📂Electrodynamics

Divergence of the Electric Field

Formulas¹

The divergence of the electric field $\mathbf{E}$ produced by a volume charge with volume charge density $\rho$ is as follows.

$$\nabla \cdot \mathbf{E} = \dfrac{1}{\epsilon_{0}} \rho ( \mathbf{r} )$$

Description

The divergence of the electric field is also referred to as the differential form of Gauss’s law. Integrating both sides yields the integral form of Gauss’s law.

Proof

Electric field produced by the volume charge

$$\mathbf{E}(\mathbf {r}) =\dfrac{1}{4\pi \epsilon_{0}} \int _\mathcal{V} \dfrac{\rho (\mathbf{r}^{\prime})}{\cR^2} \crH d\tau^{\prime}$$

Here, $\bcR=\mathbf{r}-\mathbf{r}^{\prime}$ is the separation vector. Since the divergence of the separation vector is $\nabla \cdot \left( \dfrac{1}{\cR^2}\crH \right) = 4\pi \delta^3(\bcR)$, calculating the divergence of the electric field results in the following.

$$\begin{align*} \nabla \cdot \mathbf{E} =&\ \dfrac{1}{4\pi \epsilon_{0}} \int \nabla \cdot \left( \dfrac{ \hat {\boldsymbol {\cR}} } { \cR ^2} \right) \rho ( \mathbf{r}^{\prime} ) d\tau^{\prime} \\ =&\ \dfrac{1}{4\pi \epsilon_{0}} \int 4\pi \delta ^3 (\bcR ) \rho ( \mathbf{r}^{\prime} ) d\tau^{\prime} \\ =&\ \dfrac{1}{4\pi \epsilon_{0}} 4\pi \int\delta ^3 (\boldsymbol{\mathbf{r} - \mathbf{r}^{\prime}} ) \rho ( \mathbf{r}^{\prime} ) d\tau^{\prime} \\ =&\ \dfrac{1}{\epsilon_{0}} \int \delta ^3 (\boldsymbol{\mathbf{r}^{\prime} - \mathbf{r}} ) \rho ( \mathbf{r}^{\prime} ) d\tau^{\prime} \\ =&\ \dfrac{1}{\epsilon_{0}} \rho (\mathbf{r} ) \end {align*}$$

Here, $\delta$ is the Dirac delta function. Taking the integral on both sides yields the following.

$$\int_\mathcal{V} \nabla \cdot \mathbf{E} d\tau = \dfrac{1}{\epsilon_{0}} \int _\mathcal{V} \rho (\mathbf{r})d\tau =\dfrac{1}{\epsilon_{0}}Q_{\text{in}}$$

Since $\rho$ is the volume charge density, integrating over the entire volume gives the total charge in the volume $Q_{\text{in}}$. And as known, this is the integral form of Gauss’s theorem.

■