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) → \( K : 1 \)
Once the battery is disconnected, the charge on the capacitor stays constant. Filling it with a dielectric multiplies the capacitance by \( K \), and the energy stored is \( U = Q^{2}/2C \).
\( \displaystyle U_1 = \frac{Q^{2}}{2C} \qquad U_2 = \frac{Q^{2}}{2KC} \)
\( \displaystyle \frac{U_1}{U_2} = \frac{Q^{2}/2C}{Q^{2}/2KC} = K \)
So the ratio of the energies in the two cases is \( K : 1 \).