Energy Stored In A Capacitor

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After disconnecting a charged isolated capacitor from its battery, the plate separation is doubled by hand, so what happens to the stored energy?

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For an isolated capacitor the charge $Q$ is fixed, so it is convenient to write the stored energy as $U = \frac{Q^2}{2C}$, a form presented in NCERT Class 12, Chapter 2 (Electrostatic Potential and Capacitance). Since $C = \frac{\varepsilon_0 A}{d}$, doubling the separation $d$ halves the capacitance $C$. With $Q$ constant and $C$ halved, the energy $U = \frac{Q^2}{2C}$ doubles. This extra energy comes from the mechanical work done by the hand pulling the oppositely charged, mutually attracting plates apart. The option that energy halves is wrong because it mistakenly assumes constant voltage rather than constant charge. The option that energy stays constant is wrong because it ignores the external work supplied against attraction. The option of one quarter is wrong because it treats the change as if voltage were fixed and squares the factor. Contrast this with a capacitor kept connected to a battery, where the voltage is fixed instead of the charge, and separating the plates would actually decrease the stored energy. A physical and sign check confirms the doubling here: pulling the mutually attracting plates apart requires positive external work, which must be stored as increased field energy, so the energy rising to twice its value is entirely consistent.

Energy Stored In A Capacitor

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