A solid sphere of radius R has charge q uniformly distributed over its volume. The distance from it surfce at which the electrostatic potential is equal to half of the potential at the centre is

Options

A. R
B. R/2
C. R/3
D. 2R

Answer

To solve the problem, we need to find the distance from the surface of a solid sphere of radius \( R \) with charge \( q \) uniformly distributed over its volume, at which the electrostatic potential is equal to half of the potential at the center of the sphere.

### Step-by-Step Solution:

1. **Calculate the Potential at the Center of the Sphere**:
The potential \( V \) at a point inside a uniformly charged solid sphere is given by the formula:
\[ V(r) = \frac{kq}{2R} \left( 3R^2 - r^2 \right) \]
where \( k \) is Coulomb's constant, \( q \) is the total charge, and \( R \) is the radius of the sphere. For the center of the sphere, \( r = 0 \):
\[ V_{\text{center}} = \frac{kq}{2R} \left( 3R^2 - 0 \right) = \frac{3kq}{2R} \]

2. **Determine Half of the Potential at the Center**:
We need to find a point \( P \) where the potential is half of the potential at the center:
\[ V_P = \frac{1}{2} V_{\text{center}} = \frac{1}{2} \left( \frac{3kq}{2R} \right) = \frac{3kq}{4R} \]

3. **Set Up the Equation for Potential at Point \( P \)**:
The potential at a distance \( r \) from the center of the sphere (where \( r > R \)) is given by:
\[ V(r) = \frac{kq}{r} \]
We set this equal to \( \frac{3kq}{4R} \):
\[ \frac{kq}{r} = \frac{3kq}{4R} \]

4. **Solve for \( r \)**:
Cancel \( kq \) from both sides (assuming \( kq \neq 0 \)):
\[ \frac{1}{r} = \frac{3}{4R} \]
Thus,
\[ r = \frac{4R}{3} \]

5. **Calculate the Distance from the Surface**:
The distance from the surface of the sphere to point \( P \) is given by:
\[ \text{Distance from surface} = r - R = \frac{4R}{3} - R = \frac{4R}{3} - \frac{3R}{3} = \frac{R}{3} \]

### Final Answer:
The distance from the surface at which the electrostatic potential is equal to half of the potential at the center is:
\[ \frac{R}{3} \]