An air filled parallel capacitor has capacity of 2pF. The separation of the plates is doubled and the interspaces between the plates is filled with wax. If the capacity is increased to 6pF, the dielectric constant of was is

Options

A. 2
B. 3
C. 4
D. 6

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between capacitance, area, and distance
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: Set up the initial condition
Given that the initial capacitance \( C_1 \) is 2 pF, we can write:
\[ C_1 = \frac{A \epsilon_0}{D} = 2 \text{ pF} \]

### Step 3: Modify the conditions
The problem states that the separation of the plates is doubled. Therefore, the new separation \( D' \) is:
\[ D' = 2D \]
Now, the space between the plates is filled with wax, which has a dielectric constant \( k \). The new capacitance \( C_2 \) can be expressed as:
\[ C_2 = \frac{k A \epsilon_0}{D'} \]
Substituting \( D' \):
\[ C_2 = \frac{k A \epsilon_0}{2D} \]

### Step 4: Relate the new capacitance to the old capacitance
From the initial condition, we know that:
\[ \frac{A \epsilon_0}{D} = 2 \text{ pF} \]
Substituting this into the equation for \( C_2 \):
\[ C_2 = \frac{k}{2} \cdot 2 \text{ pF} = k \text{ pF} \]

### Step 5: Set up the equation with the new capacitance value
We are given that the new capacitance \( C_2 \) is 6 pF:
\[ k = 6 \text{ pF} \]

### Step 6: Solve for the dielectric constant \( k \)
From the equation \( k = 6 \), we find that the dielectric constant of the wax is:
\[ k = 6 \]

### Final Answer
The dielectric constant of the wax is **6**.

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