The plates of a variable parallel-plate capacitor have area of 700 cm2 and the d

Question

The plates of a variable parallel-plate capacitor have area of 700 cm2 and the distance between them is 4.00 mm , and they are connected across the terminals of a battery . The capacitor is then disconnected from the battery after a certain amount of time. The two plates are then moved 0.60 cm farther apart, the charge on the plates remains constant but the potential difference changes to 100 V. In the new configuration, do you expect the energy stored in the capacitor: to increase to decrease to stay the same

Explanation / Answer

Here ,

when the plates are moved farther apart ,

as the capacitance is given as

C = Area * epsilon/d

as d is increased , C will decrease

as energy stored = 0.5 Q^2/C

as the charge is constant , decreasing C will increase energy stored

the energy stored in capacitor a) to increase

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