The capacitance of an air capacitor is `15 muF` te separation between the parallel plates is `6 mm`. A copper plate of `3 mm` thickness is introduced symmetrically between the plates. The capacitance now becomes

Options

A. `5 muF`
B. `7.5 muF`
C. `22.5 muF`
D. `30 muF`

Explanation

To solve the problem, we need to find the new capacitance of the capacitor after introducing a copper plate between the parallel plates. Let's break down the solution step by step.

### Step 1: Understand the Initial Setup
The initial capacitance \( C_0 \) of the air capacitor is given as:
\[ C_0 = 15 \, \mu F \]
The separation between the plates \( D \) is:
\[ D = 6 \, mm = 6 \times 10^{-3} \, m \]

### Step 2: Introduce the Copper Plate
A copper plate of thickness \( t \) is introduced symmetrically between the plates. The thickness of the copper plate is:
\[ t = 3 \, mm = 3 \times 10^{-3} \, m \]

### Step 3: Calculate the New Separation
After placing the copper plate, the effective distance between the plates (air gap) changes. The new separation \( D' \) between the two air gaps is:
\[ D' = D - t = 6 \, mm - 3 \, mm = 3 \, mm = 3 \times 10^{-3} \, m \]

### Step 4: Determine the Capacitance with the Copper Plate
The capacitance 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 separation between the plates.

Since we have a copper plate in between, we can treat the system as two capacitors in series:
1. The capacitance of the air gap above the copper plate \( C_1 \):
\[ C_1 = \frac{\varepsilon_0 A}{D'} \]
Where \( D' = 3 \, mm = 3 \times 10^{-3} \, m \).

2. The capacitance of the air gap below the copper plate \( C_2 \):
\[ C_2 = \frac{\varepsilon_0 A}{D'} \]
Where \( D' = 3 \, mm = 3 \times 10^{-3} \, m \).

### Step 5: Calculate the Total Capacitance
The total capacitance \( C \) of capacitors in series is given by:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]
Since \( C_1 = C_2 \):
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_1} = \frac{2}{C_1} \]
Thus,
\[ C = \frac{C_1}{2} \]

### Step 6: Substitute the Values
Now, substituting \( C_1 \):
\[ C_1 = \frac{\varepsilon_0 A}{D'} = \frac{\varepsilon_0 A}{3 \times 10^{-3}} \]
Since \( C_0 = \frac{\varepsilon_0 A}{D} \) where \( D = 6 \times 10^{-3} \):
\[ C_0 = \frac{\varepsilon_0 A}{6 \times 10^{-3}} = 15 \, \mu F \]
Thus, we can express \( C_1 \) in terms of \( C_0 \):
\[ C_1 = \frac{15 \, \mu F \cdot 6 \times 10^{-3}}{3 \times 10^{-3}} = 30 \, \mu F \]

Now substituting back into the total capacitance:
\[ C = \frac{30 \, \mu F}{2} = 15 \, \mu F \]

### Final Answer
The new capacitance after introducing the copper plate is:
\[ C = 15 \, \mu F \]