There is an air filled 1pF parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increases to 2pF. The dielectric constant of wax is

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the problem, we need to understand the relationship between capacitance, plate separation, and the dielectric constant. Let's break it down step by step.

### Step 1: Understand the formula for capacitance
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{A \epsilon_0}{d} \]
where:
- \( C \) is the capacitance,
- \( A \) is the area of the plates,
- \( \epsilon_0 \) is the permittivity of free space,
- \( d \) is the separation between the plates.

### Step 2: Initial conditions
Initially, we have an air-filled capacitor with a capacitance of 1 pF (picoFarad). Therefore, we can write:
\[ C_1 = \frac{A \epsilon_0}{d} = 1 \text{ pF} \]

### Step 3: Change in conditions
When the plate separation is doubled, the new separation \( d' \) becomes \( 2d \). The space between the plates is then filled with wax, which has a dielectric constant \( k \). The new capacitance \( C_2 \) can be expressed as:
\[ C_2 = \frac{A k \epsilon_0}{d'} \]
Substituting \( d' = 2d \):
\[ C_2 = \frac{A k \epsilon_0}{2d} \]

### Step 4: Relate the new capacitance to the old capacitance
Since we know that the new capacitance \( C_2 \) is 2 pF, we can set up the equation:
\[ \frac{A k \epsilon_0}{2d} = 2 \text{ pF} \]

### Step 5: Substitute the initial capacitance
From the initial condition, we have:
\[ \frac{A \epsilon_0}{d} = 1 \text{ pF} \]
Now, we can express \( A \epsilon_0 \) in terms of \( d \):
\[ A \epsilon_0 = 1 \text{ pF} \cdot d \]

### Step 6: Substitute into the new capacitance equation
Substituting \( A \epsilon_0 \) into the equation for \( C_2 \):
\[ C_2 = \frac{1 \text{ pF} \cdot d \cdot k}{2d} = \frac{k}{2} \text{ pF} \]
Setting this equal to the new capacitance:
\[ \frac{k}{2} = 2 \text{ pF} \]

### Step 7: Solve for the dielectric constant \( k \)
Multiplying both sides by 2 gives:
\[ k = 4 \]

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

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