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# Capacitor Puzzle

Dennise
We have two identical discharged capacitors C1 and C2 with a capacitance of 1F each.

Now charge capacitor C1 to V1 = 1 volt.
From C = Q/V, we have a charge stored in C1 of Q=1. So Q=V1=C=1.

Connect C1 directly to uncharged identical capacitor C2 in parallel. Now the total capacitance = 2C or 2F.

Since charge must be conserved, we calculate the new voltage across both C1 and C2.
From C=Q/V we have V2 = Q/C and V2 = 1/2 volts; or 1/2 the voltage of the initial disconnected C1. All is well.

We now compare the energy Ea stored in C1 initially with the combined energy Eb stored with C1 and C2 connected in parallel. Both energies should be equal right? First, the energy stored in a capacitor is given as E=(C*(V^2))/2.

For C1 initially we have Ea = (1*1^2)/2 = 1/2.

For C1 and C2 in parallel, we have Eb = 2C*(.5^2)/2 = 2*.25/2 = 2/(4*2) = 1/4.

But since charge AND energy must be conserved how can this be? We began with C1's energy = 1/2, but after C1(charged) and C2 (discharged) are connected in parallel we end up with the energy stored in C1 and C2 together of only 1/4. HALF THE ENERGY HAS DISAPPEARED!

Both capacitors and all connections are ideal and have zero resistance.

Can you explain this?
Bikerman
Well, the quick 'explanation' is that electrical energy here is potential, and PE is not conserved....

A more complete look at it:

EDIT - Oops - wasn't until I read _AVG's post that I realised I hadn't credited the above...I am very sorry...I should have cited the source and genuinely thought I had - unpardonable lapse on my part, which I must of course remedy immediately.
http://web.mit.edu/6.013_book/www/chapter11/chap11.html
_AVG_
Yes, as Bikerman showed, even in zero resistance there is still power dissipated in the wire so the energy need not be constant; however, if you're worried about the conservation of energy in general, I would guess that the energy would be lost in the form of heat (a spark or something when connecting the other capacitor would probably be observed, even under ideal conditions)
Dennise
Try again.

Remember, the capacitors, connecting wires and all else are ideal:

Zero resistance in all wires and connections. No resistance anywhere.
Ideal capacitors: zero inductance and zero resistance. Only pure capacitance

How does one account for the missing energy, or are the calculations wrong?
Dennise
To further make the problem's conditions more ideal, I forgot to mention the experiment is done in a vacuum.

Bikerman, I don't understand why potential energy isn't conserved. If a 50 lb anvil were suspended (motionless) 20 ft. above you, and then was released and fell on your head, you might revisit your statement that PE is not conserved.

Well, I guess it's time to cut to the chase anyway.

AVG .... you are close if one considered normal atmosphere conditions. Yes, some energy would be lost by heat. In a vacuum however, I believe the energy is lost in broadband radiation. In such an experiment, I'd be interesting to know the actual spectrum of that radiation.

Any guesses?
kelseymh
 Dennise wrote: To further make the problem's conditions more ideal, I forgot to mention the experiment is done in a vacuum. Bikerman, I don't understand why potential energy isn't conserved. If a 50 lb anvil were suspended (motionless) 20 ft. above you, and then was released and fell on your head, you might revisit your statement that PE is not conserved.

Potential energy isn't "conserved" because it cannot be measured absolutely. Your example above is perfect. What you're describing is the conversion of potential into kinetic energy, not conservation. How much potential you had to begin with depends on where you set your zero point.

 Quote: AVG .... you are close if one considered normal atmosphere conditions. Yes, some energy would be lost by heat. In a vacuum however, I believe the energy is lost in broadband radiation. In such an experiment, I'd be interesting to know the actual spectrum of that radiation. Any guesses?

I'd say there are three possibilities: (1) a Gaussian spread around the 1/RC frequency of the circuit, with the spread determined by the Q value; (2) white noise; (3) a Planckian (black body) spread at whatever the equivalent temperature of the circuit is.
Dennise
The problem is a fairly well known physics paradox. Here is one solution that shows the lost energy IS radiated. Charge AND energy are both conserved. Credit to the author K.T. McDonald.

http://www.hep.princeton.edu/~mcdonald/examples/twocaps.pdf
_AVG_
Wow, that is a really rigorous article. Nice, thanks for sharing it

I never thought of considering dipole radiation and applying the Larmor formula in order to solve this so called "paradox"!
andro_king
does an ideal capacitor exist..?
Dennise
No. Neither do ideal inductors, resistors, transistors, diodes, transformers and many many other components. Ideal people do not exist.

Analyzing physics, engineering and other problems very often consider ideal components and conditions to make it easier to get at the fundamental understanding of such problems. Once this has been accomplished, real world - non ideal - components are considered and their effects included in the problem. Another example when studying mechanical problems is the consideration of frictionless surfaces.