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[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.

 

[rwisz]

To solve this problem, we can use the conservation of charge and energy principles for capacitors.

First, we know that the total charge in the system must remain constant before and after the capacitors are connected. This means that the total charge on C1 and C2 before connection must equal the total charge on C1 and C2 after connection. Mathematically, this can be represented as:

Q1 + Q2 = Q1' + Q2'

Where Q1, Q2 are the initial charges on C1 and C2 respectively, and Q1', Q2' are the final charges on C1 and C2 respectively.

Next, we can use the formula Q=CV to relate the charge to the capacitance and potential difference. This means that we can rewrite the above equation as:

C1V1 + C2V2 = C1'V1' + C2'V2'

Where C1, C2 are the initial capacitances of C1 and C2 respectively, and C1', C2' are the final capacitances of C1 and C2 respectively.

Now, we can use the given information to solve for C1'. We know that the initial potential difference of C1 is 20 V, and the final potential difference is 2 V. This means that V1 = 20 V and V1' = 2 V. We also know that C2 = 33µF, V2 = 4 V, and V2' = 2 V. Plugging these values into the above equation, we get:

C1(20) + 33(4) = C1'(2) + 33(2)

Simplifying, we get:

20C1 + 132 = 2C1' + 66

Subtracting 2C1 from both sides, we get:

18C1 + 132 = 66

Subtracting 132 from both sides, we get:

18C1 = -66

Dividing both sides by 18, we get:

C1 = -3.67µF

This is not the answer we were looking for, so we need to check our work. We can see that the negative sign indicates that our initial assumption of the charge on C1 being positive was incorrect. This means that the charge on C1 must have been negative before connection. This also means that the potential difference across C1