A point charge Q is placed inside a conducting spherical shell of inner radius 3R and outer radius 5R at a distance R from the centre of the shell. The electric potential at the centre of the shell will be

Options

A. `(1)/(4pi epsi_(0)).(Q)/(R )`
B. `(1)/(4pi epsi_(0)).(5Q)/(6R)`
C. `(1)/(4pi epsi_(0)).(13Q)/(15R)`
D. `(1)/(4pi epsi_(0)).(7Q)/(9R)`

Answer

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

### Step 1: Understand the Configuration
We have a conducting spherical shell with:
- Inner radius = \( 3R \)
- Outer radius = \( 5R \)
- A point charge \( Q \) is placed at a distance \( R \) from the center of the shell.

### Step 2: Identify the Electric Field Inside the Conducting Shell
In a conductor, the electric field inside the conducting material is zero. Therefore, the electric field in the region between the inner and outer surfaces of the shell (from \( 3R \) to \( 5R \)) is also zero.

### Step 3: Induced Charges
The point charge \( Q \) will induce a charge of \( -Q \) on the inner surface of the shell (at radius \( 3R \)). This is because the electric field inside the conductor must remain zero. Consequently, a charge of \( +Q \) will be induced on the outer surface of the shell (at radius \( 5R \)).

### Step 4: Calculate the Electric Potential at the Center
The electric potential \( V \) at a point due to a point charge \( Q \) is given by the formula:
\[ V = \frac{KQ}{r} \]
where \( K = \frac{1}{4\pi \epsilon_0} \) and \( r \) is the distance from the charge to the point where the potential is being calculated.

Since we are interested in the potential at the center of the shell (point O), we need to consider the contributions from both the point charge \( Q \) and the induced charges.

1. **Potential due to the point charge \( Q \)**:
The distance from the charge \( Q \) (located at distance \( R \) from the center) to the center of the shell (point O) is \( R \). Thus, the potential at point O due to charge \( Q \) is:
\[ V_Q = \frac{KQ}{R} \]

2. **Potential due to the induced charge on the inner surface**:
The induced charge \( -Q \) on the inner surface does not contribute to the potential at the center of the shell because the electric field inside the conductor is zero.

3. **Potential due to the induced charge on the outer surface**:
The induced charge \( +Q \) on the outer surface also does not contribute to the potential at the center because it is outside the shell.

### Step 5: Final Potential Calculation
The total potential at the center O is thus solely due to the point charge \( Q \):
\[ V_O = V_Q = \frac{KQ}{R} \]

### Step 6: Express in Terms of \( \epsilon_0 \)
Substituting \( K = \frac{1}{4\pi \epsilon_0} \):
\[ V_O = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R} \]

### Conclusion
The electric potential at the center of the conducting spherical shell is:
\[ V_O = \frac{Q}{4\pi \epsilon_0 R} \]