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

Options

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

Explanation

To solve the problem step by step, we will find the distance from the surface of a uniformly charged solid sphere where the electrostatic potential is half of that at the center.

### Step 1: Understand the Problem
We have a solid sphere of radius \( R \) with a total charge \( q \) uniformly distributed throughout its volume. We need to find the distance from the surface of the sphere where the electrostatic potential \( V \) is equal to half of the potential at the center of the sphere.

### Step 2: Calculate the Potential at the Center
The potential \( V \) at the center of a uniformly charged sphere is given by the formula:
\[ V_{\text{center}} = \frac{3}{2} \frac{kq}{R} \]
where \( k \) is Coulomb's constant.

### Step 3: Set Up the Equation for the Potential at Point P
Let \( X \) be the distance from the center of the sphere to the point \( P \) where the potential is half of that at the center. According to the problem, we have:
\[ V_P = \frac{1}{2} V_{\text{center}} = \frac{1}{2} \left( \frac{3}{2} \frac{kq}{R} \right) = \frac{3}{4} \frac{kq}{R} \]

### Step 4: Calculate the Potential at Point P
For a point outside the sphere (at distance \( X \) from the center), the potential \( V_P \) is given by:
\[ V_P = \frac{kq}{X} \]

### Step 5: Equate the Two Expressions for Potential
Set the two expressions for \( V_P \) equal to each other:
\[ \frac{kq}{X} = \frac{3}{4} \frac{kq}{R} \]
Since \( kq \) is common on both sides, we can cancel it out:
\[ \frac{1}{X} = \frac{3}{4R} \]

### Step 6: Solve for X
Rearranging gives:
\[ X = \frac{4R}{3} \]

### Step 7: Find the Distance from the Surface
The distance \( Y \) from the surface of the sphere to point \( P \) is given by:
\[ Y = X - R \]
Substituting the value of \( X \):
\[ Y = \frac{4R}{3} - R = \frac{4R}{3} - \frac{3R}{3} = \frac{R}{3} \]

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