I am so much confused about capacitors

[i_island0]

I am posting my question again. I hope its very clear this time.
The circuit is in the link provided:
http://img259.imageshack.us/my.php?image=newestag1.jpg

In the circuit shown C1 was charged by some external source of emf to provide it with some initial charge qo. Next, it was connected with an initially uncharged capacitor C2 and arranged in the circuit shown. Can you please just write the equation so that i can find the charges on both the capacitors as a function of time.
thx again

 

[i_island0]

Moreover, I am thinking that before the switch was closed, the right plate of capacitor C1 and left plate of capacitor C2 had a total charge of -qo. Those plates are electrically isolated from rest of the circuit. So, the final charge on them should be same, i.e. -qo.
[from...Halliday, Resnick, Krane..4th edition, vol..2,, page 682..2nd para.. 6th line]
If we keep this point in mind, then we should say that final charges on both the capacitors is not going to be the same.
?? any corrections this time??

 

[dynamicsolo]

Moreover, I am thinking that before the switch was closed, the right plate of capacitor C1 and left plate of capacitor C2 had a total charge of -qo. Those plates are electrically isolated from rest of the circuit. So, the final charge on them should be same, i.e. -qo.

If the capacitors start out electrically neutral and a charge of +q0 is then applied to the left plate of C1, as you have been showing, the following will happen very rapidly (on a time scale of perhaps nanoseconds):

the applied charge will spread itself as uniformly as possible over the left plate of C1;

the electric field generated by that plate will attract a charge -q0 to the right plate from the conductive materials making up the right plate of C1, the connecting wire, and the left plate of C2; the potential drop across C1 will arrive quickly at q0/C1 ;

the same electric field of C1 will repel a charge in the amount +q0 to the left plate of C2; this will happen because the connected plates and wire were initially electrically neutral;

the charge of the left plate of C2 will in turn produce an electric field which will attract a charge -q0 to the right plate of C2, causing the potential drop across C2 to rapidly reach q0/C2 .

It is not necessary for the switch to be closed, as this effect is not caused by a flow of charge carriers, but by the action of the electric field from the "excess" charge. (When you get to Maxwell's Laws, you'll run across something called "displacement current". Certain actions of electric fields across open space mimic those produced by a physical current.) What you now have resembles the condition of the circuit if it started out uncharged and was allowed to be charged by the battery up to the point where the charge on each capacitor was q0.

Earthing does not enter into things in this circuit. Nothing in your diagram shows any components connected to the Earth; instead, you are simply applying some charge to an isolated set of conductive materials "linked" by electric fields across gaps.

q = CV(1 - e^(-t/CR)) + qoe^(-t/CR) where, C = C1C2/(C1 + C2)

I get this also: what you have is the sum of a "transient term",
(qo) · e^(-t/CR) , which goes to zero "in the long term", and an "asymptotic term", CV(1 - e^(-t/CR)), which goes to CV "in the long term". Again, this shows that the charge on the two capacitors will end up the same; the voltage across each one will differ:

V1 = C·V/C1 and V2 = C·V/C2 , with V1 + V2 = V .

The transient term shows that, within reason, the initial charge q0 has no effect on the equilibrium situation. This can also be seen by going back to the differential equation

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

or solving for dq/dt,

dq/dt = (V/R) - (q/RC) , again with C being the "effective capacitance".

The equilbrium solution is found by setting dq/dt = 0, giving

(V/R) = (q/RC) , or q = CV (as expected).

In terms of the properties of this differential equation, if q < CV , then dq/dt > 0 , meaning that the charge will rise asymptotically from its initial value q0 to the equilibrium value CV. Similarly, if q > CV, then dq/dt < 0 , so the charge will fall asymptotically from q0 to CV. This will apply as long as q0 is not so big that the "breakdown voltage" of the capacitors is approached or exceeded.

