A solid dielectric (K = 1) sphere of radius R is charged uniformly by a total charge Q. At what distance from the centre will the electrostatic potential, be the average of that at the centre and at the surface.

Answer

To solve the problem, we need to find the distance from the center of a uniformly charged solid dielectric sphere where the electrostatic potential is the average of the potential at the center and the potential at the surface.

### Step-by-Step Solution:

1. **Calculate the Electrostatic Potential at the Center (V_C)**:
The electrostatic potential at the center of a uniformly charged sphere is given by the formula:
\[ V_C = \frac{3}{2} \frac{kQ}{R} \]
where \( k \) is the electrostatic constant, \( Q \) is the total charge, and \( R \) is the radius of the sphere.

2. **Calculate the Electrostatic Potential at the Surface (V_S)**:
The electrostatic potential at the surface of the sphere is given by:
\[ V_S = \frac{kQ}{R} \]

3. **Calculate the Average Potential (V_A)**:
The average potential between the center and the surface can be calculated as:
\[ V_A = \frac{V_C + V_S}{2} \]
Substituting the values we calculated:
\[ V_A = \frac{\frac{3}{2} \frac{kQ}{R} + \frac{kQ}{R}}{2} \]
Simplifying this:
\[ V_A = \frac{\frac{3kQ}{2R} + \frac{2kQ}{2R}}{2} = \frac{\frac{5kQ}{2R}}{2} = \frac{5kQ}{4R} \]

4. **Set the Potential at a Distance x from the Center Equal to the Average Potential**:
The potential at a distance \( x \) from the center of the sphere (inside the sphere) is given by:
\[ V_x = \frac{kQ}{x} \]
We want to find \( x \) such that:
\[ V_x = V_A \]
Therefore, we set:
\[ \frac{kQ}{x} = \frac{5kQ}{4R} \]

5. **Solve for x**:
Canceling \( kQ \) from both sides, we get:
\[ \frac{1}{x} = \frac{5}{4R} \]
Taking the reciprocal gives:
\[ x = \frac{4R}{5} \]

6. **Final Answer**:
Thus, the distance from the center where the electrostatic potential is the average of that at the center and at the surface is:
\[ x = 0.8R \]