A solid conducting sphere of radius 'a' is surrounded by a thin uncharged concentric conducting shell of radius `2a`. A point charge q is placed at a distance `4a` from common centre of conducting sphere and shell. The inner sphere is then grounded
The potential of outer shell is

Options

A. `(q)/(32 pi epsi_(0)a)`
B. `(q)/(16 pi epsi_(0)a)`
C. `(q)/(8 pi epsi_(0)a)`
D. `(q)/(4 pi epsi_(0)a)`

Answer

To solve the problem of finding the potential on the outer shell of a system consisting of a solid conducting sphere, a concentric conducting shell, and a point charge, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Configuration**:
- We have a solid conducting sphere of radius \( a \).
- This sphere is surrounded by a thin uncharged concentric conducting shell of radius \( 2a \).
- A point charge \( q \) is placed at a distance \( 4a \) from the common center of the conducting sphere and shell.
- The inner sphere is grounded.

2. **Identify the Effects of Grounding**:
- Grounding the inner conducting sphere means that its potential is set to zero. This will influence the charge distribution on the sphere.

3. **Calculate the Potential at the Surface of the Inner Sphere**:
- The potential \( V \) at the surface of the inner sphere due to the point charge \( q \) is given by:
\[ V = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{4a} + \frac{Q_1}{a} \]
where \( Q_1 \) is the induced charge on the inner surface of the conducting sphere.

4. **Set the Potential to Zero**:
- Since the inner sphere is grounded, we set the total potential to zero:
\[ 0 = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{4a} + \frac{Q_1}{a} \]
- Rearranging gives:
\[ Q_1 = -\frac{q}{4} \]

5. **Determine the Charge on the Outer Shell**:
- By conservation of charge, the total charge on the outer shell must be equal and opposite to the charge on the inner surface. Therefore, the charge on the outer surface of the shell is:
\[ Q_{\text{outer}} = +\frac{q}{4} \]

6. **Calculate the Potential on the Outer Shell**:
- The potential \( V_0 \) at the outer shell (radius \( 2a \)) is due to the point charge \( q \) and the charge on the outer shell:
\[ V_0 = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{4a} + \frac{1}{4\pi\epsilon_0} \cdot \frac{Q_{\text{outer}}}{2a} \]
- Substituting \( Q_{\text{outer}} = \frac{q}{4} \):
\[ V_0 = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{4a} + \frac{1}{4\pi\epsilon_0} \cdot \frac{\frac{q}{4}}{2a} \]
- Simplifying:
\[ V_0 = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{4a} + \frac{q}{8a} \right) \]
\[ V_0 = \frac{1}{4\pi\epsilon_0} \cdot \frac{2q + q}{8a} = \frac{3q}{8\pi\epsilon_0 a} \]

7. **Final Expression for Potential**:
- Therefore, the potential at the outer surface of the conducting shell is:
\[ V_0 = \frac{3q}{8\pi\epsilon_0 a} \]