A 1.03 µF capacitor is connected in series with a 1.98 µF capacitor. The 1.03 µF capacitor carries a charge of + 11.1 µC on one plate, which is at a potential of 49.5 V. (a) Find the potential on the negative plate of the 1.03 µF capacitor. (b) Find the equivalent capacitance of the two capacitors.

Question

Karly Glenn

Physics

29 January, 20:34

A 1.03 µF capacitor is connected in series with a 1.98 µF capacitor. The 1.03 µF capacitor carries a charge of + 11.1 µC on one plate, which is at a potential of 49.5 V. (a) Find the potential on the negative plate of the 1.03 µF capacitor. (b) Find the equivalent capacitance of the two capacitors.

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Answers (1)

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Steven Jefferson · Answer 1

Answer: a) 38.73 V; b) 0.67 μ F

Explanation: In order to solve this problem we have to take into account the expression used for the capacitor and its conection in a circuit.

We know that C=Q/V then Q = 11,1 μC and C1=1.03 µF

V1=Q/C1=11.1 μC/1.03 * µF = 10.77 V so as the positive plate is at 49.5 V then negative plate is 38.73 V.

The capaciro equivalent for a seri connection is given by;

1/Cequi=1/c1+1/c2 = (1 / 1.03 µF) + (1 / 1,98 µF) = 0.67 µF