Question

A charge $q$ is distributed over two concentric hollow conducting sphere of radii $r$ and $R(>r)$ such that their surface charge densities are equal. The potential at their common centre is

• A. $\frac{q}{4\pi {ϵ}_0}(\frac{1}{r}+\frac{1}{R})$
• B. Zero
• C. $V_c=\frac{q}{4\pi {ϵ}_0}\left(\frac{r+R}{r^2+R^2}\right)$
• D. $\frac{q}{4\pi {ϵ}_0}\frac{(r+R)}{(r^2+R^2{)}^2}$

Answer 1

The correct option is C $V_c=\frac{q}{4\pi {ϵ}_0}\left(\frac{r+R}{r^2+R^2}\right)$

Let charge on inner sphere be $q_1$ and, on outer shell be $q_2$.

According to the given data in the problem,
$q=q_1+q_2\dots \dots \left(1\right)$

Since the surface charge density $\left(\sigma \right)$ on both the spheres are equal, so

$\sigma =\frac{q_1}{4\pi r^2}=\frac{q_2}{4\pi R^2}$

$\Rightarrow \frac{q_1}{q_2}=\frac{r^2}{R^2}........\left(2\right)$

Adding $+1$ on both sides we get,

$\Rightarrow \frac{q_1+q_2}{q_2}=\frac{r^2+R^2}{R^2}$

From equation $\left(1\right)$,

$\frac{q}{q_2}=\frac{r^2+R^2}{R^2}$

$\therefore q_2=\frac{q{R}^2}{r^2+R^2}...\left(3\right)$

Substituting this in $\left(2\right)$,

$q_1=\frac{q{r}^2}{r^2+R^2}...\left(4\right)$

The potential at the common centre is

$V_c=V_1+V_2$

$\Rightarrow V_c=\frac{1}{4\pi {ϵ}_0}[\frac{q_1}{r}+\frac{q_2}{R}]$

Substituting the values from $\left(3\right)$ and $\left(4\right)$

$\Rightarrow V_c=\frac{1}{4\pi {ϵ}_0}\left[\left(\frac{q}{r^2+R^2}\right)[\frac{r^2}{r}+\frac{R^2}{R}]\right]$

$\therefore V_c=\frac{q}{4\pi {ϵ}_0}\left(\frac{r+R}{r^2+R^2}\right)$

Hence, option (d) is the correct answer.