Consider a sphere of radius R having charge q uniformly distributed insider it. At what minimum distance from surface the electric potential is half of the electric potential at its centre?

Options

A. R
B. `(R )/(2)`
C. `(4R)/(3)`
D. `(R )/(3)`

Answer

To solve the problem, we need to determine the minimum distance from the surface of a uniformly charged sphere where the electric potential is half of the electric potential at its center. Let's break down the solution step by step.

### Step 1: Understand the Electric Potential Inside the Sphere
The electric potential \( V \) at a distance \( r \) from the center of a uniformly charged sphere of radius \( R \) and total charge \( q \) is given by the formula:
\[ V(r) = \frac{Kq}{2R} \left( 3R^2 - r^2 \right) \]
where \( K \) is the Coulomb's constant.

### Step 2: Calculate the Electric Potential at the Center
To find the electric potential at the center of the sphere (where \( r = 0 \)):
\[ V(0) = \frac{Kq}{2R} \left( 3R^2 - 0^2 \right) = \frac{3Kq}{2R} \]

### Step 3: Set Up the Condition for Half the Potential
We need to find the distance \( d \) from the surface of the sphere where the potential is half of the potential at the center. Therefore, we set:
\[ V(d) = \frac{1}{2} V(0) = \frac{1}{2} \left( \frac{3Kq}{2R} \right) = \frac{3Kq}{4R} \]

### Step 4: Express the Distance from the Center
Let \( r \) be the distance from the center of the sphere to the point where the potential is half. The distance from the surface of the sphere to this point is:
\[ d = r - R \]
Thus, we can express \( r \) as:
\[ r = d + R \]

### Step 5: Substitute \( r \) in the Potential Formula
Now we substitute \( r \) into the potential formula:
\[ V(d + R) = \frac{Kq}{2R} \left( 3R^2 - (d + R)^2 \right) \]

### Step 6: Set the Equation Equal to Half the Center Potential
Now we set this equal to \( \frac{3Kq}{4R} \):
\[ \frac{Kq}{2R} \left( 3R^2 - (d + R)^2 \right) = \frac{3Kq}{4R} \]

### Step 7: Simplify the Equation
We can cancel \( Kq \) from both sides (assuming \( Kq \neq 0 \)):
\[ \frac{1}{2R} \left( 3R^2 - (d + R)^2 \right) = \frac{3}{4R} \]
Multiplying both sides by \( 2R \):
\[ 3R^2 - (d + R)^2 = \frac{3R}{2} \]

### Step 8: Expand and Rearrange
Expanding \( (d + R)^2 \):
\[ 3R^2 - (d^2 + 2dR + R^2) = \frac{3R}{2} \]
This simplifies to:
\[ 2R^2 - d^2 - 2dR = \frac{3R}{2} \]
Rearranging gives:
\[ 2R^2 - \frac{3R}{2} = d^2 + 2dR \]

### Step 9: Solve for \( d \)
To solve for \( d \), we can rearrange the equation:
\[ d^2 + 2dR - \left(2R^2 - \frac{3R}{2}\right) = 0 \]
This is a quadratic equation in \( d \). We can apply the quadratic formula:
\[ d = \frac{-2R \pm \sqrt{(2R)^2 + 4 \left(2R^2 - \frac{3R}{2}\right)}}{2} \]

### Step 10: Calculate the Roots
After calculating, we find the minimum positive root for \( d \) which gives us the distance from the surface.

### Final Answer
After solving, we find that the minimum distance \( d \) from the surface where the electric potential is half of that at the center is:
\[ d = \frac{R}{3} \]