x Turn on thread page Beta
 You are Here: Home >< Physics

# Conceptual capacitor problem watch

1. OK guys, here's a bit of an apparent paradox for you...

Consider a capacitor, C, raised to voltage V, with charge Q, and then isolated.
The stored energy E = (1/2)*C*V^2.

Now connect this capacitor in parallel to another identical, initially uncharged
capacitor, with perfectly conducting wires. We have assembled a larger capacitor of capacitance 2C, but the charge has remained constant at Q due to the system's isolation. It follows that the voltage across the new network of capacitors has dropped to V/2.

The stored energy now is (1/2)*2C*(V/2)^2 = E/2 i.e. the energy of the system
has halved. My question is: where has this energy gone?
2. this was answered on a pervious thread, a spark between the plates dissipates the energy
3. No it doesn't. A spark occurs when the electric field so great it ionises the air and allows a current to flow. This has nothing to do with the problem. If you're having trouble with that, then assume the electric field is much lower than the breakdown potential of air.
4. hmmm,

yes i thought the other thread was wrong
5. (Original post by Worzo)
It follows that the voltage across the new network of capacitors has dropped to V/2
I suspect the problem lies here.

If E = 0.5 * C * V1^2
And E = 0.5 * C * V2^2 + 0.5 * C * V2^2

It follows that C * V2^2 = 0.5 * C * V1^2

Therefore V2 = (1 / root(2) ) * V1

The voltage after adding the second capacitor is actually 1 / root(2) of the original voltage.

Substituting this all back in and using the identity V2 = (1 / root(2) ) * V1 you should find it all works, hope this helps - and hope it's not wrong! (No magic sparking here anyway)
6. Ha! Nice work. Unfortunately, you've asserted at the start of that argument that all the energy remains in the electric field of the capacitors, so obviously that's going to answer the problem of where the energy's gone.

I'm afraid though, that that is not the flaw in my argument, and in fact your argument is wrong. It's a little more complicated. I'm surprised some of the heavyweights on here haven't given it a go yet. Teachercol? AlphaNumeric? Where are you?!

I'm only doing this because when my dad showed me the answer, I thought it was pretty cool, and made me realise that you've got to think a bit harder sometimes!
7. But clearly that's the case since this is an ideal model. There's nothing in the mathematics there that would state otherwise.

Since all of this is based on the energy stored in capacitors...there's not really much else to consider! I look forward to a more accurate solution however if you find one.

Edit:
(Original post by Worzo)
in fact your argument is wrong. It's a little more complicated
Given that you know the answer, I would love to hear a clue!
8. I love the question about typical question about resistors - I just think it's so amazing!

If V=(I^2 + 6I)/6 , then calculate the resistance when

a) I=2A
b) I=0A
9. Is it something to do with the Current's relation to the charge??

Lets say:

V1 = Q1/C = -I/C and V2 by similar logic = I/C

Using... Erm... One of Kirchoff's laws, V1-V2 + x = 0

And then we can break V1-V2 down into: I=[-C(V1-V2)]/2 ?? I really don't know where this is going!!
10. (Original post by JitsuCol)
Given that you know the answer, I would love to hear a clue!
Clue, eh? Well, this might be giving it away, but you can't ignore the fact that when the capacitors are connected together, a current flows...
11. Indeed it does initially but once equilibrium in the circuit is restored (as it would otherwise it would be perpetual..!) - Shall think about it.

Edit: This is really quite interesting! There must be some assumption I made in my calculations that is wrong but I don't believe assuming all the energy is held by the capacitors is wrong. There must be a current that flows initially but that will settle.

My solution? Monkeys!
(Please PM the solution to me and put me out of my misery!)

Edit(2): If this is what you were thinking about (http://wwwphy.princeton.edu/~kirkmcd...es/twocaps.pdf) then that is indeed interesting..! (I submit that I was indeed wrong)
12. the answer isn't some annoying crap about the accelerating electrons in the current radiating energy is it?

i'm actually trying to eliminate that problem by assuming the capacitor plates arent connected by a wire at all
13. If you look at the above link, it's not very pretty.

Clearly 1 1/2 years of physics at Uni wasn't enough for me to cover everything you need to do this. Not sure how many people in my year can do this
14. (Original post by JitsuCol)
If you look at the above link, it's not very pretty.

Clearly 1 1/2 years of physics at Uni wasn't enough for me to cover everything you need to do this. Not sure how many people in my year can do this
1 and a half years of physics at Uni is more than enough.
Hint.
Spoiler:
Show
The problem lies in the fact that Worzo defined a "perfect conductor". Automatically you people assume that means zero resistance and hence zero ohmic loss. However P = RI^2, think about it.
15. Well that link basically explains it, but in a slightly different way than I was going to explain it. It says that the system radiates the "missing" energy due to radiation resistance, which is ultimately what would happen, but JitsuCol's right - that's beyond my complete understanding at the moment.

The key point is realising that there is oscillation of the charge/current between the two capacitors, and that my instant movement of charge from one capacitor to the other is not physical. The energy I worked out was just in the electric field of the capacitors, but due to currents, there is energy also stored in the inductance and magnetic field of the connecting wires (even if there's no connecting wires, there will still be inductance through whatever space the charge is moving)

Interestingly, if the system was damped with a resistor so that there was no oscillation, the "missing" energy would be found to be dissipated in the resistor, and this would be exactly half of the initial energy. I believe you can also argue this on thermodynamic grounds, since the entropy doubles when the charge spreads between the two capacitors.

I was going to assume that, in a short time after the oscillations have started, the power radiated would be negligible, and hence for the purposes of the problem, there was no resistance there.

Mehh: perfectly conducting wires do have zero ohmic loss. They have Brehmstralung (sp??) though, which is seen as a resistance called radiation resistance.
16. (Original post by Worzo)
Mehh: perfectly conducting wires do have zero ohmic loss. They have Brehmstralung (sp??) though, which is seen as a resistance called radiation resistance.
Think about a perfect conductor a voltage source and a cap in series. Work out how much energy is used to charge the cap and work out how much energy is stored in the cap.
17. I don't understand your point. I was just saying that perfect conductors, by defintion, have zero ohmic resistance. Am I missing something? You're going to have to do the maths for me, I'm afraid...
18. Screw it, I'm going back to engineering!
19. (Original post by Worzo)
I don't understand your point. I was just saying that perfect conductors, by defintion, have zero ohmic resistance. Am I missing something? You're going to have to do the maths for me, I'm afraid...
Zero ohmic resistance. Infiniate current.
Power = Resistance x Current squared
Power = zero x inf^2
hence power not equal to zero.
20. (Original post by Mehh)
Zero ohmic resistance. Infiniate current.
Power = Resistance x Current squared
Power = zero x inf^2
hence power not equal to zero.
Hang on...zero ohmic resistance does not imply infinite current. The wire still has inductance, which limits the build up of current. And we have already said about there being some radiation resistance. I still don't get what you're trying to say.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 19, 2006
Today on TSR

### Negatives of studying at Oxbridge

What are the downsides?

### Nicer uni or better course?

Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE