# Capacitance (should be simple)

1. Feb 17, 2008

### mer584

The problem.
A 2.50uF capacitor is charged to 857V and a 6.80uF capacitor is charged to 652V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? (note that charge is conserved)

2. Relevant equations
PE= V/Q ; V=Ed; Q=CV; C= Eo (A/D); Vb-Va

Attempt:
What I started problem I used the forumla Q=CV to find the charge that is occuring in each case finding Q1 = .0021C and Q2= .0044 C. I really wasnt sure if this was even helpful or where to go from there. It appears to be a uniform feild so I know I can use V=Ed but we don't have a distance or an area in this problem.

In order to get the potential difference I know you need to work with Vb-Va and possibly the potential energy. Should I use PE= V/Q then subtract the potential energies to find the work and then the poential difference?

2. Feb 17, 2008

### Staff: Mentor

Since the two capacitors are now connected in parallel, what must be true about the voltage across them?

3. Feb 17, 2008

### belliott4488

What happens when the two caps are connected to each other? You know that some charge will flow (else the answer would be "no change") - so what's the condition for the charge to stop flowing?

Answering that should give you an equation relating the quantities you're asked solve for.

4. Feb 17, 2008

### mer584

the voltage will be in half, split between the two, or i would have thought it was the average between the two which is 754V but the answer points towards 712V

Last edited: Feb 17, 2008
5. Feb 17, 2008

### Staff: Mentor

No. How will the voltage across one capacitor compare to the voltage across the other?

6. Feb 17, 2008

### mer584

i guess I dont really know
I would have said its based on the charge and the amount of capacitance but that will just bring me back to the original voltages given in the problem...I'm clearly missing a fundamental concept here.

I do get that once the battery is turned off they will maintain the same level of charge. true?

Last edited: Feb 17, 2008
7. Feb 17, 2008

### belliott4488

Nope - you're missing a key point. You know charge will flow ... which means there's a potential difference pushing it. The charge will stop flowing when there's no more potential difference. So, what's the potential between the two connected positive plates? And for the two connected negative plates?

8. Feb 17, 2008

### Staff: Mentor

Here's a hint: Think of it as a circuit with two parallel branches. Can one branch have a greater voltage than the other if they are in parallel?

Once this clicks you'll use the relationship between the branch voltages to set up an equation for finding the voltage. (Making use of the total charge that you've found, also.)

9. Feb 17, 2008

### Biest

You might want to think about what is the total charge and total capacitance...

10. Feb 17, 2008

### mer584

OK so let me back up, because i'm fundamentally confused.
I know that the electrical potential is equal to V=Ed. which is also equal to V=PE/q. We know that E= kQ/r^2 but that we don't have a distance here so we cant use that. we also dont know d.

I'm assuming that there will be no more potential difference when the two plates are equal to each other because then Vb-Va =0 This doesnt seem right.

Can I set it up as though there is a test charge in between the two doing work..i cant really do this without a distance either i dont think

11. Feb 17, 2008

### mer584

won't it always be the same?? as in no one branch can not be greater? or is that just that the total voltage is conserved but that it can be different

Last edited: Feb 17, 2008
12. Feb 17, 2008

### Staff: Mentor

Exactly! The voltage across each capacitor will be equal. Since C = Q/V, you can set up an equation for V = V1 = V2. (You know what the total charge is, Q1 + Q2, so you can solve for V and the two charges.)

13. Feb 17, 2008

### mer584

Ok, so I completely understand how to get the different charges once I have the common voltage, and I understand why there is a common voltage. (doing better than before). But nothing I seem to set up is giving me the answer. If I want to solve for V and I know Q is .0065 the only thing i can set up is CV=CV which doesn't work.

Am I supposed to add the two capacities?

14. Feb 17, 2008

### Staff: Mentor

Since the voltage must be the same:
V1 = V2
Q1/C1 = Q2/C2

You know C1 and C2. Combine this with what you know about the total charge (since you know the total charge is conserved): Q1 + Q2 = Q(total).

15. Feb 17, 2008

### mer584

I understand Q total, I've solved for that using C1 and C2 and the two voltages given in the problem, but when solving for V total I cant just use that above equation...I know all 4 of the variables and they don't equal each other or leave me a V total to solve for. Am I misunderstanding you?

16. Feb 17, 2008

### Staff: Mentor

You have two equations that you can solve together to get Q1 and Q2 (those are the new charges on the capacitors, once they've been attached):

(1) Q1/C1 = Q2/C2
(2) Q1 + Q2 = Q(total)

Once you find Q1 and Q2, you can find V from V = Q/C.

17. Feb 17, 2008

### Biest

You can find the charges for any of the two capacitors respectively. You can use that result to find the V around the capacitor or not?

18. Feb 17, 2008

### mer584

If you are saying that Q1 and Q2 are after they are rerouted and not before then we cant use the Q total from before....that leaves us with 3 unknowns and now I'm totally confused. Can I use the original Q values at all in this case??? or are they just completely irrelevant. Sorry, something just isnt clicking here.

19. Feb 17, 2008

### belliott4488

Why not?? Remember that total charge is conserved. Where else would any charge go?

20. Feb 17, 2008

### mer584

What I know: the initial Q1 value: .0021
the initial Q2 value: .0044
the total Q value = .0066
then using the 2 equations above i can solve for Q1b = .00177
Q2b= .00481
which gives me V values of 707 and not 712 I carried it out as many digits in my calculator as I could so I guess I'm still concerned i'm doing something wrong, or just being too picky