Question

The Capacitor Paradox

A capacitor C is charged to Q, giving voltage V,
this is connected in parallel with another capacitor uncharged C.
Find the new energy stored in the system

we do this by charge conservation.
So net capacitance is 2C, and net charge is Q, hence new voltage V' = Q/2C = V/2

hence new energy is (1/2)(2C)(V/2)^2 = CV^2/4

previously it was CV^2/2

where is the missing energy?
Everything is ideal.. so no question of dissipation of energy.

Answer 1

@Vincent-Lyon.Fr @LastDayWork @AnthonyStark

Answer 2

@agent0smith

Answer 3

By moving charge you have changed the potential energy of the system.

Answer 4

care to elaborate?

Answer 5

the moving charges are stored in the second capacitor as well.. so kinetic energy must be retransformed into potential energy!

Answer 6

i found a pdf that hits this apparent paradox hard.. m yet to read.. anyone interested can go through it.. ll post my understanding once i go through it

http://arxiv.org/pdf/1309.5034.pdf

Answer 7

Will look forward to that.

Answer 8

When the capacitors are joined up, this can be done thru a high resistance in which case the lost energy is mainly dissipated as heat in the resistance. If the connection is made thru a low resistance, so there is rapid charge acceleration, much of the lost energy is dissipated as radiation.

Answer 9

The pdf proves it.. (or that's its goal), that radiation losses cannot account for the losses..

apparently from what i read so far, like you mentioned, the energy is transferred as k.E.of the electrons, and then they Go BAAMM slam into the second capacitor.. and get stuck there.. so there is loss..
only once the second capacitor starts getting a little charge and develops an E field, will it be able to decelerate those incoming charges, converting their k.E into electrostatic potential energy

what do you think?

Answer 10

I did not say that all of the lost energy is dissipated as radiation. However a practical test with a spectrum analyzer indicates that there is significant radiation when the capacitors are connected with a low resistance.

Answer 11

i see.. what you think of what i said.. ? does that make sense? :P

Answer 12

http://puhep1.princeton.edu/~kirkmcd/examples/twocaps.pdf
This pdf on the other hand, COMPLETELY accounts the missing energy for radiation losses..

since I don't know much about the math behind the electromagnetic waves, i can't understand the approach that the two pdfs have, one says its negligible the other one totally accounts for it :-/

Answer 13

When we connect a charged capacitor to an uncharged one, we join two wires which are at different potentials, thereby creating a spark. Hence we cannot expect energy to remain conserved (in the circuit).

It is similar to charging a capacitor by an ideal battery -
Work done by battery = QV
Energy stored in capacitor = QV/2

Answer 14

"we join two wires which are at different potentials, thereby creating a spark"
say what? :P

Answer 15

Work done by battery = QV
thats wrong man :O

Answer 16

battery does just QV/2 amount of work!

Answer 17

Battery pushed Q amount of charge through a potential V
Can you explain how is work done by battery = QV/2 ?? :P

Answer 18

No I won't explain.. ll give u a hint
when the battery STARTS doing the work, the pd across capacitor is zero.. and gradually starts increasing, so the work done by the battery as it starts charging starts from zero and gradually increases to max, and when u add that all up (basically integrate it), u get QV/2

you should have known that.. disappointed :-/

Answer 19

I am not asking about the work done on the capacitor..Sorry to disappoint :P

& I almost forgot to add...you should have known that.. disappointed :-/
:P

Answer 20

BTW...Am I the only one who observes sparks while connecting wires ?? o.O

Answer 21

Maybe I see things that don't exist.. >.<
I need to see a shrink..or a warlock maybe.. :P

Answer 22

Knock! Knock!
Anybody ??

Answer 23

At the instant of joining the two capacitors with a low resistance it is usual to see a spark at the point where contact is made.

Answer 24

@LastDayWork quite right.. my bad :D... and my lil comment made it even more embarrassing..
I never ever thought about that.. I always thought the total work done by the batter is the same as the energy stored in the capacitor..
so the basic idea being.. whatever you wanna call it as sparking or not.. that all the kinetic energy doesn't get efficiently converted into potential energy!

Answer 25

The electrostatic energy is converted into heat by the components' resistance (including wires).

