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 a distance `R/2` from the centre of the shell is

Options

A. a) `((q+_Q))/(4pi epsilon_(0))(2)/(R)`
B. b) `(2Q)/(4pi epsilon_(0)R)`
C. c) `(2Q)/(4 pi epsilon_(0)R)-(2q)/(4pi epsilon_(0)R)`
D. d)`(2Q)/(4pi epsilon_(0)R)+(q)/(4pi epsilon_(0)R)`

Answer

To solve the problem, we need to find the electrostatic potential at a point P located at a distance \( \frac{R}{2} \) from the center of a thin spherical conducting shell of radius \( R \) that has a charge \( q \), with another charge \( Q \) placed at the center of the shell.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a spherical conducting shell with radius \( R \) and charge \( q \).
- There is a point charge \( Q \) located at the center of this shell.
- We need to find the electrostatic potential at point P, which is at a distance \( \frac{R}{2} \) from the center.

2. **Electrostatic Potential Due to the Conducting Shell**:
- Inside a conducting shell, the electric field is zero. Therefore, the potential inside the shell is constant and equal to the potential on the surface of the shell.
- The potential \( V_1 \) due to the charge \( q \) on the shell at the surface (radius \( R \)) is given by:
\[ V_1 = \frac{q}{4 \pi \epsilon_0 R} \]
- Since the electric field inside the shell is zero, this potential remains the same at any point inside the shell, including point P.

3. **Electrostatic Potential Due to the Central Charge \( Q \)**:
- The potential \( V_2 \) due to the point charge \( Q \) at a distance \( \frac{R}{2} \) from the center is given by:
\[ V_2 = \frac{Q}{4 \pi \epsilon_0 \left(\frac{R}{2}\right)} = \frac{2Q}{4 \pi \epsilon_0 R} = \frac{Q}{2 \pi \epsilon_0 R} \]

4. **Total Electrostatic Potential at Point P**:
- The total electrostatic potential \( V \) at point P is the sum of the potentials due to the conducting shell and the central charge:
\[ V = V_1 + V_2 = \frac{q}{4 \pi \epsilon_0 R} + \frac{Q}{2 \pi \epsilon_0 R} \]
- To combine these, we can express \( V_2 \) with a common denominator:
\[ V = \frac{q}{4 \pi \epsilon_0 R} + \frac{2Q}{4 \pi \epsilon_0 R} = \frac{q + 2Q}{4 \pi \epsilon_0 R} \]

5. **Final Answer**:
- Therefore, the electrostatic potential at point P is:
\[ V = \frac{q + 2Q}{4 \pi \epsilon_0 R} \]