# Homework Help: Capacitance Question - Total Energy, Potential Difference

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1. Mar 6, 2016

### ratinatinycar

1. The problem statement, all variables and given/known data
Two identical parallel-plate capacitors, each with capacitance 10.0 μF, are charged to potential difference 50.0 V and then disconnected from the battery. They are then connected to each other in parallel with plates of like sign connected. Finally, the plate separation in one of the capacitors is doubled. (a) Find the total energy of the system of two capacitors before the plate separation is doubled. (b) Find the potential difference across each capacitor after the plate separation is doubled. (c) Find the total energy of the system after the plate separation is doubled.

2. Relevant equations
U = (1/2)CV2 and Q = CV

3. The attempt at a solution
For part (a), I know that to find U I must add (1/2)CV2 for both parallel capacitors, which would give me

U = (1/2)(10)(50)2 + (1/2)(10)(50)2
U = (10)(50)2
Add then simplify algebraically to find total energy. I know to use these values because we are considering the plates before they have separated and therefore I need to use their original capacitances and voltages.

For part (b) I am unsure if I need to use the equation V = Q/C for both capacitors now that the distance of one has been doubled. I am not sure what thought process I should have. Explaining this part in explicit detail would help me immensely.

For part (c) I assume I would go about it the same way as part (a) except using the values I have calculated in part (b).

Here is some background irrelevant to the current problem:

Physics does not come easily to me. I find the college pace to be extremely challenging to keep up with and am currently struggling immensely in my electromagnetism and thermodynamics class. I feel that I only absorb maybe about 20% of the material if I'm lucky. I am trying very hard to maintain a good attitude towards physics but I usually become overwhelmed with frustration and completely give up. I am an extremely strong math student but have found physics to be a completely different way of thinking and it has left me completely heartbroken and baffled. Please have patience with me as I struggle through this problem and understand that I am really trying my best but will need an extremely thorough explanation to really understand the sort of thought process I need to solve this problem. And if you read this far, thank you very much!

2. Mar 6, 2016

### Staff: Mentor

Hi ratinatinycar.

After the plates of one have been spaced more widely, what is the new total capacitance of the pair?

3. Mar 6, 2016

### ratinatinycar

Hello, and thank you! I understand that doubling the distance of the plates halves the capacitance, so the capacitance goes from 10.0 μF to 5.00 μF in one of the capacitors.

4. Mar 6, 2016

### Staff: Mentor

So you can calculate the total new capacitance.

Can you now finish the problem?

5. Mar 6, 2016

### ratinatinycar

Unfortunately, no. I am extremely insecure of what my thought process should be. I have done this problem before so some information regarding it I still remember, but I am afraid of knowing how to do this problem from memorization. I would really like to understand not only what I need to do, but why I need to do it. I hate to ask for a step-by-step explanation but I feel like I would benefit most from one.

6. Mar 6, 2016

### Staff: Mentor

With the new geometry, which out of Q, C, and V are you confident you know?

7. Mar 6, 2016

### ratinatinycar

I am not sure I understand the question, but I have since figured out the solution. The missing piece of information that I did not know was that charge must be conserved before and after the distance is doubled because of the closed circuit.
Q0 = QF
Q0 = C1V1 + C2V2
(C2V2)VB = (2.0 x 10-6F)(50V)=1.0 x 10-3C = QF
QF/CeqF = 1.0 x 10-3C / 15.0 x 10-6F = 66.7 V for part (b).

Thank you for all your help!

8. Mar 6, 2016

### Staff: Mentor

You knew Q because charge has nowhere to escape.

There's a typo in your working.