I am so much confused about capacitors

[i_island0]

there are two parallel plate capacitors C1 and C2 connected in series across a battery of emf V with a switch S. The capacitor C1 has some initial charge q0 while C2 is uncharged.
I want to know will the final charges in the two capacitors be same?
if yes, why?

 

[Astronuc]

What happens when C1 = C2? What happens when C1 not = C2?

Please refer to - http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capac.html

 

[i_island0]

Sir, that much i also know. My problem is not that. In the link given -- http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capac.html -- the two capacitors C1 and C2 in series... if C1 was initially charged by some amount (say qo) while C2 has no charge... and then the battery is connected. will the charges on the two capacitors be same.

 

[Astronuc]

Write the relationship of charge on C1 and C2 in terms of V1 and V2, which would equal V, i.e. V = V1 + V2. What is the criterion for Q1 = Q2?

Do not confuse initial charge with final charge. The initial charge only affects the time to achieve the final charge - it does not affect the magnitude of final charge.

Capacitors in series divide the voltage.

 

[i_island0]

ok.. i have solved it.. can you please tell me if my answer is correct.
I have assumed that initially charge on C1 is q0 and charge on C2 is zero.
I have connected the positive terminal of battery (pd V) to the positive plate of capacitor C1.
Final charge on the two capacitors that i got are:
Final charge on C1: q0 + (VC1 - q0)C2/(C1 + C2)
Final charge on C2: (VC1 - q0)C2/(C1 + C2)
is my answer correct

 

[i_island0]

Sir, thanks for your answer, but just a final word from you will give me some relief, can you please tell me if my final answer is right

 

[Shooting Star]

Write the relationship of charge on C1 and C2 in terms of V1 and V2, which would equal V, i.e. V = V1 + V2. What is the criterion for Q1 = Q2?

Do not confuse initial charge with final charge. The initial charge only affects the time to achieve the final charge - it does not affect the magnitude of final charge.

Capacitors in series divide the voltage.

Hi i_island0,

Why don't you answer the questions he has asked? Does the 2nd para convey anything to you? May be it'll help you answer the question yourself.

You have posted the same question and your answer later in the forum, and there you have flatly refused to believe the answer given by user rl.bhat.

To put your mind at rest, so that you can start afresh, your answer is wrong. And this time we don't want polite disclaimers from you.

Just think what happens between the two "inner" plates of the two capacitors when they are joined.

Best wishes.

 

[i_island0]

So, you mean final charges on both of them are same, i.e. Q1 = Q2 = V.C1C2/(C1 + C2).
But, when i write equations and solve, i get the answers that wrote earlier, i .e.

Final charge on C1: q0 + (VC1 - q0)C2/(C1 + C2)
Final charge on C2: (VC1 - q0)C2/(C1 + C2)

Where, am i going wrong.

I wrote, (assuming some charge q is provided by the battery)
V = V1 + V2
= (qo+q)/C1 + q/C2

Sir, according to you what will be the correct equation?

 

[dynamicsolo]

OK, I think you now have this problem on three separate threads... You complicate the situation for getting help if people are trying to respond to you in one of those, while you are entering replies somewhere else.

So, you mean final charges on both of them are same, i.e. Q1 = Q2 = V.C1C2/(C1 + C2).

This would be correct. The final charge on each will equal the voltage of the battery divided by the "effective" capacitance of the set of capacitors.

But, when i write equations and solve, i get the answers that wrote earlier, i .e.

Final charge on C1: q0 + (VC1 - q0)C2/(C1 + C2)
Final charge on C2: (VC1 - q0)C2/(C1 + C2)

Where, am i going wrong.

I wrote, (assuming some charge q is provided by the battery)
V = V1 + V2
= (qo+q)/C1 + q/C2

You are correct that the sum of the final voltages across each capacitor must equal the voltage of the battery. The final charge on each capacitor must be the same (why?) and will not simply be the result of adding some amount of charge to each capacitor (in part, because some of the initial charge q0 on C1 will migrate).

Also, as I remarked on one of those other threads, your expressions for the final charge here cannot be correct because the difference VC1 - q0 appears to be the difference of a voltage and a charge, which is not conceptually meaningful. It would help to see what you did to arrive at these to be able to tell you what you're doing wrong there...

 

[Shooting Star]

> VC1 - q0 appears to be the difference of a voltage and a charge

Actually, VC1 is charge.

 

[dynamicsolo]

Actually, VC1 is charge.

OK, thanks, I see now. That was left unclear on the other thread; I was interpreting it as V-sub-C1, the initial voltage on the initial charged capacitor (I asked about this over there...).

