A parallel plate capacitor with air between the plates has a capacitance C. If the distance between the plates is doubled and the space between the plates is filled with a dielectric of dielectric constant 6, then the capacitance will become.

Options

A. 3C
B. `(C )/(3) `
C. 12C
D. `(C )/(6)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the initial capacitance
The capacitance \( C \) of a parallel plate capacitor with air between the plates is given by the formula:
\[ C = \frac{\varepsilon_0 A}{d} \]
where:
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( d \) is the distance between the plates.

### Step 2: Determine the new conditions
According to the problem:
- The distance between the plates is doubled, so the new distance \( d' = 2d \).
- The space between the plates is filled with a dielectric of dielectric constant \( K = 6 \).

### Step 3: Write the formula for the new capacitance
When a dielectric is introduced, the capacitance of the capacitor changes to:
\[ C' = \frac{K \varepsilon_0 A}{d'} \]
Substituting the new distance \( d' = 2d \):
\[ C' = \frac{K \varepsilon_0 A}{2d} \]

### Step 4: Substitute the values
Now substituting \( K = 6 \):
\[ C' = \frac{6 \varepsilon_0 A}{2d} \]
This simplifies to:
\[ C' = \frac{3 \varepsilon_0 A}{d} \]

### Step 5: Relate the new capacitance to the original capacitance
From the original capacitance \( C = \frac{\varepsilon_0 A}{d} \), we can express \( C' \) in terms of \( C \):
\[ C' = 3 \left( \frac{\varepsilon_0 A}{d} \right) = 3C \]

### Conclusion
Thus, the new capacitance \( C' \) becomes:
\[ C' = 3C \]

### Final Answer
The capacitance will become \( 3C \).

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