The capacity of a parallel plate condenser is 15 ` mu ` F when the distance between its plates is 6 cm. if the distance between the plates is reduced to 2 cm, then the capacity of this prarallel condenser will be

Options

A. ` 15 mu F `
B. ` 30 mu F `
C. ` 45 mu F `
D. ` 60 mu F `

Answer

To solve the problem, we need to understand the relationship between the capacitance of a parallel plate capacitor and the distance between its plates. The capacitance \( C \) of a parallel plate capacitor is given by the formula:

\[ C = \frac{\epsilon_0 A}{D} \]

where:
- \( C \) is the capacitance,
- \( \epsilon_0 \) is the permittivity of free space (a constant),
- \( A \) is the area of the plates,
- \( D \) is the distance between the plates.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial capacitance \( C = 15 \, \mu F \)
- Initial distance between plates \( D = 6 \, cm \)
- New distance between plates \( D' = 2 \, cm \)

2. **Set up the equation for the initial capacitance:**
\[ C = \frac{\epsilon_0 A}{D} \]
Substituting the known values:
\[ 15 \times 10^{-6} = \frac{\epsilon_0 A}{0.06} \]

3. **Rearranging to find the area \( A \):**
\[ A = 15 \times 10^{-6} \times 0.06 \times \epsilon_0 \]

4. **Set up the equation for the new capacitance \( C' \):**
\[ C' = \frac{\epsilon_0 A}{D'} \]
Substituting \( D' = 0.02 \, m \):
\[ C' = \frac{\epsilon_0 A}{0.02} \]

5. **Substituting the expression for \( A \) from the first equation into the second:**
\[ C' = \frac{\epsilon_0 \left(15 \times 10^{-6} \times 0.06 \times \epsilon_0\right)}{0.02} \]

6. **Simplifying the expression:**
\[ C' = \frac{15 \times 10^{-6} \times 0.06 \times \epsilon_0^2}{0.02} \]

7. **Calculate \( C' \):**
\[ C' = 15 \times 10^{-6} \times 0.06 \times 50 \times \epsilon_0 \]
\[ C' = 15 \times 10^{-6} \times 3 = 45 \times 10^{-6} \, F \]
\[ C' = 45 \, \mu F \]

### Final Answer:
The new capacitance \( C' \) when the distance between the plates is reduced to 2 cm is \( 45 \, \mu F \).