A solid sphere of radius R is charged uniformly through out the volume. At what distance from its surface is the electric potential 1/4 of the potential at the centre?

Options

A. `(8R)/(3)`
B. `(R)/(3)`
C. `(5R)/(3)`
D. `(2R)/(3)`

Explanation

To solve the problem, we need to find the distance from the surface of a uniformly charged solid sphere where the electric potential is \( \frac{1}{4} \) of the potential at the center of the sphere.

### Step-by-Step Solution:

1. **Understand the Electric Potential Inside and Outside a Solid Sphere**:
The electric potential \( V \) at a distance \( r \) from the center of a uniformly charged solid sphere of radius \( R \) and total charge \( Q \) is given by:
- For \( r < R \) (inside the sphere):
\[ V = k \frac{Q}{2R} \left( 3 - \frac{r^2}{R^2} \right) \]
- For \( r \geq R \) (outside the sphere):
\[ V = k \frac{Q}{r} \]

2. **Calculate the Potential at the Center**:
To find the potential at the center of the sphere (\( r = 0 \)):
\[ V_{\text{center}} = k \frac{Q}{2R} \left( 3 - 0 \right) = \frac{3kQ}{2R} \]

3. **Set Up the Equation for \( \frac{1}{4} \) of the Potential at the Center**:
We need to find the distance \( d \) from the surface of the sphere where the potential is \( \frac{1}{4} V_{\text{center}} \):
\[ V = \frac{1}{4} V_{\text{center}} = \frac{1}{4} \left( \frac{3kQ}{2R} \right) = \frac{3kQ}{8R} \]

4. **Write the Expression for the Potential Outside the Sphere**:
For a point outside the sphere, at a distance \( r \) from the center:
\[ V = k \frac{Q}{r} \]
Setting this equal to \( \frac{3kQ}{8R} \):
\[ k \frac{Q}{r} = \frac{3kQ}{8R} \]

5. **Cancel \( kQ \) from Both Sides**:
\[ \frac{1}{r} = \frac{3}{8R} \]
Therefore:
\[ r = \frac{8R}{3} \]

6. **Calculate the Distance from the Surface**:
The distance from the surface of the sphere is given by:
\[ d = r - R = \frac{8R}{3} - R = \frac{8R}{3} - \frac{3R}{3} = \frac{5R}{3} \]

### Final Answer:
The distance from the surface where the electric potential is \( \frac{1}{4} \) of the potential at the center is:
\[ \frac{5R}{3} \]