-\mu_{0}\epsilon_{0} \dfrac{\partial ^2 \mathbf{A} }{\partial t^2} \nonumber \\ \implies && \left( \nabla ^2 \mathbf{A}-\mu_{0}\epsilon_{0} \dfrac{\partial ^2 \mathbf{A} }{\partial t^2} \right) -\nabla\left( \nabla \cdot \mathbf{A} +\mu_{0}\epsilon_{0} \dfrac{\partial V}{\partial t}\right) = -\mu_{0} \mathbf{J} \end{align}$$

Therefore, information about the four Maxwell’s equations is all included in $(3)$, $(4)$.