\( 10^{4}\,\mathrm{pF} \) capacitor is charged by a \( 12\,\mathrm{V} \) battery. The capacitor is disconnected from the battery and connected to another uncharged \( 10^{4}\,\mathrm{pF} \) capacitor. The electrostatic energy stored by this two capacitor system is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 3.6\times10^{-7}\,\mathrm{J} \)
When a charged capacitor is joined to an uncharged one, charge is conserved and is shared until both have the same potential; energy is not conserved, as part of it is lost as heat in the connecting wires.
\( C = 10^{4}\,\mathrm{pF} = 10^{4}\times10^{-12}\,\mathrm{F} = 10^{-8}\,\mathrm{F} \)
\( V = 12\,\mathrm{V} \)
\( Q = CV = 10^{-8}\times12 = 1.2\times10^{-7}\,\mathrm{C} \)
After connection the two capacitors are in parallel: capacitance \( 2C \), charge still \( Q \).
\( \displaystyle V' = \frac{Q}{2C} = \frac{12}{2} = 6\,\mathrm{V} \)
\( \displaystyle U' = \frac{1}{2}(2C)V'^{2} = \frac{1}{2}(2\times10^{-8})(6)^{2} = 3.6\times10^{-7}\,\mathrm{J} \)
Shortcut: For two identical capacitors the stored energy is halved: \( U = \frac{1}{2}CV^{2} = \frac{1}{2}(10^{-8})(144) = 7.2\times10^{-7}\,\mathrm{J} \), so \( U' = U/2 = 3.6\times10^{-7}\,\mathrm{J} \).