Where does the excess charge q0 > CV go, if it doesn't matter in the end? The battery acts as a sort of reservoir and takes up excess energy associated with it. In the case of the battery charging the circuit (q0 < CV), the battery "discharged" and is the source of the energy to separate charges in the circuit. If there is excess charge (q0 > CV), the effect would be to "charge" the battery, supplying energy to it (much as what happens when other circuits send current toward the positive terminal of a battery); the reduction of potential energy in the circuit would also reduce the amount of separated charge.

All of this is somewhat idealized and is neglecting a certain amount of dissipation in the wires and the battery itself. I don't quite have the conservation of charge issue resolved in the case where q0 > CV . It seems like it ought to be possible to better describe what happens; I can't say I've ever seen that situation discussed. It may be that this model doesn't adequately apply in that case.

 

[i_island0]

i think it will be qo/2 and not q0

 

[dynamicsolo]

i think it will be qo/2 and not q0

What would make the charge divide in that fashion? I believe you will need to look somewhat further into the matter of how capacitors work.

 

[i_island0]

ok.. thx.. i will look into.. i give up already..

 

[Shooting Star]

If the capacitors start out electrically neutral and a charge of +q0 is then applied to the left plate of C1, as you have been showing, the following will happen very rapidly (on a time scale of perhaps nanoseconds):

I think the formulation of the problem by the OP is a bit different. The left capacitor C1 has previously been charged by a battery so that there is +q0 on the left and –q0 on the right plate. In this condition it is just connected to C2. No batteries, no resistors yet. What happens now?

Does the +q0 and –q0 remain on C1? The +q0 has nowhere to go. (The left plate of C2 has to be at the same potential as the right plate of C1.)

 

[i_island0]

yes.. that's what i have been saying all the time. how can they go..
Now, the battery along with the resistor is connected with the capacitors C1 and C2.. what will happen next. Do you still say that the final charges on the two capacitors going to be same.

 

[dynamicsolo]

I think the formulation of the problem by the OP is a bit different. The left capacitor C1 has previously been charged by a battery so that there is +q0 on the left and –q0 on the right plate. In this condition it is just connected to C2. No batteries, no resistors yet.

The charge on C1 doesn't have to go anywhere: its presence on the plates creates an electric field in the conductive materials that will attract and repel charges elsewhere, including across the gap in C2 (look at what "displacement current" is). With the capacitors arranged as shown in the diagrams and the switch placed where it is, charging C1 can't help but affect C2.

This still doesn't essentially change what I posted on the 26th about the final arrangement of the charges, in answer to the original question.

 

[Shooting Star]

I am rather intrigued now about only the following situation:

C1 has been charged. The right plate of C1, having -q0, is connected to one of the plates of an uncharged C2. There is nothing else. Let everything settle down, so that I don't have to worry about displacement currents. As a favour to me, can you tell me what exactly is the charge distribution now?

 

[dynamicsolo]

I am rather intrigued now about only the following situation:

C1 has been charged. The right plate of C1, having -q0, is connected to one of the plates of an uncharged C2. There is nothing else. Let everything settle down, so that I don't have to worry about displacement currents. As a favour to me, can you tell me what exactly is the charge distribution now?

All right, since we've decided to just look at this situation, let's get something else clarified: was the charge +q0 applied to the left plate of C1 only, or were equal and opposite charges applied to the plates of C1 initially? That will make a difference as to what happens with C2; I feel we've been talking round in circles without agreeing as to what the exact initial configuration is (although I still say that it is irrelevant to the question that was originally asked about the final configuration).

 

[i_island0]

charge qo was applied to the left plate of C1 and -qo was applied to the right plate of C1.
Thats the configuration.

 

[dynamicsolo]

charge qo was applied to the left plate of C1 and -qo was applied to the right plate of C1.
Thats the configuration.

And is the "final" configuration that your original question was asking about the one that exists before or after the switch is closed? (I've been talking about what happens after the switch is closed because that is usually what is asked for in problems such as these. But it sure seems that we haven't all been talking about the same thing...)

 

[Shooting Star]

I am rather intrigued now about only the following situation:

C1 has been charged. The right plate of C1, having -q0, is connected to one of the plates of an uncharged C2. There is nothing else. Let everything settle down, so that I don't have to worry about displacement currents. As a favour to me, can you tell me what exactly is the charge distribution now?

