Question

A parallel plate capacitor with air between the plates has a capacitance of 3 μF. Calculate the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant \( k \).

• A. \( 72 \, \mu F \)
• B. \( 81 \, \mu F \)
• C. \( 36 \, \mu F \)
• D. \( 6 \, \mu F \)

Hint:

Capacitance increases when the distance between plates is reduced or when a dielectric is introduced.

Answer 1

The Correct Option is A

Step 1: Formula for capacitance.
The capacitance \( C \) of a parallel plate capacitor is given by: \[ C = \frac{\varepsilon A}{d} \] where \( \varepsilon \) is the dielectric constant, \( A \) is the area of the plates, and \( d \) is the distance between them.

Step 2: Effects of halving the distance and introducing the dielectric.
If the distance is reduced by half and the dielectric constant \( k \) is introduced, the capacitance becomes: \[ C' = k \times \frac{\varepsilon A}{d/2} = 2k \times C \] Substituting \( k = 6 \), we get: \[ C' = 72 \, \mu F \]