There are two concentric hollow conducting spherical shells of radii r and `R ( R gt r)`. The charge on the outer shell is Q. What charge should be given to the inner shell, so that the potential at a point P , at a distance 2R from the common centre is zero ?

Options

A. `( -Qr )/( R )`
B. `(- QR)/( r )`
C. `( - 2QR)/( r )`
D. `-Q)`

Answer

To solve the problem, we need to determine the charge that should be given to the inner shell so that the potential at point P, located at a distance of 2R from the common center, is zero.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two concentric hollow conducting spherical shells. The inner shell has a radius \( r \) and the outer shell has a radius \( R \) (where \( R > r \)).
- The outer shell has a charge \( Q \), and we need to find the charge \( q \) on the inner shell.

2. **Potential at Point P**:
- Point P is located at a distance of \( 2R \) from the center of the shells.
- The potential \( V \) at point P due to a charged shell is given by the formula:
\[ V = \frac{kQ}{d} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( d \) is the distance from the center to the point where we are calculating the potential.

3. **Calculating the Potential Due to Each Shell**:
- The potential at point P due to the outer shell (charge \( Q \)):
\[ V_{\text{outer}} = \frac{kQ}{2R} \]
- The potential at point P due to the inner shell (charge \( q \)):
\[ V_{\text{inner}} = \frac{kq}{2R} \]

4. **Setting Up the Equation for Zero Potential**:
- For the total potential at point P to be zero, we set up the equation:
\[ V_{\text{outer}} + V_{\text{inner}} = 0 \]
- Substituting the expressions for the potentials:
\[ \frac{kQ}{2R} + \frac{kq}{2R} = 0 \]

5. **Simplifying the Equation**:
- We can factor out \( \frac{k}{2R} \) (which is non-zero) from the equation:
\[ \frac{k}{2R} (Q + q) = 0 \]
- This implies:
\[ Q + q = 0 \]

6. **Finding the Charge on the Inner Shell**:
- Rearranging gives:
\[ q = -Q \]
- Therefore, the charge that should be given to the inner shell is \( -Q \).

### Final Answer:
The charge on the inner shell should be \( -Q \).