Question

If in the concentric hollow spheres of radii r and R (where r<R), the charge Q is distributed such that their surface densities are same, then the potential at their common centre is

• A. $\frac{Q(r^2+R^2)}{4{\mathrm{\pi \epsilon }}_o(r+R)}$
• B. $\frac{QR}{(r+R)}$
• C. zero
• D. $\frac{Q(r+R)}{4{\mathrm{\pi \epsilon }}_o(r^2+R^2)}$

Answer 1

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

Q is distributed
So $q_1+q_2=Q$
Given that surface densities are same.
$\Rightarrow \frac{q_1}{4{\mathrm{\pi r}}^2}=\frac{q_2}{4{\mathrm{\pi R}}^2}$
$q_1=\frac{Q{r}^2}{(r^2+R^2)}\mathrm{and}{q}_2=\frac{Q{R}^2}{(r^2+R^2)}\mathrm{Potential}\mathrm{at}\mathrm{common}\mathrm{centre}=\frac{1}{4{\mathrm{\pi \epsilon }}_o}\left(\frac{Q{r}^2}{(r^2+R^2)}+\frac{Q{R}^2}{(r^2+R^2)}\right]$
=$\frac{Q(r+R)}{4{\mathrm{\pi \epsilon }}_o(r^2+R^2)}$