Redistribution of charge in a capacitor

[ehabmozart]

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

 

[NascentOxygen]

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?

 

[ehabmozart]

U mean it's going to 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!

 

[ehabmozart]

And infact i mean, for example what if one cap. had more voltage or less capacitance.. I mean is there a rule behind that??

 

[MalachiK]

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.

 

[thisischris]

I asked a similar question here: https://www.physicsforums.com/showthread.php?t=612395

 

[NascentOxygen]

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. ✔ [/size][/color]

Sounds right! ☺[/size][/color]

 

[Ratch]

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

 

[NascentOxygen]

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.

 

[ehabmozart]

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 got to an exam tomorrow! :D

 

[Ratch]

NascentOxygen,

Not only is your approach wrong, so is your answer.

Now you got me interested. How is the approach and answer wrong?

Ratch

 

[MalachiK]

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?

 

[MalachiK]

Thinking about the energy, if the initial energy of the single cap is...

E_i = 1/2 × 0.01 × V_i²

and the final energy of the system is

E_f = 1/2 × 0.02 × V_f²

Then 0.01 V_i² = 0.02 V_f²

which leads to...

V_i = 2V_f, no?

 

[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.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

 

[Ratch]

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

 

[Ratch]

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

 

[MalachiK]

Yep, that's right! If I wasn't so stupid I'd have got exactly the same result as you.

 

[MalachiK]

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 going to 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?

 

[Ratch]

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

 

[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.

 

[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

 

[ehild]

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

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

The current:

I=\frac{dq}{dt}=\frac{Q}{RC}e^{\frac{-2t}{RC}}

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:

W=\int_0^{\infty}{I^2 Rdt}

ehild

 

[Ratch]

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

 

[ehild]

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 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

 

[Ratch]

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

 

[MalachiK]

Hmmm, but if we go along with this 'no charge is stored' idea then doesn't it follow that there can never be any device that stores charge? I mean, the universe is electrically neutral overall, isn't it? Any build up of charge in any location is going to be caused by a loss of charge somewhere else. Since charge is conserved in the universe there's only ever going to be 'separated' charge hanging around. If this is the case we, shouldn't we give up using the phrase 'stored charge' all together? What would be gained by this? In addition, you've stated several times that there's no conservation of charge in a capacitor. But if the total charge is always zero, that's still a conserved quantity right?

Also, how can the charge transfer take place in zero time? Any collection of particles is going to take some time to move one location to another. They're going to have to start moving, travel some distance and then stop. I'm not sure how these particles can fail to radiate energy when they do this, but maybe this is a failing in my understanding of electrodynamics.

Like I say, it's all a bit too non-physical for my taste. We end up with these two capacitors with exactly zero resistance separated by zero distance or something. It doesn't sound much like the situation posited in the OP.

 

[Ratch]

MalachiK,

Hmmm, but if we go along with this 'no charge is stored' idea then doesn't it follow that there can never be any device that stores charge?

I said that caps do not store a net charge. You cannot expect to take x coulombs from one cap and give it to another.

I mean, the universe is electrically neutral overall, isn't it?

I don't know.

Any build up of charge in any location is going to be caused by a loss of charge somewhere else.

Not if free charge carriers already exist somewhere.

Since charge is conserved in the universe there's only ever going to be 'separated' charge hanging around.

You have no basis to extrapolate from a cap to the universe.

If this is the case we, shouldn't we give up using the phrase 'stored charge' all together? What would be gained by this? In addition, you've stated several times that there's no conservation of charge in a capacitor. But if the total charge is always zero, that's still a conserved quantity right?

Charge is equally separated and conserved within a cap A. But you cannot give net charge away to cap B, because cap A has no net charge to give away. If you try to work problems that way, you will fail. Work those problems on an energy basis. Then you will never go wrong.

Also, how can the charge transfer take place in zero time? Any collection of particles is going to take some time to move one location to another. They're going to have to start moving, travel some distance and then stop. I'm not sure how these particles can fail to radiate energy when they do this, but maybe this is a failing in my understanding of electrodynamics.

You are trying to explain mathematical infinitesimals and infinities by a real world model. I showed what happens when the resistance is finite. You can make the resistance as small as you want, until the physical model can no longer support the values of current and velocity of the charge carriers.

Like I say, it's all a bit too non-physical for my taste. We end up with these two capacitors with exactly zero resistance separated by zero distance or something. It doesn't sound much like the situation posited in the OP.

The OP did not specify a resistance or say that any resistance was in effect. Therefore, it does seem to match what the OP asked.

