Question:
A capacitor with capacitance \(5\mu\mathrm{F}\) is charged to \(5\mu\mathrm{C}\). If the plates are pulled apart to reduce the capacitance to \(2\mu\mathrm{F}\), how much work is done ?
A capacitor with capacitance \(5\mu\mathrm{F}\) is charged to \(5\mu\mathrm{C}\). If the plates are pulled apart to reduce the capacitance to \(2\mu\mathrm{F}\), how much work is done ?
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The charge stays constant, so the work done equals the change in stored energy \(U=Q^2/2C\).
Updated On: Nov 5, 2026
- \(2.16\times10^{-6}\) J
- \(2.55\times10^{-6}\) J
- \(3.75\times10^{-6}\) J
- \(6.25\times10^{-6}\) J
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The Correct Option is C
Solution and Explanation
Charge \(Q=5\ \mu\mathrm{C}\) stays fixed (isolated capacitor).
\(U_i=\frac{Q^2}{2C_1}=\frac{25\times10^{-12}}{2\times5\times10^{-6}}=2.5\times10^{-6}\) J.
\(U_f=\frac{Q^2}{2C_2}=\frac{25\times10^{-12}}{2\times2\times10^{-6}}=6.25\times10^{-6}\) J.
\(W=U_f-U_i=3.75\times10^{-6}\) J.
Answer: (3) \(3.75\times10^{-6}\) J
\(U_i=\frac{Q^2}{2C_1}=\frac{25\times10^{-12}}{2\times5\times10^{-6}}=2.5\times10^{-6}\) J.
\(U_f=\frac{Q^2}{2C_2}=\frac{25\times10^{-12}}{2\times2\times10^{-6}}=6.25\times10^{-6}\) J.
\(W=U_f-U_i=3.75\times10^{-6}\) J.
Answer: (3) \(3.75\times10^{-6}\) J
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