Two capacitors, C1=5700pF and C2=3000pF, are connected in series to a 9.00 V battery. The capacitors are later disconnected from the battery and connected directly to each other, positive plate to positive plate, and negative plate to negative plate. What then will be the charge on each capacitor? Q1?, Q2?

Question

Physics

Mitzy

2 April, 01:09

Two capacitors, C1=5700pF and C2=3000pF, are connected in series to a 9.00 V battery. The capacitors are later disconnected from the battery and connected directly to each other, positive plate to positive plate, and negative plate to negative plate. What then will be the charge on each capacitor? Q1?, Q2?

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Answers (1)

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Francisco · Answer 1

Q1 = 23199 pC

Q2 = 12210 pC

Explanation:

C1 = 5700 pF = 5700 x 10^-12 F

C2 = 3000 pF = 3000 x 10^-12 F

Connected in series

V = 9V

Let the equivalent capacitance is C.

C = C1 x C2 / (C1 + C2) = 5700 x 3000 / (5700 + 3000) = 1965.5 pF

Let the charge is Q.

Q = C x V = 1965.5 x 10^-12 x 9 = 17689.65 x 10^-12 C

The total charge on both the capacitors = 2 x 17689.65 x 10^-12 C

= 35379.3 x 10^-12 C

Equivalent parallel capacitance, Cp = C1 + C2 = 5700 + 3000 = 8700 pF

Voltage = V = total charge / Cp = 35379.3 / 8700 = 4.07 V

Charge on 5700 pF, Q1 = C1 x V = 5700 x 4.07 = 23199 pC

charge on 3000 pF, Q2 = C2 x V = 3000 x 4.07 = 12210 pC