A 2.1-F capacitor is fully charged by a 6.0-V battery. The battery is then disconnected. The capacitor is not ideal and the charge slowly leaks out from the plates. The next day, the capacitor has lost half its stored energy. Calculate the amount of charge lost.

6.3 C

Explanation:

From the question,

Energy lost by the capacitor(half it stored energy) = 1/4CV² Equation 1

Where C = Capacitance of the capacitor, V = Volatge across the parallel plate of the capacitor.

E = 1/4CV²

Given: C = 2.1 F, V = 6.0 V

Substitute these values into equation 1

E = 1/4(2.1)(6²)

E = 18.9 J

But,

E = 1/2QV Equation 2

Where Q = amount of charge lost

Make Q the subject of the equation

Q = 2E/V Equation 3

Give: E = 18.9 J, V = 6.0 V

Q = 2(18.9)/6

Q = 6.3 C.

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