Suppose$\mathit{J}\left(r\right)$ is constant in time, so (Prob. 7.60 ) $\mathit{p}\left(r,t\right)\mathbf{=}\mathit{p}\left(r,0\right)\mathbf{+}\mathit{p}\left(r,0\right)\mathit{t}$. Show that

$\mathit{E}\left(r,t\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\int }\frac{\mathbf{p}\left(r\text{'},t\right)}{{\mathbf{r}}^{\mathbf{2}}}\hat{\mathbf{r}}\mathit{d}\mathcal{b}\mathbf{\text{'}}$

that is, Coulomb’s law holds, with the charge density evaluated at the non-retarded time.

When is constant in time, the condition is as follows:

$$\begin{gathered} \mathit{p}\left(r,t\right)\mathbf{=}\mathit{p}\left(r,0\right) \\ \mathit{J}\left(r,t\right)\mathbf{=}\mathbf{0} \end{gathered}$$

Here,$p$ is the charge density and J is the current density.

Write the expression for the time-dependent generalization of Coulomb’s law.

$\mathit{E}\left(r,t\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\int }\left[\frac{p\left(r\text{'},t_r\right)}{r^2}\hat{r}+\frac{p\left(r\text{'},t_r\right)}{cr}\hat{r}-\frac{J\left(r\text{'},t_r\right)}{c^{2}r}\right]\mathit{d}\mathcal{b}\mathbf{\text{'}}$ …… (1)

Here,${\epsilon }_0$ is the permittivity of free space, c is the speed of light.

Substitute $p\left(r,t\right)=p\left(r,0\right)$and $J\left(r,t\right)=0$in equation (1).

$$\begin{gathered} E\left(r,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int \left[\frac{p\left(r\text{'},t_r\right)}{r^2}\hat{r}+\frac{p\left(r\text{'},t_r\right)}{cr}\hat{r}-\frac{J\left(r\text{'},t_r\right)}{c^{2}r}\right]d\mathcal{b}\text{'} \\ E\left(r,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int \left[\frac{p\left(r\text{'},t_r\right)}{r^2}\hat{r}+\frac{p\left(r\text{'},t_r\right)}{cr}\hat{r}-\frac{J\left(0\right)}{c^{2}r}\right]d\mathcal{b}\text{'} \\ E\left(r,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int \left[\frac{p\left(r\text{'},t_r\right)}{r^2}\hat{r}+\frac{p\left(\left(r\text{'},t_r\right)\right)}{cr}\hat{r}\right]d\mathcal{b}\text{'} \\ E\left(r,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int \left[\frac{p\left(r\text{'},0\right)+p\left(r\text{'},0\right)t_r}{r^2}\hat{r}+\frac{p\left(r\text{'},t_r\right)}{cr}\hat{r}\right]d\mathcal{b}\text{'}.....\left(2\right) \end{gathered}$$

Here,$t_r$is the retarded time which is given as:

$t_r=t-\frac{r}{c}$

Substitute $t_r=t-\frac{r}{c}$in equation (2).

$$\begin{gathered} E\left(r,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int \left[\frac{p\left(r\text{'},0\right)+p\left(r\text{'},0\right)\left(t-{\displaystyle \frac{r}{c}}\right)}{r^2}\hat{r}+\frac{p\left(r\text{'},t_r\right)}{cr}\hat{r}\right]d\mathcal{b}\text{'} \\ E\left(r,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int \left[\frac{p\left(r\text{'},0\right)+p\left(r\text{'},0\right)\left(t\right)}{r^2}\hat{r}+\frac{p\left(r\text{'},0\right)\left({\displaystyle \frac{r}{c}}\right)}{r^2}+\frac{p\left(r\text{'},0\right)}{cr}\right]\hat{r}d\mathcal{b}\text{'} \\ E\left(r,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int \left[\frac{p\left(r\text{'},0\right)+p\left(r\text{'},0\right)\left(t\right)}{r^2}\hat{r}\right]\hat{r}d\mathcal{b}\text{'} \\ E\left(r,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int \frac{p\left(r\text{'},t\right)}{r^2}\hat{r}d\mathcal{b}\text{'} \end{gathered}$$

Therefore, it is showed that the equation$E\left(r,t\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int \frac{p\left(r\text{'},t\right)}{r^2}\hat{r}d\mathcal{b}\text{'}$ holds the coulomb’s law with the charge density evaluated at the non-retarded time.