2 identical capacitors, given potential in joules

[jendead]

Homework Statement

Capacitors A and B are identical. Capacitor A is charged so it stores 4J of energy and capacitor B is uncharged. The capacitors are then connected in parallel. The total stored energy in the capacitors is now ____.

Homework Equations

U = q^2/2C

The Attempt at a Solution

Ok, I'm not sure if this is allowed.. I already know the answer is 2J. My professor gave us an answer key with the problem done out, but I don't understand what is happening. I'm not sure where 2J went. It says 2J of energy were required to move charge from the charged capacitor to the uncharged one.

I know initial potential of A is 4J, and initial potential of B is 0. For final potential, she has U_final = q^2/2C = 1/2*q^2/2C. Where did the second 1/2 come from?

I would really appreciate some clarification.. I'm really confused.

 

[Dick]

When the two capacitors are connected in parallel the capacitance doubles, while Q remains the same.

 

[jendead]

I understand that, but I don't see how that relates here. From the look of the answer key, I don't understand where total potential is halved when being spread out over 2 capacitors.

 

[Dick]

The potential is halved because the same charge is being spread out over two capacitors. What the relation between potential and charge for a capacitor?

 

[jendead]

It's U = q^2/2C.. so if I plug in (q/2)^2/2(2C), which is halving the charge and doubling the capacitance, I'd have q^2/16C.. which still makes no sense.

Am I missing something conceptually or algebraically?

 

[Dick]

The U in that equation is energy stored the capacitor, not potential between the plates (volts). Is that your problem?

 

[jendead]

Yes, I'm trying to find the potential energy (in joules), not the voltage. I may have been unclear when I said potential.

 

[Dick]

If you understand why capacitance doubles, then why can't you just change the C in Q^2/(2C) to 2C and conclude 4J changes to 2J?

 

[jendead]

Ok, that's starting to make sense.. but why isn't Q also changed to Q/2 since it's being halved at the same time?

I apologize for being so dense - I tend to have a really hard time with these concepts sometimes.

 

[Dick]

Q is total Q, between both capacitors. Total charge doesn't change. C changes to C/2. It does change.

 

[jendead]

Ok, that makes sense. Thank you for being patient. :)