A parallel plate air capacitor has a capacity of 2 pF. If the separation between its plates is doubled and a mica sheet is introduced between its plates, its capacity becomes 6 pF. What is the dielectric constant of mica?

Options

A. 6
B. 5
C. 4
D. 3

Explanation

To solve the problem, we need to use the formula for the capacitance of a parallel plate capacitor and the effect of introducing a dielectric material.

### Step-by-step Solution:

1. **Understand the formula for capacitance:**
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon A}{d} \]
where:
- \( C \) is the capacitance,
- \( \varepsilon \) is the permittivity of the material between the plates,
- \( A \) is the area of the plates,
- \( d \) is the separation between the plates.

2. **Initial conditions:**
The initial capacitance \( C_1 \) of the air capacitor is given as:
\[ C_1 = 2 \text{ pF} = 2 \times 10^{-12} \text{ F} \]
For air, the permittivity \( \varepsilon_0 \) is used, so:
\[ C_1 = \frac{\varepsilon_0 A}{d} \]

3. **New conditions after changes:**
When the separation \( d \) is doubled, the new separation becomes \( 2d \). The introduction of the mica sheet changes the capacitance to \( C_2 \):
\[ C_2 = 6 \text{ pF} = 6 \times 10^{-12} \text{ F} \]
The capacitance with the mica sheet is given by:
\[ C_2 = \frac{\varepsilon_r \varepsilon_0 A}{2d} \]
where \( \varepsilon_r \) is the dielectric constant of mica.

4. **Set up the equations:**
From the initial condition:
\[ C_1 = \frac{\varepsilon_0 A}{d} \quad \text{(1)} \]
From the new condition:
\[ C_2 = \frac{\varepsilon_r \varepsilon_0 A}{2d} \quad \text{(2)} \]

5. **Relate the two capacitances:**
We can express \( A \) in terms of \( C_1 \) and \( d \):
\[ A = \frac{C_1 d}{\varepsilon_0} \]
Substitute \( A \) into equation (2):
\[ C_2 = \frac{\varepsilon_r \varepsilon_0 \left(\frac{C_1 d}{\varepsilon_0}\right)}{2d} \]
Simplifying this gives:
\[ C_2 = \frac{\varepsilon_r C_1}{2} \]

6. **Substituting known values:**
Now substitute \( C_1 = 2 \text{ pF} \) and \( C_2 = 6 \text{ pF} \):
\[ 6 = \frac{\varepsilon_r \cdot 2}{2} \]
Simplifying this gives:
\[ 6 = \varepsilon_r \]

7. **Conclusion:**
The dielectric constant of mica is:
\[ \varepsilon_r = 6 \]

### Final Answer:
The dielectric constant of mica is **6**.