 

[i_island0]

ok.. i see.. but can i know what makes the charge q0 migrate from capacitor C1 to capacitor C2.

 

[dynamicsolo]

ok.. i see.. but can i know what makes the charge q0 migrate from capacitor C1 to capacitor C2.

The negative plate of C1 is connected by a conductive wire to the positive plate of C2. It is inevitable that some of the charge q0 at C1 will end up at C2.

Is there anything in the problem statement, by the way, that says that q0 is less than (V·C1·C2)/(C1 + C2) ? Otherwise, there is another case that must be considered. (I suspect the problem assumes q0 is smaller, though.)

 

[i_island0]

there is no such problem in the book. I was just trying to understand the chapter by taking different configuration of capacitors. I was trying to write equations and checking the consistency of my results with the theory i read.
For a while i m assuming that VC1 > q0
Now, since you are saying its inevitable and charge must flow from negative plate of C1 to positive plate of C2; i am trying to find the charge on both capacitors as a function of time. So, in this case can you give me some ideas on how to write the equation as a function of time.

 

[Shooting Star]

The question is a very good one, and that's why there are not many replies pouring in. The thing that is tricky, as dynamicsolo has indicated, I believe, is what happens when the charge Q0 is greater than the charge that would have been attained by connecting the capacitors to the battery. Suppose it is a huge charge, and you've got a puny battery?

Perhaps now more people will try to answer the question. Be patient. The correct answer you will get, that I pomise.

 

[i_island0]

yes.. i m facing a big trouble regarding that.. i am trying since last one week various ways.

 

[dynamicsolo]

Now, since you are saying its inevitable and charge must flow from negative plate of C1 to positive plate of C2; i am trying to find the charge on both capacitors as a function of time. So, in this case can you give me some ideas on how to write the equation as a function of time.

You are not going to be able to write such an equation for the problem in its present form. The idealized capacitors in this problem have no resistance, so it is impossible to compute currents in such an imaginary circuit. As a result, there is no sensible way to describe the charge on the capacitors as a function of time. (You can work out the final charges, but only because those just depend on the battery voltage and the individual capacitances. The "equilibrium" or steady-state situation can be described, but there's no way to talk about how the circuit gets there.)

If you want to deal with the problem in terms of more realistic behavior and be able to come up with functions of time which describe the charges on the capacitors, you'll need to add a resistor to the circuit. (There, I just did it: put a resistor with resistance R anywhere along the circuit.) Now you can use Kirchhoff's circuit laws to set up an equation relating the voltages in the battery, resistor, and capacitors. You will have a differential equation you can now work with, subject to the initial conditions q_C1(0) = q0 and q_C2(0) = 0.

 

[i_island0]

i did that too.. and i ended up with the result that i wrote earlier. I have put a resistor as u said. and the equation that i wrote is:
(qo + q)/C1 + q/C2 + (dq/dt)R - V = 0
Is this equation right?

 

[Shooting Star]

Why is q0 there in the differential eqn? You'll have to put in q0 as an initial value after integration.

 

[dynamicsolo]

... the equation that i wrote is:
(qo + q)/C1 + q/C2 + (dq/dt)R - V = 0
Is this equation right?

You don't put in the q0 to start with: you would add that later as an initial value that the solution for q(t) must satisfy. As I think about this, though, I wonder whether you have given yourself additional trouble by putting a charge on one of the capacitors. It seems there would now be two time scales in the problem. The charge on the plate of C1 connected to the plate of C2 would spread itself extremely rapidly over the linked conductors. This would happen much faster than the battery would be able to move charges through the circuit.

It seems like you would want to just treat this like a very brief transient condition, which would really start the remainder of the journey to equilibrium as if the capacitors began with initial charges such that the two plates and connecting wire be at an equipotential. In any event, the differential equation would be

V - R(dq/dt) - q/C1 - q/C2 = 0 ,

which you would solve using the appropriate initial condition. This seems a bit of a mess (which may explain why such a problem doesn't turn up in introductory E&M courses).

For the purpose of answering the original question, the final configuration can be solved just from the requirement that the charges on both capacitors have to end up the same. The description of how the charges flow seems not so simple...

 

[Shooting Star]

>
For the purpose of answering the original question, the final configuration can be solved just from the requirement that the charges on both capacitors have to end up the same. The description of how the charges flow seems not so simple...
>

What would be the magnitude of that charge?