Ratch

 

[MalachiK]

But clearly the question is a real world question. It says that there's this capacitor that's connected to another capacitor. Whyfore this assumption that the question concerns some utterly non-physical idealisation?

To take your free charge example... wouldn't removing some of these free charges and putting them some place else represent a net reduction in charge in the original location? Isn't that what happens when we charge a capacitor? We have the delocalised electrons in the metal of one plate moved around to the other plate. I don't understand how this is different. What if I get some charge from some other place and put it on one of the capacitor plates? Does it store charge then? Or is that plate now just storing separated charge in concert with wherever I got the charge from? I honestly don't understand how this thinking makes anything clearer.

Anyway, the only way to do a complete analysis from the point of view of energy transfer would be to take into account all of the energy losses. I don't see how you can invoke energy conservation and they ignore the losses that must take place when we have a non uniform current flowing between two points. But each time I suggest a reason why there might be additional energy losses to account for you propose some reasons that I don't understand to say that they don't happen.

I'm not saying your solution is wrong within the limitations that it sets itself - I just don't think that it's valid to ignore radiative losses. Maybe I'm too hung up on this 'real world model', but if we stop thinking about charge carriers moving around then we're not talking about two capacitors that have been connected together anymore - we're talking about some mathematical novelty that hasn't got anything to do with the reality of how this system behaves.

In any event, it's given me something to think about.

 

[ehild]

Electric charges can be separated, and it is possible to give one plate of a capacitor some charge Q while the other plate stays neutral or has some charge different from -Q, although capacitors in electric circuits usually have equal and opposite charges on the plates.

The model of connecting a charged and an uncharged capacitor with zero resistance wire would mean infinitely great current flowing for infinitely short time. It does not happen in real life. The wire has resistance, and it has got also inductance. The capacitors and the wires connecting them make a loop. The current flowing in the loop changes in time, causing a varying magnetic field around. Varying magnetic field causes changing electric field, it causes magnetic field... the electric energy of the capacitors radiates away.

The real situation is quite complex, and the system can loose energy in different ways. If the resistance of the wire dominates, the lost energy transforms to heat, warming up the wire. The inductance of the wire makes a resonant circuit with the capacitors and the stored energy would oscillate between the inductor and the capacitors, and at the same time dissipating on the resistor. If the resistance is very small, radiation would be the energy-dissipating process. ehild

 

[Ratch]

Malachik,

But clearly the question is a real world question. It says that there's this capacitor that's connected to another capacitor. Whyfore this assumption that the question concerns some utterly non-physical idealisation?

It is real world if a finite resistance is assumed. It is an idealization if no resistance is assumed.

To take your free charge example... wouldn't removing some of these free charges and putting them some place else represent a net reduction in charge in the original location?

Where else? Outside the capacitor? What location? The capacitor?

Isn't that what happens when we charge a capacitor?

You don't charge a capacitor, you energize it.

We have the delocalised electrons in the metal of one plate moved around to the other plate.

You add some amount of electrons to one plate and remove the same amount from the other plate. They are not necessary the same electrons.

I don't understand how this is different.

Different from what?

What if I get some charge from some other place and put it on one of the capacitor plates? Does it store charge then? Or is that plate now just storing separated charge in concert with wherever I got the charge from? I honestly don't understand how this thinking makes anything clearer.

That is what I have been trying to tell you. You can't do that. Whatever voltage you use to put coulombs on one plate will remove the same number of coulombs from the opposite plate. Net charge is zero, remember? You can't have 4 coulombs on one plate and 2 coulombs on the opposite plate. Or how about 4 coulombs on each plate? How would you do that?

Anyway, the only way to do a complete analysis from the point of view of energy transfer would be to take into account all of the energy losses.

Which I did with the resistor.

I don't see how you can invoke energy conservation and they ignore the losses that must take place when we have a non uniform current flowing between two points. But each time I suggest a reason why there might be additional energy losses to account for you propose some reasons that I don't understand to say that they don't happen.

You are going beyond the scope of the problem if you suggest any significant loses occur from things other than resistive elements.

I'm not saying your solution is wrong within the limitations that it sets itself - I just don't think that it's valid to ignore radiative losses. Maybe I'm too hung up on this 'real world model', but if we stop thinking about charge carriers moving around then we're not talking about two capacitors that have been connected together anymore - we're talking about some mathematical novelty that hasn't got anything to do with the reality of how this system behaves.

The problem definition sets the limits. Ignore radiation unless you are given more info from the problem. No one said charge carriers don't move. But their quantity stays in balance. For instance, the amount of charge that exits from one plate of a capacitor is match by the same amount of charge that enters the opposite plate.

