# I am so much confused about capacitors

1. Dec 20, 2007

### 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?

2. Dec 20, 2007

### Staff: Mentor

3. Dec 20, 2007

### 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.

4. Dec 20, 2007

### Staff: Mentor

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.

5. Dec 20, 2007

### 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)

6. Dec 20, 2007

### i_island0

Sir, thx 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

7. Dec 21, 2007

### Shooting Star

Hi i_island0,

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.

Last edited: Dec 21, 2007
8. Dec 22, 2007

### 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?

9. Dec 23, 2007

### 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.

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.

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...

Last edited: Dec 23, 2007
10. Dec 23, 2007

### Shooting Star

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

Actually, VC1 is charge.

11. Dec 23, 2007

### dynamicsolo

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...).

12. Dec 23, 2007

### i_island0

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

13. Dec 23, 2007

### dynamicsolo

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.)

14. Dec 24, 2007

### 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.

Last edited: Dec 24, 2007
15. Dec 24, 2007

### 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.

16. Dec 24, 2007

### i_island0

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

17. Dec 24, 2007

### dynamicsolo

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.

18. Dec 24, 2007

### 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?

19. Dec 24, 2007

### Shooting Star

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

20. Dec 24, 2007

### dynamicsolo

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...