Question

A parallel plate capacitor has a capacitance of \( 10 \, \mu\text{F} \). What will be its capacitance when the distance between the plates is halved?

• A. \( 20 \, \mu\text{F} \)
• B. \( 5 \, \mu\text{F} \)
• C. \( 10 \, \mu\text{F} \)
• D. \( 40 \, \mu\text{F} \)

Hint:

The capacitance of a parallel plate capacitor is inversely proportional to the distance between the plates. Reducing the distance increases the capacitance.

Answer 1

The Correct Option is A

Step 1: Use the formula for capacitance of a parallel plate capacitor The capacitance \( C \) of a parallel plate capacitor is given by: \[ C = \epsilon_0 \frac{A}{d} \] where: - \( \epsilon_0 \) is the permittivity of free space, - \( A \) is the area of the plates, - \( d \) is the distance between the plates. Step 2: Understand the effect of halving the distance Since the capacitance is inversely proportional to the distance between the plates, halving the distance \( d \) will double the capacitance. Step 3: Calculate the new capacitance Given the initial capacitance is \( 10 \, \mu\text{F} \), when the distance between the plates is halved, the new capacitance will be: \[ C_{\text{new}} = 2 \times 10 \, \mu\text{F} = 20 \, \mu\text{F} \] Answer: Therefore, the new capacitance is \( 20 \, \mu\text{F} \). So, the correct answer is option (1).