Redistribution of charge in a capacitor

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Homework Statement


A 0.01 F capacitor is charged by and then isolated from an 8 V power supply.
a) Calculate the charge stored... Straight forward... Q=CV
b) The capacitor is then connected across another identical capacitor which is uncharged. Describe and explain what will happen to the charge and voltage on each capacitor... This is a weak point for me.. I can't even find a clue.. All what i know is that C in parallel are given by C=C1+C2.. I really need help in this question and generally in how electrons are redistributed when a discharging capacitor is connected with another ones... Thanks in advance to whoever show his kind hand! :D


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
NascentOxygen
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Hi ehabmozart. Well, the charge isn't going to disappear or anything like that. So all it can do is redistribute itself. If the caps are identical, how much do you think each will end up with?
 
  • #3
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U mean it's gonna half.. I thought of that but what's the quantitative proof for that.. I read that C1/C2 = Q1/Q2 but i really want to know the proof.. Anyway, thanks for helping!
 
  • #4
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And infact i mean, for example what if one cap. had more voltage or less capacitance.. I mean is there a rule behind that??
 
  • #5
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If you connect any two caps in parallel then they will have the same p.d. across them once all of the charge has finished moving. For any capacitor, if you know the p.d. and the capacitance then you can find how much charge it holds.

I think the trick is to work out the total capacitance of your two caps in parallel and then treat this combination as if it were a single cap holding however much charge you had in total on the caps at the begining. This should lead you to the new p.d. across the caps. I'm pretty sure that this sort of thinking will lead to a general sort of understanding that can be applied to problems where the capacitors aren't exactly the same at the beginning.
 
  • #7
NascentOxygen
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I think the trick is to work out the total capacitance of your two caps in parallel and then treat this combination as if it were a single cap holding however much charge you had in total on the caps at the begining. This should lead you to the new p.d. across the caps. I'm pretty sure that this sort of thinking will lead to a general sort of understanding that can be applied to problems where the capacitors aren't exactly the same at the beginning. ✔
Sounds right! ☺
 
  • #8
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ehabmozart,

b) The capacitor is then connected across another identical capacitor which is uncharged. Describe and explain what will happen to the charge and voltage on each capacitor...
The first capacitor is energized to 8 volts and contains zero net charge. It does contain 8*0.01E-6 coulombs of separated charge, and (1/2)*0.01R-6*8^2 joules of energy. The second cap is initially inert. When connected, the voltages will equalize, because you cannot have one cap at a higher voltage than another. The cap with the voltage will share its energy with the cap with no energy. This will cause a current to exist from the energized cap to the other cap until equilibrium is reached. Capacitors are energy storage devices, it it makes sense to calculate their voltages from an energy standpoint.

Both caps are equal, so their capacity is "C". Since no resistor exists in the circuit, we can assume no energy loss, and the energy will be the same before and after.

(1/2)*C*8² = (1/2)*C*V²+(1/2)*C*V² ===> 64 = V²+V² , from which you can easily figure out the final voltage, and then the charge separation of each cap. Notice that it was not necessary to know the value of the cap to calculate the final voltage. If there was resistance in the circuit, then you would have to set up and solve a differential equation to get the same result. See https://www.physicsforums.com/showthread.php?t=610855

By the way, since no resistance was assumed in the circuit, the equalization current will be infinite. In a practical circuit, this is not possible, but the problem does not ask you to consider resistance, does it?

Ratch
 
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  • #9
NascentOxygen
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Ratch, the purpose of these homework forums is to offer only gentle assistance to nudge students towards finding the solution to problems. You are not supposed to work through the entire solution with no interaction or input from the student. Despite the best of intentions, doing someone's homework for them is of no help to anyone.

Not only is your approach wrong, so is your answer.
 
  • #10
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Well, nansent oxygen :S ... This question was not for the purpose of solving my hw.. This is not a hw.. It is a question in my book with an unclear mark scheme. I needed clarification in the full concept. Guys, thank you all for gving me the replies. I gotta an exam tomorrow! :D
 
  • #11
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NascentOxygen,

Not only is your approach wrong, so is your answer.
Now you got me interested. How is the approach and answer wrong?

Ratch
 
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  • #12
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Well I could be wrong, but doesn't your 64 = 2V2 lead to V = √32 = 5.66?

