The capacitance of a parallel plate capacitor is `12 muF`. If the distance between the plates is doubled and area is halved, then new capacitance will be

Options

A. `8 muF`
B. `6 muF`
C. `4 muF`
D. `3 muF`

Answer

To solve the problem, we need to use the formula for the capacitance of a parallel plate capacitor, which is given by:

\[ C = \frac{\varepsilon_0 \cdot A}{d} \]

where:
- \( C \) is the capacitance,
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( d \) is the distance between the plates.

### Step 1: Identify the initial conditions
Given:
- Initial capacitance \( C_1 = 12 \, \mu F \)
- Let the initial area be \( A \)
- Let the initial distance be \( d \)

Using the formula for capacitance, we can express the initial capacitance as:

\[ C_1 = \frac{\varepsilon_0 \cdot A}{d} \]

### Step 2: Determine the new conditions
According to the problem:
- The distance between the plates is doubled: \( d' = 2d \)
- The area of the plates is halved: \( A' = \frac{A}{2} \)

### Step 3: Calculate the new capacitance
Using the new values in the capacitance formula, we have:

\[ C_2 = \frac{\varepsilon_0 \cdot A'}{d'} \]

Substituting the new area and distance:

\[ C_2 = \frac{\varepsilon_0 \cdot \left(\frac{A}{2}\right)}{2d} \]

This simplifies to:

\[ C_2 = \frac{\varepsilon_0 \cdot A}{2 \cdot 2d} = \frac{\varepsilon_0 \cdot A}{4d} \]

### Step 4: Relate the new capacitance to the initial capacitance
From the initial capacitance formula, we know:

\[ C_1 = \frac{\varepsilon_0 \cdot A}{d} \]

Thus, we can express \( C_2 \) in terms of \( C_1 \):

\[ C_2 = \frac{C_1}{4} \]

### Step 5: Substitute the value of \( C_1 \)
Now substituting the value of \( C_1 \):

\[ C_2 = \frac{12 \, \mu F}{4} = 3 \, \mu F \]

### Final Answer
The new capacitance \( C_2 \) is \( 3 \, \mu F \).

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