1. Two identical 46.10 µF capacitors are connected with superconducting wire thr

Question

1. Two identical 46.10 µF capacitors are connected with superconducting wire through a switch, S, which is initially open. A charge of 51.64 µC is initially stored on the capacitor on the left; the capacitor on the right is initially uncharged. The two capacitors, conducting wires, and switch are all contained inside a rigid box composed of adiabatic walls. How much electrical energy is initially stored inside the box?

2. The switch now is closed and charge begins to flow from the capacitor on the left to the capacitor on the right. Once equilibrium is established again, how much electrical energy is stored inside the box?

Explanation / Answer

1)let the two identical capacitors be C1 andC2.The capacitors C1 and C2 areconnected in series therefore the equivalent capacitance is (1/C) = (1/C1) + (1/C2) C1 = C2 = 46.1 µF = 46.10 *10-6 F or (1/C) = (1/46.10) + (1/46.10) = (1 + 1/46.10) =(2/46.10) or C = (46.10/2) = 23.05 µF when the two capacitors are connected in series the chargeflowing through the two capacitors is same therefore the charge onthe two capacitors is q = 51.64 µC the electrical energy which is initially stored insidethe box is E = (1/2)CV2 = (1/2)C * (q/C)2 = (1/2) *(q2/C) 2)When equilibrium is established again the electrical energywhich is stored inside the box is same.This is because the chargeon the two capacitors is same when equilibrium is reached as thetwo capacitors are in series.

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