A parallel plate capacitor of capacitance C connected across a supply of V volts has energy U. A slab of dielectric constant 5 is used to fill the gap between the plates without disconnecting the supply. The energy stored by the capacitor will be now :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 5 U
The capacitor stays connected to the supply, so the potential difference stays V. Filling the gap with a dielectric multiplies the capacitance by K.
\( C' = KC = 5C \)
\( \displaystyle U = \frac{1}{2}CV^{2} \)
\( \displaystyle U' = \frac{1}{2}(5C)V^{2} \)
\( U' = 5U \)