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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.

Respuesta :

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

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