Question

The capacitance of a parallel plate capacitor is 1.5 µF. If the distance between the plates is halved and the space between the plates is filled with a medium of dielectric constant 3, then the new capacitance is:

• A. 4.5 µF
• B. 1.5 µF
• C. 9 µF
• D. 6 µF

Hint:

The capacitance increases when the distance between the plates is reduced or when a dielectric medium is introduced between the plates.

Answer 1

The Correct Option is C

Step 1:
The capacitance of a parallel plate capacitor 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, and \( d \) is the separation between the plates.
Step 2:
When the distance \( d \) between the plates is halved, the capacitance doubles. Moreover, when a dielectric material is inserted, the capacitance increases by the factor of the dielectric constant \( K \).
Step 3:
The new capacitance is: \[ C' = K \times \frac{C}{2} \] where \( K = 3 \) and \( C = 1.5 \, \mu F \). \[ C' = 3 \times \frac{1.5}{2} = 9 \, \mu F \]