Question

A capacitor C1 =5μF is connected across a 20 V battery. The battery is removed and the capacitor is reconnected in parallel to another capacitor C2 =6μF, which is uncharged initially. What are the final charges and potential differences for the capacitors?

Answer 1

For the 4.2 µF capacitor, the charge on it is 151.2 µC. After the capacitors are reconnected in parallel, the voltage across each capacitor is approximately 34.29 V for C1 and C2, and approximately 68.57 V for C3.

Step-by-step explanation:

When the 8.4 µF, 8.4 µF, and 4.2 µF capacitors are connected in series across a 36.0 V potential difference, the charge on each capacitor can be determined using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. Therefore, the charge on the 4.2 µF capacitor is Q3 = (4.2 µF)(36.0 V) = 151.2 µC (microcoulombs).

When the capacitors are disconnected from the potential difference and reconnected in parallel, the voltage across each capacitor in the parallel combination will be the same. Since the positive plates are connected together, the total charge on the capacitors is considered and divided equally among them. Therefore, the voltage across each capacitor in the parallel combination is the same and can be determined using the formula V = Q/C, where V is the voltage, Q is the charge, and C is the capacitance. The voltage across each of the capacitors is V1 = Qtotal/C1 = (151.2 µC + 151.2 µC)/(8.4 µF) ≈ 34.29 V, V2 = Qtotal/C2 = (151.2 µC + 151.2 µC)/(8.4 µF) ≈ 34.29 V, and V3 = Qtotal/C3 = (151.2 µC + 151.2 µC)/(4.2 µF) ≈ 68.57 V.