Question

A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point P which is at a distance $\frac{R}{2}$ from the centre of the shell is :

• A. $\frac{2Q}{4\pi {ϵ}_{0}R}-\frac{2q}{4\pi {ϵ}_{0}R}$
• B. $\frac{2Q}{4\pi {ϵ}_{0}R}+\frac{q}{4\pi {ϵ}_{0}R}$
• C. $\frac{(q+Q)}{4\pi {ϵ}_{0}R}+\frac{2}{R}$
• D. $\frac{2Q}{4\pi {ϵ}_{0}R}$

Answer 1

The correct option is B $\frac{2Q}{4\pi {ϵ}_{0}R}+\frac{q}{4\pi {ϵ}_{0}R}$

$\text{Step 1: Potential due to charge}Q$

Potential at P due to charge $Q=$$V_1=\frac{Q}{4\pi {ϵ}_{o}r}=\frac{2Q}{4\pi {ϵ}_{o}R}$

$\text{Step 2: Potential due to sphere}$

Potential at P due to sphere $=V_2=\frac{q}{4\pi {ϵ}_{o}R}$ which is the same for all points inside the shell.

$\text{Step 3: Net potential at point P}$

As potential is scalar quantity, so net potential at a point will be sum of potentials due to all the charge configurations.

Therefore, $V=V_1+V_2=\frac{2Q}{4\pi {ϵ}_{o}R}+\frac{q}{4\pi {ϵ}_{o}R}$

Hence Option(B) is correct.