If the distance between the plates of parallel plate capacitor is halved and the dielectric constant of dielectric is doubled, then its capacity will

Options

A. increase by `16` times
B. increase by `4` times
C. increase by `2` times
D. remain the same

Explanation

To solve the problem, we need to analyze how the capacitance of a parallel plate capacitor changes when the distance between the plates is halved and the dielectric constant is doubled.

### Step-by-Step Solution:

1. **Understand the Formula for Capacitance**:
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{A \epsilon_0 K}{d} \]
where:
- \( A \) = area of the plates
- \( \epsilon_0 \) = permittivity of free space
- \( K \) = dielectric constant
- \( d \) = distance between the plates

2. **Initial Conditions**:
Let's denote the initial capacitance as \( C_1 \):
\[ C_1 = \frac{A \epsilon_0 K}{d} \]

3. **New Conditions**:
According to the problem:
- The distance \( d \) is halved: \( d' = \frac{d}{2} \)
- The dielectric constant \( K \) is doubled: \( K' = 2K \)

4. **Calculate the New Capacitance**:
Now, we can find the new capacitance \( C_2 \) using the new values:
\[ C_2 = \frac{A \epsilon_0 K'}{d'} \]
Substituting \( K' \) and \( d' \):
\[ C_2 = \frac{A \epsilon_0 (2K)}{\frac{d}{2}} = \frac{A \epsilon_0 (2K) \cdot 2}{d} = \frac{4A \epsilon_0 K}{d} \]

5. **Relate New Capacitance to Initial Capacitance**:
We can express \( C_2 \) in terms of \( C_1 \):
\[ C_2 = 4 \cdot \frac{A \epsilon_0 K}{d} = 4C_1 \]

6. **Conclusion**:
Therefore, the new capacitance \( C_2 \) is 4 times the initial capacitance \( C_1 \):
\[ C_2 = 4C_1 \]

### Final Answer:
The capacitance will increase by a factor of 4.

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