91Ó°ÊÓ

Consider the given expression.

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

Here, $A$and $\mathrm{λ}$are constant.

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

$E=-∇V$ …â¶Ä¦(2)

Hence, the electric filed is calculated as follows:

$$\begin{gathered} E=-∇V \\ =-\frac{∂}{∂r}\left(A\frac{e^{-\mathrm{λ}r}}{r}\right) \\ =-A\left[\frac{-r{e}^{-\mathrm{λ}r}\left(-\mathrm{λ}\right)-e^{-\mathrm{λ}r}\left(1\right)}{r^2}\right]\stackrel{Áåœ}{r} \\ =-A\left[\frac{-r{e}^{-\mathrm{λ}r}-e^{-\mathrm{λ}r}}{r^2}\right]\stackrel{Áåœ}{r} \end{gathered}$$

Solve further.

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

$=A{e}^{-\mathrm{λ}r}\left(1+r\mathrm{λ}\right)\frac{\stackrel{Áåœ}{r}}{r^2}$

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

The Gauss’s law in different way[S1]

$$\begin{gathered} ∇.E=\frac{1}{{\mathrm{Î\mu }}_0}\mathrm{ÒÏ} \\ \mathrm{ÒÏ}={\mathrm{Î\mu }}_0∇.E \end{gathered}$$ …… (4)

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

$$\begin{gathered} \mathrm{ÒÏ}={\mathrm{Î\mu }}_0∇.\left(A{e}^{-\mathrm{λ}r}\left(1+r\mathrm{λ}\right)\frac{r}{r^2}\right) \\ ={\mathrm{Î\mu }}_0\left[\left(A{e}^{-\mathrm{λ}r}\left(1+\mathrm{λ}\right)\right)\left(∇.\frac{\stackrel{Áåœ}{r}}{r^2}\right)+\left(\frac{\stackrel{Áåœ}{r}}{r^2}\right).∇\left(A{e}^{-\mathrm{λ}r}\left(1+r\mathrm{λ}\right)\right)\right] \end{gathered}$$

But $∇.\frac{\stackrel{Áåœ}{r}}{r^2}=4{\mathrm{πδ}}^3\left(\mathrm{r}\right)$,

Therefore,

$$\begin{gathered} \mathrm{ÒÏ}={\mathrm{Î\mu }}_0\left[\left(A{e}^{-\mathrm{λ}r}\left(1+r\mathrm{λ}\right)\right)\left(4{\mathrm{πδ}}^3\left(\mathrm{r}\right)\right)+\left(\frac{\stackrel{Áåœ}{r}}{r^2}\right)∇.\left(A{e}^{-\mathrm{λ}r}\left(1+r\mathrm{λ}\right)\right)\right] \\ ={\mathrm{Î\mu }}_0\left[A4{\mathrm{πδ}}^3\left(\mathrm{r}\right)+\left(\frac{\stackrel{Áåœ}{\mathrm{r}}}{{\mathrm{r}}^2}\right)∇.\left({\mathrm{Ae}}^{-\mathrm{λ°ù}}\left(1+\mathrm{°ùλ}\right)\right)\right] \end{gathered}$$

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

Write the expression for the total charge.

$$\begin{gathered} Q=∫\mathrm{ÒÏ}d\mathrm{Ï„} \\ =∫A{\mathrm{Î\mu }}_0\left[4{\mathrm{πδ}}^3\left(\mathrm{r}\right)-\frac{{\mathrm{λ}}^2}{\mathrm{r}}{\mathrm{e}}^{-\mathrm{λ°ù}}\right]d\mathrm{Ï„} \\ =∫A{\mathrm{Î\mu }}_{0}4{\mathrm{πδ}}^3\left(\mathrm{r}\right)d\mathrm{Ï„}-∫A{\mathrm{Î\mu }}_0\frac{{\mathrm{λ}}^2}{\mathrm{r}}{\mathrm{e}}^{-\mathrm{λ°ù}}\mathrm{»åÏ„} \\ =4{\mathrm{Ï€Î\mu }}_0\mathrm{A}∫{\mathrm{δ}}^3\left(\mathrm{r}\right)\mathrm{»åÏ„}-{\mathrm{´¡Î\mu }}_0{\mathrm{λ}}^2∫\frac{{\mathrm{e}}^{-\mathrm{λ°ù}}}{\mathrm{r}}\mathrm{»åÏ„} \end{gathered}$$

Simplify further,

$$\begin{gathered} Q=4{\mathrm{Ï€Î\mu }}_0\mathrm{A}\left(1\right)-{\mathrm{´¡Î\mu }}_0{\mathrm{λ}}^2∫\frac{{\mathrm{e}}^{-\mathrm{λ°ù}}}{\mathrm{r}}\left(4{\mathrm{Ï€°ù}}^2\mathrm{dr}\right) \\ =4{\mathrm{Ï€Î\mu }}_0\mathrm{A}-4{\mathrm{Ï€´¡Î\mu }}_0{\mathrm{λ}}^2∫{\mathrm{re}}^{-\mathrm{λ°ù}}\mathrm{dr} \\ =4{\mathrm{Ï€Î\mu }}_0\mathrm{A}-4{\mathrm{Ï€´¡Î\mu }}_0{\mathrm{λ}}^2{∫}_0^{∞}{\mathrm{re}}^{-\mathrm{λ°ù}}\mathrm{dr} \\ =4{\mathrm{Ï€Î\mu }}_0\mathrm{A}-4{\mathrm{Ï€´¡Î\mu }}_0{\mathrm{λ}}^2{\left[-\frac{{\mathrm{re}}^{-\mathrm{λ°ù}}}{\mathrm{λ}}-\frac{{\mathrm{e}}^{-\mathrm{λ°ù}}}{{\mathrm{λ}}^2}\right]}_0^{∞} \end{gathered}$$

Also,

$$\begin{gathered} Q=4{\mathrm{Ï€Î\mu }}_0\mathrm{A}-4{\mathrm{Ï€´¡Î\mu }}_0{\mathrm{λ}}^2\left(\frac{1}{{\mathrm{λ}}^2}\right) \\ =4{\mathrm{Ï€Î\mu }}_0\mathrm{A}-4{\mathrm{Ï€´¡Î\mu }}_0 \\ =0 \end{gathered}$$

Thus, the total charge Q is 0.