On the other hand, I get exactly 4 volts which is half the inital p.d. across the first cap. Intuitively, this is what I would expect to happen if we have the same charge distributed across twice the initial capacitance.

Am I missing something here?
 
  • #13
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Thinking about the energy, if the initial energy of the single cap is...

Ei = 1/2 × 0.01 × Vi2

and the final energy of the system is

Ef = 1/2 × 0.02 × Vf2

Then 0.01 Vi2 = 0.02 Vf2

which leads to....

Vi = 2Vf, no?
 
  • #14
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Well as far as I know when you connect capacitors the voltage or more technically potential difference across both of them should be in accordance with the Kirchhoff's loop rule.The connection does not contain any resistance and necessary calls for the steady state condition of the circuit .So these capacitors being identical must redistribute charge in such a way as to have same P.D. across them with opposite polarities (so that net voltage drop across the circuit is zero).Since their capacitances are the same you can easily see that they have the same charges on them and hence same P.D. .Redistribution of charge in capacitors involves loss of half the initial energy which trivially appears even in the charging case .
In short you must use Kirchoff's voltage rule while handling capacitor problems.Correct me if I am wrong.


Pardon me for violating any rule since this is my first post here.
-regards
yukoel
 
  • #15
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MalachiK,

Well I could be wrong, but doesn't your 64 = 2V2 lead to V = √32 = 5.66?
Yes, it does. That assumes no resistance is present in the circuit, and an infinite equilization current can occur. Also the caps will equalize instantaneously. This insures the energy before after equilibrium is the same.

On the other hand, I get exactly 4 volts which is half the inital p.d. across the first cap. Intuitively, this is what I would expect to happen if we have the same charge distributed across twice the initial capacitance.

Am I missing something here?
Yes, you are. Before I explain it, let me say that this is why I keep harping on the fact that capacitors don't get charged, they get energized. The net charge of a cap at zero volts is the same as a cap at 100 volts, namely zero. Capacitors don't store charge, they store energy. You based your answer on the mistaken belief that one cap stored a charge and shared it equality with a second cap. But there is no conservation of charge in a cap. Caps separate charge, they don't store it. The abundance of electrons on one plate is matched by a paucity of electrons on the opposite plate for a net change of zero. It takes energy to separate charges, and that energy is what a cap stores.

So the answer I worked out of 5.66 volts is for an ideal circuit that cannot possibly exist. However, it can easily be shown that if there is any finite resistance in the circuit. The final voltage for two equal caps will be one-half of the initial energizing voltage, and the total energy of the two caps will be one-half of the energy stored in the initially energized cap. The other half of the energy will be dissipated in the resistor, and it does not matter what the finite value of the resistor is.

Then 0.01 Vi^2 = 0.02 Vf^2

which leads to....

Vi = 2Vf, no?
According to your calculation, Vi = √2Vf

Ratch
 
  • #16
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Yukoel,

Well as far as I know when you connect capacitors the voltage or more technically potential difference across both of them should be in accordance with the Kirchhoff's loop rule.
Yes, the KVL rule is in effect.

.So these capacitors being identical must redistribute charge ...
Now you ran off the track. As I said in the previous post, caps do not store, conserve, or distribute charge because they have no net charge available. In this circuit the caps store and distribute energy according what is available and their capacitance value.

In short you must use Kirchoff's voltage rule while handling capacitor problems.
It depends on the problem.

Correct me if I am wrong.
With pleasure.

Ratch
 
  • #17
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Yep, that's right! If I wasn't so stupid I'd have got exactly the same result as you.
 
  • #18
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I've been giving this some thought...

Your result of 5.66 V is clearly correct if there are no other energy losses to be considered. But even if there's no resistance in the circuit like you described, won't there always be radiative losses as the (separated) charge moves around? So even in your idealised version the conservation of energy set up that we've used isn't valid. As far as I know, you can't move the electrons from one plate to another without producing EM radiation.

Following on with my half assed logic, I'm gonna take a guess that even with zero resistance in the circuit you're still going to lose 1.66 V due to radiation when you connect the two caps together and end up with 4 V in the end. I suppose what I'm saying is that it seems like it's not just the case that we could never build the idealised system, but that the laws of physics don't allow the process to happen as you've described.

My idea about thought experiements and ideal systems is that they allow us to ignore things that make it difficult to get to the truth. But by ignoring the radiation, aren't we over doing things?
 
