A parallel plate capacitor has plate separation d and capacitance `25 muF`. If a metallic foil of thickness `(2)/(7)d` is introduced betwenn the plates, the capacitance would become

Options

A. `25 muF`
B. `35 muF`
C. `(125)/(7)muF`
D. `(175)/(2)muF`

Answer

To solve the problem, we need to find the new capacitance of a parallel plate capacitor after introducing a metallic foil of thickness \( \frac{2}{7}d \) between the plates. The initial capacitance is given as \( 25 \mu F \).

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- The initial capacitance \( C_0 \) of a parallel plate capacitor is given by the formula:
\[ C_0 = \frac{A \epsilon_0}{d} \]
- Here, \( A \) is the area of the plates, \( \epsilon_0 \) is the permittivity of free space, and \( d \) is the separation between the plates.

2. **Introducing the Metallic Foil**:
- A metallic foil of thickness \( t = \frac{2}{7}d \) is introduced between the plates.
- When a metallic conductor is placed between the plates of a capacitor, the electric field inside the conductor becomes zero.

3. **Calculating the Effective Distance**:
- The effective distance between the plates where the electric field exists is now reduced to:
\[ d' = d - t = d - \frac{2}{7}d = \frac{5}{7}d \]

4. **Finding the New Capacitance**:
- The new capacitance \( C \) can be calculated using the modified distance \( d' \):
\[ C = \frac{A \epsilon_0}{d'} \]
- Substitute \( d' \) into the equation:
\[ C = \frac{A \epsilon_0}{\frac{5}{7}d} = \frac{7A \epsilon_0}{5d} \]

5. **Relating New Capacitance to Initial Capacitance**:
- From the initial capacitance \( C_0 = \frac{A \epsilon_0}{d} = 25 \mu F \), we can express \( A \epsilon_0 \) as:
\[ A \epsilon_0 = 25d \]
- Substitute this into the new capacitance equation:
\[ C = \frac{7 \cdot 25d}{5d} = \frac{7 \cdot 25}{5} = 35 \mu F \]

### Final Answer:
The new capacitance after introducing the metallic foil is \( 35 \mu F \).

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