Question

Two concentric spheres kept in air have radii 'R' and 'r'. They have similar charge and equal surface charge density ${}^{\prime }{\sigma }^{\prime }$. The electric potential at their common centre is. $({ϵ}_0=$ permittivity of free space):

• A. $\frac{\sigma (R+r)}{{ϵ}_0}$
• B. $\frac{\sigma (R-r)}{{ϵ}_0}$
• C. $\frac{\sigma (R+r)}{2{ϵ}_0}$
• D. $\frac{\sigma (R+r)}{4{ϵ}_0}$

Answer 1

The correct option is A $\frac{\sigma (R+r)}{{ϵ}_0}$

Let the charge of the bigger and the smaller sphere be $Q$ and $q$, respectively and both are of similar nature.

As the surface charge density of bigger sphere is equal to that of smaller sphere.

$\therefore$ $\sigma =\frac{Q}{4\pi R^2}=\frac{q}{4\pi r^2}$

Electric potential at point O $V_o=\frac{Q}{4\pi {ϵ}_{o}R}+\frac{q}{4\pi {ϵ}_{o}r}$Cancel

Or $V_o=\frac{QR}{4\pi {ϵ}_o{R}^2}+\frac{qr}{4\pi {ϵ}_o{r}^2}$

Or $V_o=\frac{\sigma R}{{ϵ}_o}+\frac{\sigma r}{{ϵ}_o}$

$⟹$ $V_o=\frac{\sigma (r+R)}{{ϵ}_o}$