The capacity of an isolated sphere is increased n times when it is enclosed by an earthed concentric sphere. The ratio of their radii is

Options

A. `(n^(2))/(n-1)`
B. `(n)/(n-1)`
C. `(2n)/(n+1)`
D. `(2n+1)/(n+1)`

Answer

To solve the problem, we need to find the ratio of the radii of two concentric spheres, given that the capacitance of an isolated sphere is increased \( n \) times when it is enclosed by an earthed concentric sphere.

### Step-by-Step Solution:

1. **Understand the Capacitance of a Sphere**:
The capacitance \( C \) of an isolated sphere of radius \( A \) is given by the formula:
\[ C = 4 \pi \epsilon_0 A \]
where \( \epsilon_0 \) is the permittivity of free space.

2. **Capacitance of the Concentric Spheres**:
When the isolated sphere of radius \( A \) is enclosed by a larger earthed sphere of radius \( B \), the capacitance \( C' \) of the system is given by:
\[ C' = \frac{4 \pi \epsilon_0}{\frac{1}{A} - \frac{1}{B}} = \frac{4 \pi \epsilon_0 AB}{B - A} \]

3. **Given Condition**:
According to the problem, the capacitance of the isolated sphere is increased \( n \) times when it is enclosed by the earthed sphere. Thus, we can write:
\[ C' = n \cdot C \]
Substituting the expressions for \( C' \) and \( C \):
\[ \frac{4 \pi \epsilon_0 AB}{B - A} = n \cdot (4 \pi \epsilon_0 A) \]

4. **Simplifying the Equation**:
We can cancel \( 4 \pi \epsilon_0 \) from both sides:
\[ \frac{AB}{B - A} = nA \]

5. **Rearranging the Equation**:
Cross-multiplying gives:
\[ AB = nA(B - A) \]
Expanding the right side:
\[ AB = nAB - nA^2 \]
Rearranging terms:
\[ nA^2 = nAB - AB \]
Factoring out \( AB \):
\[ nA^2 = AB(n - 1) \]

6. **Finding the Ratio of Radii**:
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ nA = B(n - 1) \]
Rearranging gives:
\[ \frac{B}{A} = \frac{n}{n - 1} \]

### Final Answer:
The ratio of the radii \( \frac{B}{A} \) is:
\[ \frac{B}{A} = \frac{n}{n - 1} \]