A 800 pF capacitor is charged by a 100 V battery. After some time, the battery is disconnected. The capacitor is then connected across another 800 pF capacitor. The electrostatic energy stored in the combination will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2 × 10−6 J
Once the battery is disconnected, the charge is conserved. It spreads over the two capacitors until they are at a common potential.
\( \displaystyle Q = CV = 800\times10^{-12}\,\mathrm{F} \times 100\,\mathrm{V} = 8\times10^{-8}\,\mathrm{C} \)
\( \displaystyle V_{\mathrm{common}} = \frac{Q}{C + C} = \frac{8\times10^{-8}}{1600\times10^{-12}}\,\mathrm{V} = 50\,\mathrm{V} \)
\( \displaystyle U = \frac{1}{2}\,(C + C)\,V_{\mathrm{common}}^{2} \)
\( \displaystyle U = \frac{1}{2} \times 1600\times10^{-12} \times (50)^{2}\,\mathrm{J} \)
\( \displaystyle U = 2\times10^{-6}\,\mathrm{J} \)
The initial energy was \( \frac{1}{2}CV^{2} = 4\times10^{-6}\,\mathrm{J} \); half of it is lost as heat and radiation while charge flows between the capacitors.