A charged capacitor of capacitance ‘C’ has energy ‘U’. A dielectric slab with dielectric constant ‘K’ is inserted between its plates, keeping it connected across a potential difference of ‘V’ volt. Now,
A. New capacitance is K times C.
B. New energy stored by the capacitor is K times U.
C. New potential difference across the capacitor is K times V.
D. New charge on the capacitor is \( 1/K \) times the original charge Q.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A and B only
The capacitor stays connected to the source, so its potential difference stays fixed at V while the dielectric multiplies the capacitance by K.
\( C' = KC \)
\( V' = V \)
(A) New capacitance is K times C. — Correct. Filling the space between the plates with a dielectric of constant K gives \( C' = KC \).
(B) New energy stored by the capacitor is K times U. — Correct. With V fixed, the energy \( \frac{1}{2}CV^{2} \) grows in the same ratio as C.
\( \displaystyle U' = \frac{1}{2}C'V^{2} = K\left(\frac{1}{2}CV^{2}\right) = KU \)
(C) New potential difference across the capacitor is K times V. — Incorrect. The source holds the potential difference at V. It would change (to V/K) only if the capacitor had been disconnected first.
(D) New charge on the capacitor is \( 1/K \) times the original charge Q. — Incorrect. The source supplies extra charge, so the charge rises to K times Q.
\( Q' = C'V = KCV = KQ \)
NCERT (2026-27):
Exercises – …
