A metal foil of negligible thickness is introduced between the two plates of a capacitor at the centre. The capacitance of capacitor will be

Options

A. same
B. double
C. half
D. k times

Explanation

To solve the problem of finding the capacitance of a capacitor when a metal foil of negligible thickness is introduced between its plates, we can follow these steps:

### Step 1: Understand the Configuration
We have a parallel plate capacitor with plates separated by a distance \( d \). When a metal foil is introduced between the plates, it effectively divides the capacitor into two capacitors in series.

### Step 2: Analyze the Effect of the Metal Foil
The metal foil acts as a conductor and will have a dielectric constant \( k \) that approaches infinity. This is because a perfect conductor can be considered to have an infinite ability to polarize and shield electric fields.

### Step 3: Determine the New Distance Between Plates
Since the thickness of the metal foil is negligible, we can consider the distance between the plates to be effectively halved. Therefore, the distance between the plates of each of the two new capacitors becomes \( d/2 \).

### Step 4: Use the Capacitance Formula
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A}{d} \]
where:
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( d \) is the distance between the plates.

### Step 5: Calculate the Capacitance of Each Section
With the introduction of the metal foil, we now have two capacitors in series, each with a capacitance of:
\[ C_1 = \frac{\varepsilon_0 A}{d/2} = \frac{2\varepsilon_0 A}{d} \]
Since the foil is a perfect conductor, it does not contribute any additional capacitance.

### Step 6: Combine the Capacitances
For two capacitors in series, the total capacitance \( C \) is given by:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]
Since both capacitors have the same capacitance \( C_1 \):
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_1} = \frac{2}{C_1} \]
Thus, we have:
\[ C = \frac{C_1}{2} = \frac{2\varepsilon_0 A}{d} \cdot \frac{1}{2} = \frac{\varepsilon_0 A}{d} \]

### Conclusion
The capacitance of the capacitor with the metal foil introduced is the same as the original capacitance:
\[ C = \frac{\varepsilon_0 A}{d} \]