Question

A charge Q is distributed over two concentric conducting thin spherical shells radii $r$ and $R(R>r)$. If the surface charge densities on the two shells are equal, the electric potential at the common centre is

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{(R+r)}{2(R^2+r^2)}Q$
• B. $\frac{1}{4\pi {\epsilon }_0}\frac{2(R+r)}{(R^2+r^2)}Q$
• C. $\frac{1}{4\pi {\epsilon }_0}\frac{(R+2r)Q}{2(R^2+r^2)}$
• D. $\frac{1}{4\pi {\epsilon }_0}\frac{(R+r)}{(R^2+r^2)}Q$

Answer 1

The correct option is D $\frac{1}{4\pi {\epsilon }_0}\frac{(R+r)}{(R^2+r^2)}Q$

Let $\sigma$ be the surface charge density of the shells.

Charge on the inner shell, $Q_1=\sigma 4\pi r^2$
Charge on the outer shell,$Q_2=\sigma 4\pi R^2$
$\therefore \text{Total charge,}Q=\sigma 4\pi (r^2+R^2)$
$\Rightarrow \sigma =\frac{Q}{4\pi (r^2+R^2)}$

Potential at the common centre,
$V_C=\frac{K{Q}_1}{r}+\frac{K{Q}_2}{R}(\text{where}K=\frac{1}{4\pi {\epsilon }_0})$
$V_C=\frac{K\sigma 4\pi r^2}{r}+\frac{K\sigma 4\pi R^2}{R}=K\sigma 4\pi (r+R)$

$V_C=\frac{KQ4\pi (r+R)}{4\pi (r^2+R^2)}=\frac{1}{4\pi {\epsilon }_0}\frac{(r+R)Q}{(r^2+R^2)}$