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 analyze the relationship between the capacitance of a parallel plate capacitor, the dielectric constant, and the changes made to the capacitor.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the initial separation between the plates be \( D \).
- Let the area of the plates be \( A \).
- The initial capacitance \( C_1 \) without the dielectric is given by the formula:
\[ C_1 = \frac{A \epsilon_0}{D} \]
where \( \epsilon_0 \) is the permittivity of free space.

2. **Change in Conditions**:
- The separation between the plates is doubled, so the new separation \( D' = 2D \).
- A dielectric (wax) is inserted between the plates, which has a dielectric constant \( k \).

3. **Capacitance with Dielectric**:
- The new capacitance \( C_2 \) with the dielectric inserted and the new separation is given by:
\[ C_2 = \frac{k A \epsilon_0}{D'} \]
Substituting \( D' = 2D \):
\[ C_2 = \frac{k A \epsilon_0}{2D} \]

4. **Given Condition**:
- According to the problem, the new capacitance \( C_2 \) is twice the initial capacitance \( C_1 \):
\[ C_2 = 2 C_1 \]

5. **Set Up the Equation**:
- Substitute the expressions for \( C_1 \) and \( C_2 \):
\[ \frac{k A \epsilon_0}{2D} = 2 \left(\frac{A \epsilon_0}{D}\right) \]

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

7. **Solve for \( k \)**:
- Multiply both sides by 2:
\[ k = 4 \]

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

---