Question

A parallel plate condenser is charged by connecting it to a battery. The battery is disconnected and a glass slab is introduced between the plates. Then,

• A. potential increases
• B. electric intensity incrases
• C. energy decreases
• D. capacity decreases

Answer 1

The correct option is C energy decreases

$\to$ When the capacitor is charged and battery is removed the charge present on capacitor remains constant. (Law of conservation of charge)
$\to$ We know $C=\frac{{\epsilon }_{0}A}{d}$
When slab is introduced, $C=\frac{k{\epsilon }_{0}A}{d}$
$C$ increases and $Q$ is constant.
We know: $Q=CV$

$\therefore$ $V$ has to decrease.
We know energy store in capacitor $=\frac{1}{2}QV$
$\therefore$ as V decreases, energy decreases.