A highly conducting sheet of aluminium foil of negligible thickness is placed between the plates of a parallel plate capacitor. The foil is parallel to the plates. If the capacitance before the insertion of foil was `10 mu F`, its value after the insertion of foil will be.

Options

A. `20 mu F`
B. `10 mu F`
C. `5 mu F`
D. Zero

Answer

To solve the problem of finding the capacitance of a parallel plate capacitor after inserting a highly conducting sheet of aluminum foil, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Setup:**
- We have a parallel plate capacitor with an initial capacitance \( C_0 = 10 \, \mu F \).
- The distance between the plates is \( d \) and the area of the plates is \( A \).

2. **Capacitance Formula:**
- The capacitance of a parallel plate capacitor is given by the formula:
\[ C = \frac{\epsilon_0 A}{d} \]
- Here, \( \epsilon_0 \) is the permittivity of free space.

3. **Inserting the Conducting Foil:**
- When a highly conducting sheet (aluminum foil) of negligible thickness is placed between the plates, it effectively divides the capacitor into two capacitors in series.
- Each of these capacitors will have a distance of \( \frac{d}{2} \) between their plates.

4. **Capacitance of Each Section:**
- The capacitance of each section (let's call them \( C_1 \) and \( C_2 \)) is:
\[ C_1 = C_2 = \frac{\epsilon_0 A}{\frac{d}{2}} = \frac{2 \epsilon_0 A}{d} \]
- Since both sections are identical, we have \( C_1 = C_2 = 2C_0 \).

5. **Total Capacitance in Series:**
- The total capacitance \( C \) of two capacitors in series is given by:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]
- Substituting \( C_1 \) and \( C_2 \):
\[ \frac{1}{C} = \frac{1}{2C_0} + \frac{1}{2C_0} = \frac{2}{2C_0} = \frac{1}{C_0} \]
- Therefore, we find that:
\[ C = C_0 \]

6. **Final Result:**
- Since the initial capacitance \( C_0 = 10 \, \mu F \), the capacitance after the insertion of the foil remains:
\[ C = 10 \, \mu F \]

### Final Answer:
The value of the capacitance after the insertion of the foil will be \( 10 \, \mu F \).