The gap of a parallel plate capacitor can be adjusted. In position A, the capaci

Question

The gap of a parallel plate capacitor can be adjusted. In position A, the capacitance is 3.0 nF, and in position B, it is 2.7 nF. The capacitor is charged by a 12 V battery while in position A. The battery is disconnected, and the capacitor is then moved to position B. (a) How much charge is on the original capacitor (in position A)? (b) How much charge is on the capacitor while it’s in position B? Hint: think about where the charge could have gone? (c) What is the voltage across it in position B?

Explanation / Answer

A) charge q = C*V

in position A

charge qA = C*V = 3*10^-9*12 = 36*10^-9 C

B) when the battery is disconnected then charge on plates remains same

charge in position B is qB = qA = 36*10^-9 C

C) voltage V = qB/C = (36*10^-9)/(2.7*10^-9) = 13.33 V

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