Question

The potential in a region varies as $V\left( r \right) = \dfrac{q}{{4\pi {\varepsilon _0}}}{e^{ - \alpha r}}$. Then the charge enclosed inside a space of radius $R$ is __________.

Answer 1

Hint: Here, you are given variation of potential as a function of $r$ and you are asked to find the charge that is enclosed in a sphere of radius $R$. If somehow you can find the electric field as a function of $r$, you can use Gauss’s law in order to find the charge enclosed. So, you need to find the electric field from the given potential.

Complete step by step answer:
Let us first derive a relation between electric field and potential.Let the electric field and potential at ${\mathbf{r}}$ be equal to ${\mathbf{E}}$ and $V$. If a point charge $q$ is displaced from ${\mathbf{r}} \to {\mathbf{r}} + d{\mathbf{r}}$, work will be done equal to force dot displacement. $dW = {\mathbf{F}}.d{\mathbf{r}}$. The force on the charged particle will be ${\mathbf{F}} = q{\mathbf{E}}$ which will give $dW = q{\mathbf{E}}.d{\mathbf{r}}$.

The change in potential energy will be $dU = - dW = - q{\mathbf{E}}.d{\mathbf{r}}$.
You know that the potential is given as,
$dV = \dfrac{{dU}}{q} \\
\Rightarrow dV = \dfrac{{ - q{\mathbf{E}}.d{\mathbf{r}}}}{q} \\
\Rightarrow dV = - {\mathbf{E}}.d{\mathbf{r}}\\
\Rightarrow dV = - \overrightarrow E .d\overrightarrow r $
Here, we are dealing with radial distribution and hence the direction of electric field and the vector $r$ would be the same and eventually the angle between them would be zero. Therefore, we have, $dV = - Edr \to E = - \dfrac{{dV}}{{dr}}$

Let us substitute the value of potential in the above obtained equation.
$E = - \dfrac{{d\left( {\dfrac{q}{{4\pi {\varepsilon _0}}}{e^{ - \alpha r}}} \right)}}{{dr}}$ .
Here, $\dfrac{q}{{4\pi {\varepsilon _0}}}$ and $\alpha $ are constants.
$
E = - \dfrac{q}{{4\pi {\varepsilon _0}}}\dfrac{{d\left( {{e^{ - \alpha r}}} \right)}}{{dr}}
\Rightarrow E = - \dfrac{q}{{4\pi {\varepsilon _0}}}\left( { - \alpha } \right){e^{ - \alpha r}} \\
\Rightarrow E = \dfrac{{q\alpha }}{{4\pi {\varepsilon _0}}}{e^{ - \alpha r}} \\ $
Gauss’s law is given as the,
$\oint {\overrightarrow E .d\overrightarrow A } = \dfrac{Q}{{{\varepsilon _0}}}$, that is, flux through a closed surface is equal to the total charge enclosed divided by ${\varepsilon _0}$.
\[
Q = {\varepsilon _0}\oint {\overrightarrow E .d\overrightarrow A } \\
\Rightarrow Q = {\varepsilon _0}\int {EdA} \\
\Rightarrow Q = {\varepsilon _0}E\int {dA} \\
\Rightarrow Q = {\varepsilon _0}E4\pi {R^2} \\
\Rightarrow Q = 4\pi {\varepsilon _0}{R^2}E \\
\Rightarrow Q = 4\pi {\varepsilon _0}{R^2}\dfrac{{q\alpha }}{{4\pi {\varepsilon _0}}}{e^{ - \alpha R}} \\
\therefore Q = q\alpha {R^2}{e^{ - \alpha R}} \]

Therefore, the total charge enclosed is given as \[Q = q\alpha {R^2}{e^{ - \alpha R}}\].

Note: The point to be noted here, is that we wrote $\int {EdA = E\int {dA} } $. The reason for this is the electric field at any point on the sphere of radius $R$ will be the same and therefore $E$ at $r = R$ can be treated as constant. Always keep in mind the relation between electric field and potential and also the Gauss’s Law.