A planar capacitor of capacitance 1 mu F is wired to a battery providing a poten

Question

A planar capacitor of capacitance 1 mu F is wired to a battery providing a potential difference of 1000 V across the capacitor plates. Once charged the capacitor is disconnected from the battery. What is the potential difference across the capacitor plates after it has been disconnected? If the distance between the capacitor plates is increased by a factor 10. Give the final stored energy? Discuss what would happen is the capacitor remains connected when the distance between the plates is changed.

Explanation / Answer

Here

C = 1 uF

V = 1000 V

1) potential difference across the capacitor = 1000 V

2) when the capacitor plates are seperated by 10 times distance initially

capacitance will be 1/10 times

and the charge will remain constant

final energy stored = 0.5 * Q^2/C

final energy stored = 0.5 * (1000 * 1 *10^-6)^2/(1 *10^-6/10)

final energy stored = 5 J

the final energy stored is 5 J

3)if it remain connected , the potential across the capacitor will be same.

and the capacitance of the capacitor will be 1/10 times

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