A) Two capacitors(C 1 = 7.5 ? F , C 2 =19.6 ? F are charged individually to(V 1

Question

A)   Two capacitors(C1 = 7.5 ?F, C2 =19.6 ?F are charged individually to(V1 = 14.1 V, V2 =3.4 V. The two capacitors are then connected togetherin parallel with the positive plates together and the negativeplates together. Calculate the final potential difference acrossthe plates of the capacitors once they are connected.

Ok i got this part Q1=7.0252*10^-5 and V=10V

B)   Calculate the amount of charge that flows from onecapacitor to the other when the capacitors are connected together.and after.

Ok I also found this part online but it does not make any sense to me. Could you break it down barney style so i can undertand the concept behind the answer. I know you subtract the charge from q= C2*V2 from the charge you get from  when both compacitors are connected to the same circuit (2.509*10^-4 C - 1.242*10^-4 C = 1.267*10^-4 C ) but this does not make sense to me.

Explanation / Answer

PART (B)

at first there is a charge of q2 on the capacitor #2:

q2 = C2 V2 = 19.6 * 3.4 = 66.64 uC

final charge on the capacitor #2:

q = C2 V = 19.6 * 10 = 196 uC

q - q2 = 196 - 66.64 = 129 uC

Therefore a charge of 129uC must flow. :)

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