Question

A capacitor (4.30 μF ) is connected in a parallel arrangement with a second capacitor (1.50 μF ) and in series with a 12-V battery

A)The battery is then removed, leaving the two capacitors isolated.

If the smaller capacitor's capacitance is now doubled, by how much does the charge on the larger capacitor change?

Express your answer using two significant figures.

B)By how much does the charge on the smaller capacitor change?

Express your answer using two significant figures.

C)By how much does the voltage across the larger capacitor change?

Express your answer using two significant figures.

D)By how much does the voltage across the smaller capacitor change?

Express your answer using two significant figures.

Answer 1

Q1 = 4.3 uf, Q2 = 1.5 uf V = 12 v

When the capacitors are fully charged the total charge

Q = CV = (4.3 +1.5 ) x 12 = 69.6 uC

Q1 = 4.3 x12 = 51.6 uC

Q2 = 1.5 x12 = 18 uC

When the battery is disconnected charge remains constant.

When Q2 is doubled the charge gets re-distributed

Q1/Q2 = C1/C2 = 4.3/3

3Q1 = 4.3 Q2

Q1 + Q2 = Q = 69.6

Q1 = 50 uC,   Q2 = 19.6 uC

Change in Q1 = 51.6 - 50.0 = 1.6 uC

Change in Q2 = 19.6 - 18.0 = 1.6 uC

Total new capacitance C = 4.3 + 3.0

Total Charge Q = 69.6

Voltage across each capacitor = Q/C = 69.6/7.3 = 9.53 V

Change in Voltage across each capacitor = 12-9.53 = 2.47 V