A parallel plate capacitor whose capacitance C is 13.0 pF is charged by a battery to a potential difference V = 13.0 V between its plates. What is the potential energy of the capacitor ?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.098 nJ
The energy stored in a charged capacitor is the work done in charging it.
\( \displaystyle U = \frac{1}{2}\,C V^{2} \)
\( C = 13.0\,\mathrm{pF} = 13.0\times10^{-12}\,\mathrm{F} \)
\( V = 13.0\,\mathrm{V} \)
\( \displaystyle U = \frac{1}{2} \times 13\times10^{-12} \times (13)^{2} = \frac{1}{2} \times 13\times10^{-12} \times 169 \)
\( \displaystyle U = 1098.5\times10^{-12}\,\mathrm{J} \approx 1.098\times10^{-9}\,\mathrm{J} = 1.098\,\mathrm{nJ} \)
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 – …
