An air-filled parallel-plate capacitor is attached to a battery with a voltage V

Question

An air-filled parallel-plate capacitor is attached to a battery with a voltage V. While attached to the battery, the area of the plates is increased 5 times and the separation of the plates is reduced 5 times.

1. By what factor Velectricf/Velectrici, will the potential difference across the plates change if, once the capacitor was charged by the battery, it was disconnected from the battery while the area and separation were changed as above?

2. By what factor Uelectricf/Uelectrici, will the potential energy stored in the capacitor change if, once the capacitor was charged by the battery, it was disconnected from the battery while the area and separation were changed as above?

Explanation / Answer

capacitance C = Q/V

V = Q/C

C = e0*A/d

with the battery attached

potential remains same

capacitance increases by 25 times

vi = v

with battery disconnected

charge remains same

potential decreases by = reduced by 25 times

Vf = V/25

Vf/Vi = 1/25

-------------------

Ui = (1/2)*Ci*Vi^2 = (1/2)*25*C*V^2

Uf = (1/2)*25*C*(V/25)^2 = Ui/25

Uf/Ui = 1/25

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