When a parallel plate capacitor is filled with wax after separation between plates is doubled, its capacitance becomes twice. What is the dielectric constant of wax?

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to find the dielectric constant (k) of wax when a parallel plate capacitor is filled with wax after the separation between the plates is doubled, and its capacitance becomes twice.

### Step-by-Step Solution:

1. **Understand the Capacitance Formula**:
The capacitance (C) of a parallel plate capacitor without a dielectric is given by the formula:
\[ C = \frac{A \epsilon_0}{d} \]
where:
- \( A \) = area of the plates
- \( \epsilon_0 \) = permittivity of free space
- \( d \) = separation between the plates

2. **Capacitance with Dielectric**:
When a dielectric material is inserted between the plates, the capacitance becomes:
\[ C' = \frac{k A \epsilon_0}{d} \]
where \( k \) is the dielectric constant of the material.

3. **Initial Conditions**:
Let the initial capacitance be \( C \):
\[ C = \frac{A \epsilon_0}{d} \]

4. **New Conditions**:
According to the problem, the separation between the plates is doubled, so the new separation \( d' = 2d \). The capacitance with the dielectric (wax) becomes:
\[ C' = \frac{k A \epsilon_0}{2d} \]

5. **Capacitance Becomes Twice**:
It is given that the new capacitance \( C' \) is twice the original capacitance \( C \):
\[ C' = 2C \]

6. **Substituting Values**:
Substitute the expressions for \( C \) and \( C' \):
\[ \frac{k A \epsilon_0}{2d} = 2 \left( \frac{A \epsilon_0}{d} \right) \]

7. **Simplifying the Equation**:
Cancel \( A \epsilon_0 \) from both sides:
\[ \frac{k}{2} = 2 \]

8. **Solving for k**:
Multiply both sides by 2 to isolate \( k \):
\[ k = 4 \]

### Conclusion:
The dielectric constant of wax is \( k = 4 \).