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 of radius \( R \) where the electric potential is 1.5 times the potential at the surface of the sphere. Let's break down the solution step by step.

### Step 1: Understand the potential inside and on the surface of the sphere
The potential \( V \) at a distance \( r \) from the center of a uniformly charged solid insulating sphere of radius \( R \) and total charge \( Q \) is given by:

\[ V(r) = \frac{kQ}{2R} \left( 3 - \frac{r^2}{R^2} \right) \quad \text{for } r \leq R \]

where \( k \) is Coulomb's constant.

The potential at the surface of the sphere (when \( r = R \)) is:

\[ V(R) = \frac{kQ}{R} \]

### Step 2: Set up the equation based on the given condition
According to the problem, the potential at a point inside the sphere is 1.5 times the potential at the surface. Therefore, we can write:

\[ V(r) = 1.5 \times V(R) \]

Substituting the expression for \( V(R) \):

\[ V(r) = 1.5 \times \frac{kQ}{R} \]

### Step 3: Equate the two expressions for potential
Now, we can set the two expressions for potential equal to each other:

\[ \frac{kQ}{2R} \left( 3 - \frac{r^2}{R^2} \right) = 1.5 \times \frac{kQ}{R} \]

### Step 4: Simplify the equation
We can cancel \( kQ \) from both sides (assuming \( kQ \neq 0 \)):

\[ \frac{1}{2R} \left( 3 - \frac{r^2}{R^2} \right) = 1.5 \]

Multiplying both sides by \( 2R \):

\[ 3 - \frac{r^2}{R^2} = 3R \]

### Step 5: Solve for \( r^2 \)
Rearranging the equation gives:

\[ -\frac{r^2}{R^2} = 3R - 3 \]

Multiplying through by -1:

\[ \frac{r^2}{R^2} = 3 - 3R \]

### Step 6: Find the value of \( r \)
Now, we can express \( r^2 \):

\[ r^2 = R^2 (3 - 3R) \]

To find \( r \), we take the square root:

\[ r = R \sqrt{3 - 3R} \]

### Step 7: Determine the point inside the sphere
Since \( R \) is the radius of the sphere, the only valid solution for \( r \) must be less than or equal to \( R \). The only point where the potential is 1.5 times the surface potential occurs at the center of the sphere when \( R = 0 \).

### Conclusion
Thus, the point inside the sphere where the potential is 1.5 times the potential at the surface is at the center of the sphere.

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

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