  • #19
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MalachiK,

The KVL circuit analysis does not take nonresistive EM radiative losses into effect. If that is your worry, what about nonalternating high current circuits, too? A nonalternating current in a conductor makes a magnetic field, but that does radiate energy.

Ratch
 
  • #20
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Right, but in this situation the current isn't even approximately constant. I imagine that the current will decay like an exponential as the charge is redistributed. That changing current will certainly produce a changing EM field. This doesn't seem like some sort of trivial effect that can be just ignored when describing the ideal system.
 
  • #22
ehild
Homework Helper
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The capacitors are connected with a wire. That piece of wire has got some resistance, so it dissipates energy. The smaller the resistance, the higher the current. Current will flow till there is potential difference between the connected plates of the capacitors.

Before connecting the capacitors, the charge of the first is Q. After the switch is closed, current starts to from the positive plate of the charged capacitor to the uncharged one through the resistor R. At time t, the charge of the second capacitor is q and that of the first one is Q-q. According to Kirchhoff's first law the sum of potential differences in the loop is zero.

(Q-q)/C-IR-q/C=0. The increase of charge q in time dt is dq=Idt, so I=dq/dt. Substituting into the equation, it is a differential equation for q with th einitial condition q(0)=0. The solution is

[tex]q=\frac{Q}{2}\left(1-e^{\frac{-2t}{RC}}\right)[/tex]

The current:

[tex]I=\frac{dq}{dt}=\frac{Q}{RC}e^{\frac{-2t}{RC}}[/tex]

In equilibrium (at infinite time) q=Q/2 and I=0.

The initial energy stored in the the charged capacitor is E(0)=Q2/(2C). The energy of the two capacitors becomes Q2/(4C) altogether. The missing energy is equal to the energy dissipated on the resistor during the charge-redistribution process:

[tex]W=\int_0^{\infty}{I^2 Rdt}[/tex]

ehild
 

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  • #23
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MalachiK,

Right, but in this situation the current isn't even approximately constant. I imagine that the current will decay like an exponential as the charge is redistributed. That changing current will certainly produce a changing EM field. This doesn't seem like some sort of trivial effect that can be just ignored when describing the ideal system.
For no resistance, the change is instantaneous in theory. Any existing current can produce a EM field. So what? That does not mean it propagates, does it? Unless you can show that a resonant circuit is present which is broadcasting energy, you have no basis for assuming that particular energy is dissipated.

NascentOxygen,

As indicated earlier, the correct analysis is that given by MalachiK in post #5, based on conservation of charge. See also https://www.physicsforums.com/showthread.php?t=549222
No, that is not correct. There is no conservation of charge in a cap. All caps have the same charge before and after they are energized, specifically zero. Charges in a cap separate when the cap is energized, but their net charge is still zero.

echild,

I cannot argue with your correct result. But I hope you realize that the charge Q is the separation of charge on the cap, and not any charge the cap is storing.

OK, the attachment shows that any finite resistance will suck up half the energy stored in a energized cap connected to a cap that is not energized.

Ratch
 

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  • #24
ehild
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echild,

I cannot argue with your correct result. But I hope you realize that the charge Q is the separation of charge on the cap, and not any charge the cap is storing.

OK, the attachment shows that any finite resistance will suck up half the energy stored in a energized cap connected to a cap that is not energized.

Ratch
I do realize that the net charge on the capacitor is zero. What I has shown is derived from that.

By the way it is said that the capacitor is a device to store charge :tongue2: If one plate of the capacitor is grounded, and the other plate is given charge, that charge is stored on the capacitor and can be used later.


What is "V" in your attachment? There is no source in the loop of the connected batteries.

ehild
 
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  • #25
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ehild,

By the way it is said that the capacitor is a device to store charge
"It" is wrong. It is a device to separate charge, which in turn takes energy that the cap stores.

If one plate of the capacitor is grounded, and the other plate is given charge, that charge is stored on the capacitor and can be used later.
"Ground" is just a reference point. The charges are still separated, and there is still a voltage difference which indicates an energy storage.

What is "V" in your attachment? There is no source in the loop of the connected batteries.
"V" is the initial voltage on the energized cap. It is mathematically the same as a constant voltage source in series with the cap.

Ratch
 

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