And is the "final" configuration that your original question was asking about the one that exists before or after the switch is closed? (I've been talking about what happens after the switch is closed because that is usually what is asked for in problems such as these. But it sure seems that we haven't all been talking about the same thing...)

No, no, we have been talking about the same thing all the while and the final configuration with the resistor included is not difficult to arrive at. This "new" problem suddenly struck me.

 

[Rainbow Child]

Let me offer I hand!

First of all for the "final configuration"$$^1$$ i_island0 is RIGHT, the charges are not the same!

For start let the maths talk.
The charge of every capacitor is symbolized by $$q_1(t),\,q_2(t)$$ and the current $$i=\frac{d\,q}{d\,t}$$ where $$q$$ is the charge "running" in the circuit. Thus at every instant we have

$$q_1(t)=q_o+q, \quad q_2(t)=q \quad (1)$$

since the left plate of $$C_1$$ has initially charge $$q_o$$ and the capacitor $$C_2$$ is initially uncharged. Thus the voltage at every instant for each capacitor is

$$V_1=\frac{q_1}{C_1}\Rightarrow V_1=\frac{q_o+q}{C_1},\, \quad V_2=\frac{q_2}{C_1}\Rightarrow V_2=\frac{q}{C_2}$$

Applying Kirchoff's 2nd law we have the ODE

$$q'(t)+\frac{q(t)}{\tau}=\frac{V}{R}-\frac{q_o}{\tau_1},\quad q(0)=0$$ with $$\left\{\begin{array}{l}\tau=R\,C \\ \tau_1=R\,C_1\end{array}, \quad \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$$

The solution of the above ODE is

$$q(t)=(C\,V-q_o\,\frac{\tau}{\tau_1})(1-e^{-t/\tau})$$

Now from (1) we have

$$q_1(t)=q_o+(C\,V-q_o\,\frac{\tau}{\tau_1})(1-e^{-t/\tau}),\quad q_2(t)=(C\,V-q_o\,\frac{\tau}{\tau_1})(1-e^{-t/\tau})$$

Thus for

$$t=0\Rightarrow q_1(0)=q_o,\,q_2(0)=0$$ initial conditions and for

$$t\rightarrow \infty \Rightarrow q_1(\infty)=C\,V+q_o\,\frac{C}{C_2},\, q_2(\infty)=C\,V-q_o\,\frac{C}{C_1}$$

meaning that the final charges are different.

Now let physics talk!

To make things easier let's have one the capacitor $$C_1$$ with initial charge $$q_o$$.

If $$q_o=C_1\,V$$ what would have happend? Nothing at all! There would be no current and the charge would stay $$q_1(t)=q_o$$

If $$q_o=\frac{1}{2}\,C_1\,V$$ what then? In this case the current would terminate it's existence when the two voltages become equal. The final charge would be $$q_1(t)=C_1\,V$$, meaning that we had a "flow" of charge $$q=\frac{1}{2}\,C_1\,V$$

If for the problem at hand we have $$q_o=C_1\,V$$, then nothing is happening. The initial charges are the final ones!

Thus, never mix the charge of the capacitor and the charge "running" in the circuit

$$\hline$$

$$^1$$ To set the problem in the correct basis, we must say that we do two things simultaneously. We close the switch S and connect the two capacitors at the same time. If not we have the "new" problem. It needs a post of it's own

 

[i_island0]

Applying Kirchoff's 2nd law we have the ODE

$$q'(t)+\frac{q(t)}{\tau}=\frac{V}{R}-\frac{q_o}{\tau},\quad q(0)=0$$ with $$\tau=R\,C, \quad \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$$

:

don't you think that the differential equation would be:
(dq/dt) R + (q + qo)/C1 + q/C2 - V = 0

 

[dynamicsolo]

For start let the maths talk.
The charge of every capacitor is symbolized by $$q_1(t),\,q_2(t)$$ and the current $$i=\frac{d\,q}{d\,t}$$ where $$q$$ is the charge "running" in the circuit.
...Thus at every instant we have

$$q_1(t)=q_o+q, \quad q_2(t)=q \quad (1)$$
...

