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 find the new capacitance of a parallel plate capacitor when the distance between the plates is doubled and the area is halved, we can follow these steps:

### 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 distance between the plates.

### Step 2: Identify the initial values
From the problem, we know:
- Initial capacitance \( C = 12 \, \mu F \)
- Let the initial area be \( A \) and the initial distance be \( d \).

### Step 3: Determine the new values
According to the problem:
- The new distance \( d' = 2d \) (the distance is doubled).
- The new area \( A' = \frac{A}{2} \) (the area is halved).

### Step 4: Substitute the new values into the capacitance formula
Now, we can calculate the new capacitance \( C' \):
\[ C' = \frac{A' \epsilon_0}{d'} \]
Substituting the new values:
\[ C' = \frac{\left(\frac{A}{2}\right) \epsilon_0}{2d} \]

### Step 5: Simplify the expression
Now simplifying the expression:
\[ C' = \frac{A \epsilon_0}{2 \cdot 2d} = \frac{A \epsilon_0}{4d} \]
This shows that:
\[ C' = \frac{C}{4} \]
where \( C \) is the original capacitance.

### Step 6: Calculate the new capacitance
Since the original capacitance \( C = 12 \, \mu F \):
\[ C' = \frac{12 \, \mu F}{4} = 3 \, \mu F \]

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

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