A point charge q is placed inside a conducting spherical shell of inner radius 2R and outer radius 3R at a distance of R fro the centre of the shell. The electric potential at the centre of shell will (potential at infinity is zero).

Options

A. `(q)/(2R)`
B. `(4q)/(3R)`
C. `(5q)/(6R)`
D. `(2q)/(3R)`

Answer

To find the electric potential at the center of a conducting spherical shell with a point charge \( q \) placed inside, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Configuration**:
- We have a point charge \( q \) placed at a distance \( R \) from the center of a conducting spherical shell. The inner radius of the shell is \( 2R \) and the outer radius is \( 3R \).

2. **Understand the Charge Distribution**:
- The point charge \( q \) will induce a charge of \( -q \) on the inner surface of the conducting shell. Consequently, the outer surface of the shell will have a charge of \( +q \) to maintain overall neutrality.

3. **Calculate the Electric Potential Due to the Point Charge**:
- The electric potential \( V \) due to a point charge at a distance \( r \) is given by:
\[ V = k \frac{q}{r} \]
- At the center of the shell (which is at distance \( R \) from the charge), the potential due to the point charge \( q \) is:
\[ V_q = k \frac{q}{R} \]

4. **Calculate the Electric Potential Due to the Inner Surface of the Shell**:
- The inner surface of the shell has a charge of \( -q \) and is at a radius of \( 2R \). The potential due to this charge at the center is:
\[ V_{\text{inner}} = -k \frac{q}{2R} \]

5. **Calculate the Electric Potential Due to the Outer Surface of the Shell**:
- The outer surface of the shell has a charge of \( +q \) and is at a radius of \( 3R \). The potential due to this charge at the center is:
\[ V_{\text{outer}} = k \frac{q}{3R} \]

6. **Total Electric Potential at the Center of the Shell**:
- The total electric potential \( V_{\text{total}} \) at the center of the shell is the sum of the potentials due to the point charge, the inner surface, and the outer surface:
\[ V_{\text{total}} = V_q + V_{\text{inner}} + V_{\text{outer}} \]
- Substituting the values we calculated:
\[ V_{\text{total}} = k \frac{q}{R} - k \frac{q}{2R} + k \frac{q}{3R} \]

7. **Combine the Terms**:
- To combine the terms, we can factor out \( kq \):
\[ V_{\text{total}} = kq \left( \frac{1}{R} - \frac{1}{2R} + \frac{1}{3R} \right) \]
- Finding a common denominator (which is \( 6R \)):
\[ V_{\text{total}} = kq \left( \frac{6}{6R} - \frac{3}{6R} + \frac{2}{6R} \right) = kq \left( \frac{5}{6R} \right) \]

8. **Final Result**:
- Therefore, the electric potential at the center of the shell is:
\[ V_{\text{total}} = \frac{5kq}{6R} \]