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A charge of 20μC is placed on the positive plate of an isolated parallel-plate capacitor of capacitance 10μF.

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A charge of 20μC is placed on the positive plate of an isolated parallel-plate capacitor of capacitance 10μF. Calculate the potential difference developed between the plates.

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1 Answer

+1 vote
by (60.9k points)

∴ Given that

Capacitance = 10μF

Charge = 20μc

∴ The effective charge = (20 - 0)/2 = 10μF

∴ C = q/V => V = q/C =10/10 = 1V

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