A parallel-plate capacitor of area, A, and separation, d, is connected to a powe

Question

A parallel-plate capacitor of area, A, and separation, d, is connected to a power supply. You turn on the power supply and the capacitor is fully charged to the voltage, V. Still connected to the operating power supply, the two plates are pressed together so that their separation is reduced to the new separation, d'. How is the new voltage across the plates, V', related to the original voltage, V? How is the new capacitance, C', related to the original capacitance, C? How is the new charged on the plates, Q', related to the original charge, Q? A parallel-plate capacitor of area, A, and separation, d, is connected to a power supply. You turn on the power supply and the capacitor is fully charged to the voltage, V. You disconnect the power supply from the capacitor. The two plates are then pressed together so that their separation is reduced to the new separation, d'. How is the new charged on the plates, Q', related to the original charge, Q? How is the new capacitance, C', related to the original capacitance, C? How is the new voltage across the plates. V', related to the original voltage. V?

Explanation / Answer

capacitance of a parallel plate capacitor=epsilon*area/distance between plates

epsilon=electric permitivity of the medium

if C is capacitance and V is volatage across the plates,

then magnitude of charge on each plate=Q=C*V

part a.

as original power supply is still connected, new voltage V'=original voltage V

hence V'=V

part b.

as capacitance is inversely proportional to distance between plates (area and the medium being unchanged)

, here distance is decreased , hence capaictance will increase.

hence C' > C

part c.

as voltage remains constant and capacitance increases, net charge will also increase as charge=capacitance*voltage

hence Q' > Q

part d.

here as the original source is disconnected, no new charge can be supplied to the plates.

hence net charge on plate will remain constant.

hence Q'=Q

part e.

as capacitance is inversely proportional to distance between plates (area and the medium being unchanged)

, here distance is decreased , hence capaictance will increase.

hence C' > C

part f.

voltage=charge/capacitance

as charge remains constant and capacitance increases, voltage will decrease.

hence V' <V

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