A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :

Capacitor PYQ — JEE Main 2022 · 29 Jun · Shift 1 · Q56 with Solution

View written solutionFree

Correct answer: A

Capacitor connected to a battery
Since the capacitor remains connected across a battery, the voltage $V$ remains constant.
Energy of a capacitor at constant voltage
The energy stored is $U = \frac{1}{2}CV^2$ So, when $V$ is constant, $U \propto C$
Capacitance with dielectric
For a parallel plate capacitor fully filled with dielectric constant $K$ , $C = K C_0$ where $C_0$ is the capacitance without dielectric.
Initial and final capacitances
Initially, dielectric constant is $K_1 = 10$ : $C_1 = 10C_0$ Finally, dielectric constant is $K_2 = 15$ : $C_2 = 15C_0$
Compare energies
Since $U \propto C$ , $\frac{U_2}{U_1} = \frac{C_2}{C_1} = \frac{15}{10} = 1.5$

Thus, $U_2 = 1.5U_1$ So the energy increases by $50\%$
Check options
- A: increase by $50\%$ ✅
- B: decrease by $15\%$ ❌
- C: increase by $25\%$ ❌
- D: increase by $33\%$ ❌

Therefore, the correct answer is A.