Ratch

 

[Ratch]

ehild,

Electric charges can be separated, and it is possible to give one plate of a capacitor some charge Q while the other plate stays neutral or has some charge different from -Q, although capacitors in electric circuits usually have equal and opposite charges on the plates.

How would you put different amounts of charge of the plates of a cap? Inductive methods, perhaps?

Ratch

 

[MalachiK]

... it is possible to give one plate of a capacitor some charge Q while the other plate stays neutral or has some charge different from -Q, although capacitors in electric circuits usually have equal and opposite charges on the plates.
ehild

That's what's been bothering me. I mean, a capacitor doesn't have to be a sealed unit. I can just get two plates that I've independantly charged up with different amounts of +ve and -ve charge (opposite charges in each metal plate) and put them next to each other. BAM! I've got a capacitor with a different charge on each plate. That works, right?

This whole infinite current in infintesimal time is all too clever for me. I'm quite simple and infinites make my head hurt!

 

[ehild]

That's what's been bothering me. I mean, a capacitor doesn't have to be a sealed unit. I can just get two plates that I've independantly charged up with different amounts of +ve and -ve charge (opposite charges in each metal plate) and put them next to each other. BAM! I've got a capacitor with a different charge on each plate. That works, right?

It is right. Even a single metal sphere is a capacitor and you can give charge to a metal sphere... Just rub a comb and touch to it.

This whole infinite current in infintesimal time is all too clever for me. I'm quite simple and infinites make my head hurt!

It makes my head hurt, too. That is why I would include the resistance of the connecting wire. With that, everything is simple and all right.

ehild

 

[Ratch]

MalachiK,

That's what's been bothering me. I mean, a capacitor doesn't have to be a sealed unit. I can just get two plates that I've independantly charged up with different amounts of +ve and -ve charge (opposite charges in each metal plate) and put them next to each other. BAM! I've got a capacitor with a different charge on each plate. That works, right?

What are you going to use for positive charges? Protons?

Ratch

 

[ehild]

You can make a glass rod positively charged by rubbing it with cats fur. You touch it to a metal plate: That will be positively charged, too, as some electrons of the metal go over to the glass.

ehild

 

[Ratch]

ehild,

You can make a glass rod positively charged by rubbing it with cats fur.

Correct, by the absence of electrons, not by any positive charges.

Ratch

 

[ehild]

Absence of electrons leaves positively charged particles behind. Yes, because of the excess protons.
Positively charged particles are made in ion guns. Positively charged alpha particles can arise in radioactive decay. There is a material testing method using positrons, also positively charged particles.
There are both positively and negatively charged particles in the world, and they can be isolated.

ehild

 

[Ratch]

ehild,

Absence of electrons leaves positively charged particles behind. Yes, because of the excess protons.

Yes, but the protons are locked into the ionic core of the molecule and cannot move like the electrons can. They certainly do not move in capacitors.

Positively charged alpha particles can arise in radioactive decay.

Yes.

There is a material testing method using positrons, also positively charged particles

A positron is a antimater particle. How are they made and controlled?

There are both positively and negatively charged particles in the world, and they can be isolated.

Yes, ions in electrochemistry and holes in p-type semiconductor. But not in capacitors.

Ratch

 

[ehild]

See http://en.wikipedia.org/wiki/Positron_emission .

Well, what do you think, can an isolated capacitor have net positive charge? You have two metal plates and remove electrons from both of them. Is it possible?

I think you mix "charge" with charged particle. Charge is a property. Anything can have some charge, positive or negative.

The charged particles - electron, proton, positron, ions, - all carry some charge, integer multiple of the elementary charge. In different materials, different particles are the main charge carriers. In metals or semiconductors the main charge carriers are free electrons. In fluids and gases, the charge carriers are mainly ions. But there is ionic conductions also in solids, and ionic conduction is the main conduction process in insulators like glass or in ionic crystals.

ehild

 

[Ratch]

ehild,

Well, what do you think, can an isolated capacitor have net positive charge? You have two metal plates and remove electrons from both of them. Is it possible?

According to the link you gave, it would have to be a radioactive capacitor. But even if you could do it by some electrostatic induction method of taking away electrons, the sea of electrons present in metallic conductors would rush in and neutralize your efforts. That is why holes only exist in semiconductors. It is certainly not practical for a working circuit.

Ratch

 

[MalachiK]

ehild,
According to the link you gave, it would have to be a radioactive capacitor.
Ratch

So? Why not paint one of the plates with potassium-40? Everyone should have a hobby.