The function of a dielectric in a capacitor is
- to reduce the effective potential on plates.
- to increase the effective potential on plates.
- to decrease the capacity of capacitance.
- to reduce the plate area of the capacitor.
Solution
A dielectric material inserted between capacitor plates polarizes under the influence of the electric field, creating induced dipoles that generate an internal field opposing the original external field.
The net electric field becomes $E_{\text{net}} = E_0 / \kappa$, where $\kappa$ is the dielectric constant. Since potential difference is given by $V = E \cdot d$, this reduction in field strength lowers the potential difference across the plates for a fixed charge $Q$.
Capacitance is defined as $C = Q/V$, so a decrease in $V$ results in an increase in capacitance, which may be expressed as $C = \kappa C_0$.
Option A correctly identifies the reduction of effective potential, while options B, C, and D are inconsistent with the physical behavior.
Final answer: $\boxed{A}$
Asked in: MHT CET 2025 (26 April Shift 2)