A solid insulating sphere of radius R is given a charge Q. If at a point inside the sphere the potential is 1.5 times the potential at the surface, this point will be

Options

A. at a distance of `2R//3` from the centre
B. at the centre
C. at a distance of `2R//3` from the surface
D. data insufficient

Answer

To solve the problem, we need to find the point inside a solid insulating sphere where the potential is 1.5 times the potential at the surface of the sphere. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a solid insulating sphere of radius \( R \) with a total charge \( Q \). We need to find a point inside this sphere where the potential \( V \) is 1.5 times the potential at the surface of the sphere.

2. **Potential at the Surface**:
The potential \( V_R \) at the surface of the sphere (radius \( R \)) can be calculated using the formula:
\[ V_R = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant.

3. **Potential Inside the Sphere**:
The potential \( V_r \) at a distance \( r \) from the center of the sphere (where \( r < R \)) is given by:
\[ V_r = \frac{kQ}{2R} \left(3 - \frac{r^2}{R^2}\right) \]

4. **Setting Up the Equation**:
According to the problem, we have:
\[ V_r = 1.5 \times V_R \]
Substituting the expressions for \( V_r \) and \( V_R \):
\[ \frac{kQ}{2R} \left(3 - \frac{r^2}{R^2}\right) = 1.5 \times \frac{kQ}{R} \]

5. **Simplifying the Equation**:
We can cancel \( kQ \) from both sides (assuming \( Q \neq 0 \)):
\[ \frac{1}{2R} \left(3 - \frac{r^2}{R^2}\right) = 1.5 \times \frac{1}{R} \]
Multiplying both sides by \( 2R \):
\[ 3 - \frac{r^2}{R^2} = 3 \]

6. **Solving for \( r^2 \)**:
Rearranging gives:
\[ -\frac{r^2}{R^2} = 0 \]
This implies:
\[ r^2 = 0 \]
Therefore, \( r = 0 \).

7. **Conclusion**:
The point where the potential is 1.5 times the potential at the surface of the sphere is at the center of the sphere.

### Final Answer:
The point will be at the **center of the sphere**.

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