Question

The function of a dielectric in a capacitor is

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

Hint:

Dielectric constant $K$ always increases capacitance by factor $K$ ($C_{new} = KC_{old}$).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A dielectric is an insulating material placed between capacitor plates. We need to identify its primary effect on the electrical parameters of the capacitor.

Step 2: Detailed Explanation:
When a dielectric with constant \( K \) is inserted between the plates of a charged isolated capacitor (with constant charge \( Q \)):
1. The capacitance increases: \( C_{new} = K C_0 \).
2. The electric field decreases: \( E_{new} = \frac{E_0}{K} \) because of induced opposite charges on the dielectric surface.
3. Since \( V = E \times d \), the potential difference across the plates decreases: \( V_{new} = \frac{V_0}{K} \).
This reduction in potential for a given charge allows more charge to be stored for the same potential limit, effectively increasing the "capacity".

Step 4: Final Answer:
The function of a dielectric is to reduce the effective potential between the plates.