 

[i_island0]

But at some time 't' can i say that the charge was same on both the capacitors.
for eg. q/C1 + q/C2 + (dq/dt)R - V = 0
the solution for this comes out to be:
q = CV(1 - e^(-t/CR)) + qoe^(-t/CR)
where, C = C1C2/(C1 + C2)
But, is it conforming that initial charge on C1 is qo and that on C2 is zero.

Moreover, we should take care of the charge that is supplied by the battery to the capacitors

 

[Shooting Star]

That will come out of the differential eqn. For an RC circuit, it'll only happen when t becomes infinitely large. In practice, that means a few multiples of RC.

 

[i_island0]

but i was talking about at time t = 0. AT that time are we making sure that charge on C2 is zero.

 

[Shooting Star]

But at some time 't' can i say that the charge was same on both the capacitors.
for eg. q/C1 + q/C2 + (dq/dt)R - V = 0

But how do you know that until you solve the equation? Isn't it just an assumption that at some t, the q on both the caps are equal?

 

[i_island0]

And that time is infinity when q on both caps are equal.
I asked my teacher regarding this. and he was saying that its not necessary that both the capacitors have the same final charge. so i m more confused about this ..

 

[dynamicsolo]

And that time is infinity when q on both caps are equal.
I asked my teacher regarding this. and he was saying that its not necessary that both the capacitors have the same final charge. so i m more confused about this ..

I think we're going to have to clarify what the original problem is. The two capacitors of possibly different capacitances C1 and C2 are in series with a battery of emf V. In this situation as it stands, the battery and the capacitors are idealized, as is the connecting wire; this means the circuit contains no resistances. This will make it impossible to describe how current behaves in this circuit.

Going back and looking again, I see that it's never made clear how this switch S works. I am presuming that somehow it is helping to keep the charge q0 on capacitor C1, while keeping C2 uncharged. If C1 and C2 are supposed to be in series, it's a bit hard to see how that would happen: how would C2 be kept isolated? (A drawing would be helpful.) If this were the situation, it would probably mean that q0 does equal V·C1 .

Something seems incomplete about the description of this problem. Certainly, you cannot model the dynamic situation with what little detail is given here.

Are you sure the capacitors are in series and not in parallel? That would explain your instructor's remark (and how C1 is kept isolated from C2 by the switch).

 

[i_island0]

I am putting the new circuit here in:
http://img175.imageshack.us/my.php?image=newcircuitdp8.jpg

The initial conditions given are:
t = 0, q_C1(0) = qo and q_C2(0) = 0.

Now, i want to find the charges on the capacitor as a function of time.
For that i wrote the equation:
q/C1 + q/C2 + (dq/dt)R - V = 0

whose solution comes out to be:
q = CV(1 - e^(-t/CR)) + qoe^(-t/CR) where, C = C1C2/(C1 + C2)

& dq/dt = (V/R) e^(-t/CR) - (qo/CR) e^(-t/CR)

Thus, if i put t = 0 now in the solution i will get q = qo (i.e. I am indirectly saying that the initial charges on both the capacitors are same, i.e. qo). Also the current is then: (CV - qo)/CR

Now, this solution is possible only if I assume that just after closing the switch the charges from C1 flows to C2 very rapidly. But in that case the initial charges on the capacitors will not be qo.

 

[i_island0]

Are you very sure that for the circuit that i am showing now will have the same final charges, or should be treated as in series.
The textbooks are saying that if the charges on two capacitors are same then we call it in series and not ...when the two capacitors are arranged in the configuration that i am showing is definitely guarantying that they are in series.

 

[Shooting Star]

There cannot be just a +q0 on C1. Try to ascribe the proper values of charge to each plate. I always had assumed that, if nothing else, you had meant +q0 and -q0 on the plates of C1.

 

[i_island0]

yes of course .. -qo is on the other plate

 

[rl.bhat]

You can put the charge + qo on C1 in two ways. Take a positively charged rod and touch it to one of the plates of the capacitor. In that case inner and outer surfaces of the caapcitors are positive. In the second capacitor, inner surface is negative (due to induction) and outer surface is positive. In the second method, connect the capacitor across a voltage source. In that case one plate will be positive and other plate will be negative. In the given circuit which one is the case?

 

[i_island0]

In the given circuit, i have charged the capacitor using external DC supply of lesser emf such that when we solve this question, this capacitor C1 can accept charge more than qo.

 

[Shooting Star]

Even before closing the switch, do you think nothing will happen to cap C2?

 

[i_island0]

no nothing will happen. Thats what we have seen in one of my other questions regarding earthing. Only when the charges are on the outer plate of the capacitor, then only something can happen. But here i m taking the charge q0 to be in the inner plates of capacitor. (qo on +ive plate and -qo on the negative plate)