Question:
A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is conneced to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) _____
A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is conneced to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) _____
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Energy before: \(\tfrac12CV^2\). After sharing charge with an equal uncharged capacitor the common voltage is V/2; final energy is half the initial energy.
Updated On: Oct 31, 2026
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Correct Answer: 6
Solution and Explanation
Step 1: Initial energy \(U_i=\tfrac12CV^2=\tfrac12(60\times10^{-12})(20)^2=12\times10^{-9}\) J \(=12\) nJ.
Step 2: Charge \(Q=CV=1200\) pC is shared between two equal capacitors in parallel; common voltage \(V'=\dfrac{Q}{2C}=10\) V.
Step 3: Final energy \(U_f=\tfrac12(120\times10^{-12})(10)^2=6\) nJ.
Step 4: Energy lost \(=12-6=6\) nJ.
Answer: 6
Step 2: Charge \(Q=CV=1200\) pC is shared between two equal capacitors in parallel; common voltage \(V'=\dfrac{Q}{2C}=10\) V.
Step 3: Final energy \(U_f=\tfrac12(120\times10^{-12})(10)^2=6\) nJ.
Step 4: Energy lost \(=12-6=6\) nJ.
Answer: 6
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