How Do You Calculate Capacitor C1's Value After Connection?

[Qwerty459]

Homework Statement

Capacitor C1 is charged so that potential difference between its plates is 20 V. Another capacitor C2=33µF has potential difference of 4 V between its plates. After plates of the capacitors that carry the charge of opposite sign were connected the potential difference became 2 V. Find capacitance C1.

Homework Equations

Q=CV
U=(1/2)QV=(1/2)CV²

The Attempt at a Solution

I know that the charge on C2 is initially 122µC and that the potential energy on C2 is initially 2.64*10^-4 J.

The answer to the problem is supposed to be 11µF, but I don't know how I am supposed to figure this out having been given only the initial potential for C1.

 

[cupid.callin]

and that the potential on C2 is initially 2.64*10^-4 J.

You mean potential energy.
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Take charge on C1 as (C1)20 ... write charges on the 2 capacitors after they are connected, take charge on any1 capacitor as Q1 ... so on other will be net charge left - Q1.
keep in mind that charges will cancel out as opposite charge plates are connected.

Now potential of both capacitors is 2V
write eqn Q=CV for 2 capacitors and eliminate Q1
you'll get C1

 

[Qwerty459]

You mean potential energy.
_____________________________

Take charge on C1 as (C1)20 ... write charges on the 2 capacitors after they are connected, take charge on any1 capacitor as Q1 ... so on other will be net charge left - Q1.
keep in mind that charges will cancel out as opposite charge plates are connected.

Now potential of both capacitors is 2V
write eqn Q=CV for 2 capacitors and eliminate Q1
you'll get C1

Yes, Potential Energy.

 

[lswtech]

Treat the two connected plate as something parallel (as they should hold the same potential)

And by conservation of charges, compute the results. Note that we have "cancellation" of charges as we are connecting to the opposite side.