A parallel plate capacitor is charged to a certain potential and the charging battery is then disconnected. Now, if the plates of the capacitor are moved apart then:
Medium
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When a capacitor is charged and then disconnect battery, first the capacitance will remain constant and then slowly discharge according to time constant.
The potential between the plates is $V=\frac{q d}{A \epsilon_{0}},$ so when increasing the between plate, the voltage will increase.
The energy stored in capacitor is $W=\frac{1}{2} C V^{2} .$ Thus, the electrostatic energy stored in the
capacitor also increases.
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