1. Understand the charge distribution on the conducting shell

A conducting spherical shell of radius $R$ has net charge $q$. A point charge $Q$ is placed at the center.

Because of electrostatic induction:

• charge induced on the inner surface $= -Q$
• charge induced on the outer surface $= q+Q$

This is because the total charge on the conductor must remain $q$.

2. Find potential at point $P$ inside the shell

Point $P$ is at distance $R/2$ from the center, so it lies inside the shell.

Potential at $P$ is the algebraic sum of potentials due to:

• central charge $Q$
• inner spherical surface charge $-Q$
• outer spherical surface charge $q+Q$

3. Potential due to central charge $Q$ at $r=R/2$

$V_1 = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R/2} = \frac{2Q}{4\pi\varepsilon_0 R}$

4. Potential due to inner surface charge $-Q$

The inner surface is a spherical shell of radius $R$. For any point inside a uniformly charged spherical shell, potential is constant and equal to the surface potential:

$V_2 = \frac{1}{4\pi\varepsilon_0}\frac{-Q}{R}$

5. Potential due to outer surface charge $q+Q$

Similarly, for the outer spherical shell:

$V_3 = \frac{1}{4\pi\varepsilon_0}\frac{q+Q}{R}$

6. Add all contributions

$V_P = V_1 + V_2 + V_3$

$V_P = \frac{2Q}{4\pi\varepsilon_0 R} + \frac{-Q}{4\pi\varepsilon_0 R} + \frac{q+Q}{4\pi\varepsilon_0 R}$

$V_P = \frac{2Q - Q + q + Q}{4\pi\varepsilon_0 R}$

$V_P = \frac{2Q+q}{4\pi\varepsilon_0 R}$

So,

$V_P = \frac{2Q}{4\pi\varepsilon_0 R} + \frac{q}{4\pi\varepsilon_0 R}$

7. Match with options

This matches Option C.

8. Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

So they agree.