The electric potential of some configuration is given by the expression

$\mathbf{V}\left(r\right)\mathbf{=}\mathbf{A}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{\lambda r}}}{\mathbf{r}}$

Where $\mathit{A}$and $\mathit{\lambda }$are constants. Find the electric field$\mathit{E}\left(r\right)$, the charge density$\mathit{\rho }\left(r\right)$,and the total charge$\left(Q\right)$.

Consider the given expression.

$\mathbf{V}\left(r\right)\mathbf{=}\mathbf{A}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{\lambda r}}}{\mathbf{r}}$ …… (1)

Here, $A$and $\lambda$are constant.

Write the representation of electric filed is gradient of a scalar potential.

$E=-\nabla V$ ……(2)

Hence, the electric filed is calculated as follows:

$$\begin{gathered} E=-\nabla V \\ =-\frac{\partial }{\partial r}\left(A\frac{e^{-\lambda r}}{r}\right) \\ =-A\left[\frac{-r{e}^{-\lambda r}\left(-\lambda \right)-e^{-\lambda r}\left(1\right)}{r^2}\right]\stackrel{⏜}{r} \\ =-A\left[\frac{-r{e}^{-\lambda r}-e^{-\lambda r}}{r^2}\right]\stackrel{⏜}{r} \end{gathered}$$

Solve further.

$E=A\left(r\lambda e^{-\lambda r}+e^{-\lambda r}\right)\frac{\stackrel{⏜}{r}}{r^2}$ …… (3)

$=A{e}^{-\lambda r}\left(1+r\lambda \right)\frac{\stackrel{⏜}{r}}{r^2}$

Thus, the electric field is$E=A{e}^{-\lambda r}\left(1+r\lambda \right)\frac{\stackrel{⏜}{r}}{r^2}$ .

The Gauss’s law in different way[S1]

$$\begin{gathered} \nabla .E=\frac{1}{{\epsilon }_0}\rho \\ \rho ={\epsilon }_0\nabla .E \end{gathered}$$ …… (4)

Substitute $E=A{e}^{-\lambda r}\left(1+r\lambda \right)\frac{\stackrel{⏜}{r}}{r^2}$in equation (4).

$$\begin{gathered} \rho ={\epsilon }_0\nabla .\left(A{e}^{-\lambda r}\left(1+r\lambda \right)\frac{r}{r^2}\right) \\ ={\epsilon }_0\left[\left(A{e}^{-\lambda r}\left(1+\lambda \right)\right)\left(\nabla .\frac{\stackrel{⏜}{r}}{r^2}\right)+\left(\frac{\stackrel{⏜}{r}}{r^2}\right).\nabla \left(A{e}^{-\lambda r}\left(1+r\lambda \right)\right)\right] \end{gathered}$$

But $\nabla .\frac{\stackrel{⏜}{r}}{r^2}=4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)$,

Therefore,

$$\begin{gathered} \rho ={\epsilon }_0\left[\left(A{e}^{-\lambda r}\left(1+r\lambda \right)\right)\left(4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)\right)+\left(\frac{\stackrel{⏜}{r}}{r^2}\right)\nabla .\left(A{e}^{-\lambda r}\left(1+r\lambda \right)\right)\right] \\ ={\epsilon }_0\left[A4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)+\left(\frac{\stackrel{⏜}{\mathrm{r}}}{{\mathrm{r}}^2}\right)\nabla .\left({\mathrm{Ae}}^{-\mathrm{\lambda r}}\left(1+\mathrm{r\lambda }\right)\right)\right] \end{gathered}$$

Since$\left(e^{-\lambda r}\left(1+r\lambda \right)\right)\left(4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)\right)=4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)$,

Write the expression for the total charge.

$$\begin{gathered} Q=\int \rho d\tau \\ =\int A{\epsilon }_0\left[4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)-\frac{{\mathrm{\lambda }}^2}{\mathrm{r}}{\mathrm{e}}^{-\mathrm{\lambda r}}\right]d\tau \\ =\int A{\epsilon }_{0}4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)d\tau -\int A{\epsilon }_0\frac{{\mathrm{\lambda }}^2}{\mathrm{r}}{\mathrm{e}}^{-\mathrm{\lambda r}}\mathrm{d\tau } \\ =4{\mathrm{\pi \epsilon }}_0\mathrm{A}\int {\mathrm{\delta }}^3\left(\mathrm{r}\right)\mathrm{d\tau }-{\mathrm{A\epsilon }}_0{\mathrm{\lambda }}^2\int \frac{{\mathrm{e}}^{-\mathrm{\lambda r}}}{\mathrm{r}}\mathrm{d\tau } \end{gathered}$$

Simplify further,

$$\begin{gathered} Q=4{\mathrm{\pi \epsilon }}_0\mathrm{A}\left(1\right)-{\mathrm{A\epsilon }}_0{\mathrm{\lambda }}^2\int \frac{{\mathrm{e}}^{-\mathrm{\lambda r}}}{\mathrm{r}}\left(4{\mathrm{\pi r}}^2\mathrm{dr}\right) \\ =4{\mathrm{\pi \epsilon }}_0\mathrm{A}-4{\mathrm{\pi A\epsilon }}_0{\mathrm{\lambda }}^2\int {\mathrm{re}}^{-\mathrm{\lambda r}}\mathrm{dr} \\ =4{\mathrm{\pi \epsilon }}_0\mathrm{A}-4{\mathrm{\pi A\epsilon }}_0{\mathrm{\lambda }}^2{\int }_0^{\infty }{\mathrm{re}}^{-\mathrm{\lambda r}}\mathrm{dr} \\ =4{\mathrm{\pi \epsilon }}_0\mathrm{A}-4{\mathrm{\pi A\epsilon }}_0{\mathrm{\lambda }}^2{\left[-\frac{{\mathrm{re}}^{-\mathrm{\lambda r}}}{\mathrm{\lambda }}-\frac{{\mathrm{e}}^{-\mathrm{\lambda r}}}{{\mathrm{\lambda }}^2}\right]}_0^{\infty } \end{gathered}$$

Also,

$$\begin{gathered} Q=4{\mathrm{\pi \epsilon }}_0\mathrm{A}-4{\mathrm{\pi A\epsilon }}_0{\mathrm{\lambda }}^2\left(\frac{1}{{\mathrm{\lambda }}^2}\right) \\ =4{\mathrm{\pi \epsilon }}_0\mathrm{A}-4{\mathrm{\pi A\epsilon }}_0 \\ =0 \end{gathered}$$

Thus, the total charge Q is 0.