A 400 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 400 pF capacitor. The electrostatic energy lost in the process is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 4 \times 10^{-6}\,\mathrm{J} \)
When the charged capacitor is connected to the uncharged one, the charge is conserved and shared until both reach a common potential; the energy is not conserved (it is lost as heat and radiation in the connecting wires).
\( C = 400\,\mathrm{pF} = 400\times10^{-12}\,\mathrm{F} \)
\( V = 200\,\mathrm{V} \)
\( \displaystyle U_i = \frac{1}{2}CV^{2} = \frac{1}{2} \times 400\times10^{-12} \times (200)^{2} = 8\times10^{-6}\,\mathrm{J} \)
\( \displaystyle V_{\mathrm{common}} = \frac{CV}{C + C} = \frac{200}{2} = 100\,\mathrm{V} \)
\( \displaystyle U_f = \frac{1}{2}(2C)V_{\mathrm{common}}^{2} = \frac{1}{2} \times 800\times10^{-12} \times (100)^{2} = 4\times10^{-6}\,\mathrm{J} \)
\( \displaystyle \Delta U = U_i - U_f = 4\times10^{-6}\,\mathrm{J} \)
Shortcut: When a charged capacitor shares its charge with an identical uncharged capacitor, exactly half of the stored energy is lost: \( \Delta U = \frac{1}{2} \times 8\times10^{-6} = 4\times10^{-6}\,\mathrm{J} \).
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
