A capacitor of capacitance C is first charged by a battery of V volts. It is disconnected from the battery and is filled with a dielectric of dielectric constant K. The ratio of energy stored in the two cases, respectively will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $K : 1$
Given:
Capacitance: $C$
Battery voltage: $V$
Dielectric constant: $K$
Case 1: Capacitor charged by battery
Energy stored: $U_1 = \frac{1}{2} C V^2$
Case 2: Capacitor disconnected from battery and filled with dielectric
New capacitance: $C' = K C$
Charge remains the same: $Q = C V$
Energy stored: $U_2 = \frac{Q^2}{2 C'} = \frac{(C V)^2}{2 K C} = \frac{C V^2}{2 K}$
Ratio of energies: $\frac{U_1}{U_2} = \frac{\frac{1}{2} C V^2}{\frac{C V^2}{2 K}} = K$
Answer: Ratio of energies $U_1 : U_2 = K : 1$