Question

A charge $Q$ is placed in front of a conducting sphere at a distance $2r$ from center of sphere as shown in the figure. Potential at point $P$ on the surface of sphere due to induced charges is

• A. $\frac{kQ}{6r}$
• B. $\frac{kQ}{2r}$
• C. $\frac{kQ}{3r}$
• D. \N

Answer 1

The correct option is A $\frac{kQ}{6r}$

Potential at $C$ due to $Q$ and induced charges is $V_C=\frac{kQ}{2r}$ [Potential at $C$ due to induced charges is zero]
Since Field is zero inside sphere $V_C=V_P$
Where $V_P=V_Q+V_{Q_{in}}$
$V_{Q_{in}}$ is potential due to induced charges.
$\frac{kQ}{2r}=\frac{kQ}{3r}+V_{Q_{in}}$
$\Rightarrow$ $V_{Q_{in}}=\frac{kQ}{6r}$