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 step by step, we will follow the concepts of electric potential due to a uniformly charged solid sphere.

### Step 1: Understanding the Electric Potential at the Center of the Sphere
The electric potential \( V \) at the center of a uniformly charged solid sphere of radius \( R \) is given by the formula:
\[ V_{\text{center}} = \frac{3}{2} \frac{kQ}{R} \]
where \( k \) is Coulomb's constant and \( Q \) is the total charge of the sphere.

### Step 2: Setting Up the Condition for the Potential at a Distance \( x \)
We need to find the distance \( x \) from the surface of the sphere where the electric potential is one-fourth of the potential at the center:
\[ V = \frac{1}{4} V_{\text{center}} = \frac{1}{4} \left( \frac{3}{2} \frac{kQ}{R} \right) = \frac{3}{8} \frac{kQ}{R} \]

### Step 3: Electric Potential at a Point Outside the Sphere
For a point outside the uniformly charged sphere (at a distance \( r + x \) from the center), the electric potential is given by:
\[ V = \frac{kQ}{r + x} \]
where \( r \) is the radius of the sphere and \( x \) is the distance from the surface.

### Step 4: Equating the Two Expressions for Potential
We set the expression for potential at distance \( x \) equal to \( \frac{3}{8} \frac{kQ}{R} \):
\[ \frac{kQ}{R + x} = \frac{3}{8} \frac{kQ}{R} \]
We can cancel \( kQ \) from both sides (assuming \( Q \neq 0 \)):
\[ \frac{1}{R + x} = \frac{3}{8R} \]

### Step 5: Cross-Multiplying to Solve for \( x \)
Cross-multiplying gives:
\[ 8R = 3(R + x) \]
Expanding the right side:
\[ 8R = 3R + 3x \]
Subtracting \( 3R \) from both sides:
\[ 5R = 3x \]
Dividing by 3:
\[ x = \frac{5R}{3} \]

### Step 6: Finding the Distance from the Surface
Since \( x \) is the distance from the center of the sphere to the point where the potential is \( \frac{3}{8} \frac{kQ}{R} \), the distance from the surface of the sphere is:
\[ \text{Distance from the surface} = x - R = \frac{5R}{3} - R = \frac{5R}{3} - \frac{3R}{3} = \frac{2R}{3} \]

### Final Answer
Thus, the distance from the surface of the sphere where the electric potential is one-fourth of the potential at the center is:
\[ \frac{2R}{3} \]