Please Help Me with Physics?

Question

I would appreciate any helpful input on this physics problem. I am really confused on this one. Thanks for any help :)

Capacitors C1 = 4.0 µF and C2 = 2.0 µF are charged as a series combination across a 95 V battery. The two capacitors are disconnected from the battery and from each other. They are then connected positive plate to positive plate and negative plate to negative plate. Calculate the resulting charge on each of the capacitors (in µC).

I've gotten 84.44 for C1 and 42.22 or 168.89 for C2 and both are the wrong answer :(

Answer 1

Remember that in all parts of a series circuit the current is the same, and that the sum of the voltage drops in a series circuit equals the applied voltage. If C1 and C2 are series connected, they both have the same amount of charge and the voltage across them has to add up to 95V. The voltage across a capacitor is
v=Q/C where Q is cuolombs and C is Farads. So
Q/C1 + Q/C2 = 95
(Q*C2/C1*C2) + (Q*C1/C1C2) = 95
Q*C2 + Q*C1 = 95*C1*C2
Q = (95*C1*C2)/(C1 + C2) = 126.66µC
This is the charge on *each* capacitor. When they are disconnected from the circuit and connected in parallel, there is a total charge of twice that (253.33µC) to be shared between them.
In a parallel circuit, the voltage across each capacitor is the same, so Q1/C1 = Q2/C2 and
C2/C1 = Q2/Q1. That is, the ratio of the charge(s) on the capacitor(s) is in direct proportion to the magnitudes of the capacitances. In this case, C1 will have twice the charge of C2, and the sum of the charges must be 253.33µC, so the charge on C2 is 253.33/3 = 84.4µC and the charge on C1 is
168.8µC.
You seem to have had the charge(s) attached to the wrong capacitors ☺

HTH

Doug

Answer 2

OK, so you have the capacitors in series across a 95 V battery. Say that Capacitor C1 will have a voltage drop across it of V1 and Capacitor C2 will have a voltage drop across it of V2.

V1 + V2 = 95 V.

Now we know that the charge on C1 will equal the charge on C2. So that means Q1 = Q2. We also know that Q = C*V.

So if Q1 = Q2 then we can rewrite this as C1*V1 = C2*V2

Let's solve for V2 and substitute that expression into the voltage equation:

if C1*V1 = C2*V2 then

V2 = C1*V1/C2

and V1 + (C1*V1/C2) = 95 V.

V1* (1 + (C1/C2)) = 95 V.

V1 = 95/(1 + (C1/C2)) volts

We know C1 = 4 microF and C2 = 2 microF so

V1 = 95/(3) volts and since C = Q * V, then Q = C/V

Q1 = 4 * 10^-6 Farads / (95/3) volts = 1.263 * 10^-7 C.
Q2 is the same.

Now connect them in parallel to each other. The voltage drop across both capacitors must now be equal and the charge will redistribute to achieve that.

In parallel V1 = V2 so Q1/C1 = Q2/C2.

We also know that the total charge across the capacitors must remain the same because there is no path to leak charge off to any location. So Q = Q1 + Q2 = 2 * 1.26 * 10^-7 C = 2.526 * 10^-7

From the voltage equation we can solve for one of the Qs in terms of the other and substitute it into the charge equation.

Q1 = Q2 *(C1/C2)

So the charge equation becomes...

Q2* (C1/C2) + Q2 = 2.526 * 10^-7

Q2 * ((C1/C2) + 1) = 2.526 * 10^-7

Q2 * (3) = 2.526 * 10^-7

Q2 = 8.421 * 10^-8 C

You can then solve for Q1.