A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

Options

A. decreases
B. remains unchanged
C. becomes infinite
D. increases

Explanation

To solve the problem of how the capacitance of a capacitor changes when a sheet of aluminum foil is introduced between its plates, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Capacitance**:
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\epsilon_0 A}{d} \]
where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( A \) is the area of one of the plates,
- \( d \) is the separation between the plates.

2. **Introducing Aluminum Foil**:
When a sheet of aluminum foil (which is a conductor) is introduced between the plates of the capacitor, it will affect the electric field between the plates. However, since the thickness of the aluminum foil is negligible, we can consider it as having no significant thickness.

3. **Effect of the Aluminum Foil**:
The aluminum foil will create two additional surfaces that will have equal and opposite charges induced on them. This effectively divides the capacitor into two capacitors in series, each with a distance of \( d/2 \) (since the thickness of the foil is negligible).

4. **Capacitance with Aluminum**:
The capacitance of the system can be thought of as:
\[ C' = \frac{\epsilon_0 A}{d - t} \]
where \( t \) is the thickness of the aluminum foil. Since \( t \) is negligible (approaching zero), we can say:
\[ C' = \frac{\epsilon_0 A}{d} \]

5. **Conclusion**:
Since the capacitance formula remains the same and the thickness of the aluminum foil does not affect the distance significantly, the capacitance of the capacitor remains unchanged. Therefore, the answer is:
\[ \text{Capacitance remains unchanged.} \]