Answer 26

"only once the second capacitor starts getting a little charge and develops an E field, will it be able to decelerate those incoming charges, converting their k.E into electrostatic potential energy"

I do not understand that sentence; it follows the common misconception that current is like a flow of water in an empty pipe.

Answer 27

"only once the second capacitor starts getting a little charge and develops an E field, will it be able to decelerate those incoming charges, converting their k.E into electrostatic potential energy"

I do not understand that sentence; it follows the common misconception that current is like a flow of water in an empty pipe.

Answer 28

@Vincent-Lyon.Fr The question assumes, resistance less wires.. superconducting wires!

Answer 29

Well, that's a bit far-fetched, but even so, the circuit's loop has a self inductance and since no damping occurs, then the system will oscillate.
The variable current will radiate an electromagnetic wave. This wave will carry away the "missing" energy, leaving the final state with less electrostatic energy than initially present.

Answer 30

@Vincent-Lyon.Fr .. yes but i posted two links above.. one of which proves that the radiation cannot account for missing energy but it is because all the kE is not converted back to pE, while the other one proves it does. And like @LastDayWork mentioned, another explanation is sparking, but again, in ideal situation, there must be a way to NOT have sparks (if i did the whole experiment in vacuum ? :D)
i liked the explanation of the first pdf which discusses this in great detail and proves, how all k.E of the electrons is not efficiently converted back to p.E
(also self inductance effects can be easily neglected, cause in real world applications too the self inductance does not matter, so surely it cannot take away HALF the amount of total energy)

Answer 31

And when i mean self inductance does not matter, i mean self inductance of a circuit loop!

Answer 32

@Mashy
I know it's funny to talk about sparking in vacuum, but while closing the circuit we must consider the energy loss due to transient state, albeit it exists for a very small time. I used the term 'spark' to avoid this discussion.

As @Vincent-Lyon.Fr already stated
"only once the second capacitor starts getting a little charge and develops an E field, will it be able to decelerate those incoming charges, converting their k.E into electrostatic potential energy"
This ^^ is (probably) a misconception.

"Well, that's a bit far-fetched, but even so, the circuit's loop has a self inductance and since no damping occurs, then the system will oscillate. The variable current will radiate an electromagnetic wave. This wave will carry away the "missing" energy, leaving the final state with less electrostatic energy than initially present."
I once tried to do a similar ^^ analysis, but a simple counter-example proved it incorrect (in ideal state).

Consider two capacitors (as shown in the diagram)


Now, change the capacitance of any one capacitor (by changing the distance b/w them ; or by introducing a dielectric). The energy of the circuit will remain conserved.

Note that I am working with the ideal state. Implies no resistance, inductance etc.

Answer 33

"The energy of the circuit will remain conserved."
I should rather say that "total energy will remain conserved" in the sense that NO EM wave will be (mathematically) predicted. We (obviously) do have to consider the work done in moving the plates and/or introducing the dielectric.

My point is that no energy will be lost in the transient state. Hence, we can't talk about energy loss due to oscillations or the 'flow concept' in classical EM theory.

The only difference is that the circuit remain closed during our analysis thereby eliminating the possibility of 'spark'.

BTW, if you have the luxury of time, you can work with a more complicated circuit (by introducing a battery, inductor and resistor). As long as you are working with the ideal concepts, you won't find any paradox (unless you use switches ;)

Now, why do sparking (always) carry away a 'definite" amount of energy, I couldn't find an answer for it (yet)..

Answer 34

@roadjester & @ybarrap may enjoy this feed..

Answer 35

"only once the second capacitor starts getting a little charge and develops an E field, will it be able to decelerate those incoming charges, converting their k.E into electrostatic potential energy"

what i am trying to state here what i understand of how k.e gets converted to p.e
example
if i throw a +ve charge opposite to E field, then the charge decelerates and gains p.E

simiarly when a capacitor is being charged, shouldn't something similar happen?
consider the case when the battery charges the capacitor
the battery uses its chemical energy to transform it into k.e of the electrons, now these electrons k.e has to be converted into p.e of the capacitor.
BUT initially that is not possible cause there is No E field in the capacitor to decelerate it. Therefore, the k.e of electron will be lost.. (maybe thermal vibration of the capacitor plate or something)
but only once sufficient E field has built up, will it be able to decelerate the incoming electrons and convert it into potential energy?
what am i missing?

