Question

we have two capasity c1=c2=c
c1 has charged q and c2 is empty
after connection with same poles the total energy
e1+e2=1/4e+1/4e=1/2e
there is no resistance or droping energy stuff .so where is the half of energy?

Answer 1

It is lost anyway as wires cannot have zero resistance.

If wires were supraconductors, the inductance of the closed circuit would make the current oscillate indefinitely within the circuit, converting electrostatic energy to magnetic energy and back.

Answer 2

From the original question:
C1 = C2 = C so the capacitors themselves have the same capacitance.
Initially:
C1 is charged to some potential q so it has potential energy.
C2 is uncharged so it has 0 potential energy.
\[U_1=\frac{1}{2}\frac{(q_1)^2}{C}\]\[U_2=\frac{1}{2}\frac{(q_2)^2}{C}=\frac{1}{2}\frac{0^2}{C}=0\]\[U_{total}= U_1 + U_2=U_1+0\]When you connect these two capacitors "with the same poles" (in parallel) 1/2 the charge from q1 will move to the plates of q2 (due to repulsion) and you end up with:
\[U_1=\frac{1}{2}\frac{(\frac{1}{2}q_1)^2}{C}=U_2=\frac{1}{2}\frac{(\frac{1}{2}q_1)^2}{C}\]\[U_{total}=U_1 + U_2\]
Mathematically, both capacitors now have 1/2 of the total potential energy that C1 originally had...not 1/4th.

Answer 3

I am looking at it and yur just proved it is 1/4 of the original energy which is wrong but I dont know where

Answer 4

I guess I didn't explain it well enough.

U started out as 1/2 * something. It ended up (for each capacitor) as 1/4 of something. That doesn't mean U went down to 1/4...it's still just half of what it was originally. Basically, U went from 1/2 to 1/4...so it's cut in half.

Answer 5

The 1st explaination make sense joining two cappacitor together will form a dynamic stituation which the charges oscillate which makes the new V rms = √1/2 V,you can't simply divide the charge in half

Answer 6

The first explanation says energy is lost in this situation which is incorrect. Then it goes on to talk about superconductivity which is not even mentioned in the original question.

The fact is that there is effectively no potential energy lost when you take a charged capacitor and place it in parallel (or even series) with another charged capacitor. Based on the question there is nothing else in the circuit to consider.

Why exactly do you think the charges will oscillate? Once they spread out among the plates there's absolutely no reason for them to move again.

Answer 7

electrons is like other paticles it has potential energy and kinetic energy,a stable charge capacitor has only potential energy however once it is linked to another capacitor all the electron will start to accelerate gaining kinetic energy and rush to the other capacitor making it over charged once the kinetic energy is gone,because the second capacito has higher potential now it will start rush back.I dont think you can fully duplicate this in real life stuation since it will energy always lost in heat and electromagetic radiation.

Answer 8

I mostly agree with what you're saying but all of that happens almost instantaneously and if there's any oscillation at all you probably can't even measure it. At some point (again, almost instantaneously) an equilibrium will be reached and the oscillations will stop.

In the end the total potential energy before and after the exchange will effectively be the same. Note that I said effectively...there will always be *negligible* losses due to heat, radiation, etc. but for physics calculations those are almost always ignored.

Answer 9

After stability whats going on for the half an energy ?
Meantime I should mention that T=RC=0 ,so oscillation is not going to occur .
This is so interesting problem.

Answer 10

I dont konw your got this problem from a book or come up yourself,the mathematics clearly shows once it stablize half the engergy is lost in this case it is not negligible it is very similar to dividing other materials like sand water..half the potential is lost.I dont understand whats T=RC=stands for but I think that unless the energy is dispersed somehow a oscillation will occur and in ideal state it will never stop

Answer 11

I was basing all of this on the equations I first posted and after I did this again on paper I see that I made a stupid math error. I also did some more research and found this which supports what you're saying:

http://www.google.com/url?sa=t&rct=j&q=&esrc=s&frm=1&source=web&cd=2&ved=0CFMQFjAB&url=http%3A%2F%2Fciteseerx.ist.psu.edu%2Fviewdoc%2Fdownload%3Fdoi%3D10.1.1.167.3034%26rep%3Drep1%26type%3Dpdf&ei=nAivT467DpKA6QH_mNnCCQ&usg=AFQjCNH25Zkbfjfuz1j6MvfoyDqNaebdcQ&sig2=vyU56X6Y4WD2Lz12L1oNFA

Good job shadow :) I'll go hide my head in shame...it was a good discussion though.

Answer 12

@mahmit2012: you said "Meantime I should mention that T=RC=0 ,so oscillation is not going to occur"

Do not mistake T=RC (time constant) of an RC circuit with \(T_0=2\pi \sqrt{LC}\) for an LC oscillating circuit.

If circuit has some resistance, then energy will be converted into heat by the resistor and finally radiated in the environment.
If circuit has literally no resistance (superconductor), the oscillation will create variable electric and magnetic fields that will create an electromagnetic wave that will carry away the energy.

So, in both cases, the lost energy will end up in radiation.