Question

The function of a dielectric in a capacitor is to

• A. reduce the plate area of the capacitor.
• B. to decrease the capacitance.
• C. reduce the effective potential on plates.
• D. increase the effective potential on plates.

Hint:

The dielectric field opposes the applied field, so the net potential falls.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When a dielectric is placed between the plates of a charged capacitor, the molecules polarise. The induced charges on the surface of the dielectric create a field opposite to the applied field.

Step 2: Effect:
The net field inside is reduced, so the potential difference for the same charge falls. This is why the capacitance \(C=\dfrac QV\) rises by the factor \(K\).

Step 3: Check the options:
(A) Plate area is not changed by a dielectric. (B) It increases the capacitance, not decreases. (D) It reduces the effective potential, not increases it. So (C) is correct.

Final Answer:
A dielectric reduces the effective potential across the plates. \[ \boxed{C} \]