Answer 36

"if i throw a +ve charge opposite to E field, then the charge decelerates and gains p.E..simiarly when a capacitor is being charged, shouldn't something similar happen?"

There a significant difference, when a capacitor is being charged, the field develops b/w the plates of the capacitor. This alters the potential gradient of the circuit thereby reducing the amount of current.

The electrons are NOT in direct interaction with the field (created b/w the plates) of the capacitor, hence we CANNOT bring the concept of PE of a charge in a field here.

Answer 37

ok.. besides sparking what else is there? :P
also if you made a second capacitor infinite capacitance all the energy is lost

its like one tank filled with water, connected to another HUGEEEEEEEEEEEEEEEEEEEEE tank, then the potential energy of water is almost gone (cause water level is almost zero).. where does the energy disappear there?
so there it is like water goes and hits the tank, and so its k.e gets dissipated? so thats what he meant by saying, not to think of electric current as a flow of water in pipe eh.. hmmph!

Answer 38

"also if you made a second capacitor infinite capacitance all the energy is lost "

The million dollar Q's is 'how' to make the second capacitor of infinite capacitance ??

Answer 39

But u say energy is lost in the transient state
lost as WHAT?.. EM WAVE is the only way it can dissipate energy.. but EM wave cannot carry that much energy right? atleast thats what the pdf proves.. can u check the first pdf .. calculations?

Answer 40

we are in an ideal world my boy... :D..
infinite meaning.. one huge capacitance, and one extremely small capacitance.. in that case almost all energy is lost.. that's what i meant..

Answer 41

"But u say energy is lost in the transient state..lost as WHAT?.. EM WAVE is the only way it can dissipate energy.. but EM wave cannot carry that much energy right?"

I am not in a position to answer this ^^
BTW, spark is also a combination of EM Waves and Heat.
I'll read the pdf(s) to see if they make any sense..

Also, by 'how' to get infinite capacitance, I was expecting an answer related to the process one can use to achieve it.

One way is to use an ideal conductor as dielectric ; but it will cause sparking..or maybe an explosion ;D

Answer 42

"One way is to use an ideal conductor as dielectric ; but it will cause sparking..or maybe an explosion ;D"
That.. that is the reason i give u medal.. not cause its clever (that was a pathetic attempt of making an infinite capacitance).. but that was hilarious :D

Answer 43

BTW, spark is also a combination of EM Waves and Heat... u can't have Heat.. Heat needs medium..(i hope u meant Infrared which is part of EM)

so it can only be EM waves..

Answer 44

@ybarrap
are you building a capacitor or something? :D :D!! .. are you ever gonna finish typing the post man? :D

Answer 45

Here are my 2 cents - a subjective perspective

Just looked the documentation on this subject but I haven't seen anyone approach this from a thermodynamic point-of-view. It think it might helpful to look at thermodynamics to understand why so much energy is lost when two equal-sized capacitors are involved vs. why so little energy is lost when one is of infinite capacitance.

We know that in metals electrons will move in such a way to to make the electric field in the metal zero. In the case of this problem, with two equal capacitors, after all the charges have moved and everything is in equilibrium, the 2nd Law of Thermodynamics would say that it would be impossible for charges to go back on their own to where they were with all electrons on one capacitor and none on the second , because processes tend to go from low entropy to high entropy. In other words, it would take work to move the electrons to their original configuration. This work required is the work lost when the charges redistribute themselves between capacitors.

We can use the Joule Expansion to consider a process in which no work was expended to understand how all energy was transferred from one small capacitor to one with infinite capacitance- http://en.wikipedia.org/wiki/Joule_expansion. In this way, there is a physical process that seems analogous to our current one. In the Joule Expansion, a gas empties into a vacuum and thus does no work as it expands -- similar to how electrons do no work as they redistribute onto an infinite-sized capacitor.

Answer 46

Scratch that idea.

@Mashy reminds me that in fact that ALL energy is lost when the second is an infinite capacitor:
$$
\Delta E=-\cfrac{C_{\infty}\times E_{original}}{C_1+C_{\infty}}\rightarrow -E_{original}
$$
Which is contrary to the Joule Expansion.

Must have been very tired last night!