Thus, never mix the charge of the capacitor and the charge "running" in the circuit

I suppose if you start with such a premise, you can arrive at such a conclusion. But there still seems to be resistance to answering a question I already raised: why would a charge placed on one capacitor, made of conductive material and directly connected by conductive wire to a second capacitor, also made of conductive material, have no influence on the charges in the second capacitor? How could applying charge to C1 not produce charge separation in C2 when the two capacitors are connected even before the switch is closed? That situation is unstable and C2 would not be starting out uncharged before the switch is closed.

But this still doesn't matter to the charge configuration after the switch is closed and the capacitors reach equilibrium with the battery. There still seems to be an insistence that somehow the applied charge q0 "knows" it is applied charge and doesn't count toward what is happening on the plates in the capacitors. What is the physical explanation for counting it separately from whatever else is happening on the plates in the two capacitors? This description of the capacitor charges

$$q_1(t)=q_o+q, \quad q_2(t)=q \quad$$

treats q0 as if it somehow has nothing to do with what is otherwise happening in the circuit, so it is not surprising that it just rides out the charging of the capacitors and so naturally ends up in the mathematical solution. (Of course the charges on the two capacitors end up different if you claim that.) C1 seems to have been permanently set to have a charge q0 greater than the charge on C2 and nothing appears able to change that, which seems to have no physical justification...

 

[i_island0]

i checked the equilibrium condition using Gauss law.. and according to my calculation it is in equilibrium already, do you mind checking that too.

 

[Shooting Star]

But there still seems to be resistance to answering a question I already raised: why would a charge placed on one capacitor, made of conductive material and directly connected by conductive wire to a second capacitor, also made of conductive material, have no influence on the charges in the second capacitor? How could applying charge to C1 not produce charge separation in C2 when the two capacitors are connected even before the switch is closed? That situation is unstable and C2 would not be starting out uncharged before the switch is closed.

Hi dynamicsolo,

This was the situation I was interested in. I wanted to know what charges on each capacitor would be before the switch was closed. In fact, I just want to know what happens when C1 has q0 and -q0, and is connected to C2 which is uncharged. This is different from the problem that the OP had posted.

 

[Rainbow Child]

I suppose if you start with such a premise, you can arrive at such a conclusion. But there still seems to be resistance to answering a question I already raised: why would a charge placed on one capacitor, made of conductive material and directly connected by conductive wire to a second capacitor, also made of conductive material, have no influence on the charges in the second capacitor? How could applying charge to C1 not produce charge separation in C2 when the two capacitors are connected even before the switch is closed? That situation is unstable and C2 would not be starting out uncharged before the switch is closed.

Due to the length of my post you may not have noticed:

...
$$\hline$$

$$^1$$ To set the problem in the correct basis, we must say that we do two things simultaneously. We close the switch S and connect the two capacitors at the same time. If not we have the "new" problem. It needs a post of it's own

But this still doesn't matter to the charge configuration after the switch is closed and the capacitors reach equilibrium with the battery. There still seems to be an insistence that somehow the applied charge q0 "knows" it is applied charge and doesn't count toward what is happening on the plates in the capacitors. What is the physical explanation for counting it separately from whatever else is happening on the plates in the two capacitors? ...

The physical expanation is quite simple. From the definition of the capacity and the definition of the current.

1. Capacity: $$C=\frac{q}{V}$$ The charge $$q$$ is the absolute value of the charge "sitting" on of the plates of the capacitor.
2. Current: $$i=\frac{d\,q}{d\,t}$$ The charge $$q$$ represents the charge $$d\,q$$, passing through the cross-section of the wire at the time interval $$d\,t$$.

This description of the capacitor charges

$$q_1(t)=q_o+q, \quad q_2(t)=q \quad$$

treats q0 as if it somehow has nothing to do with what is otherwise happening in the circuit, so it is not surprising that it just rides out the charging of the capacitors and so naturally ends up in the mathematical solution. (Of course the charges on the two capacitors end up different if you claim that.) C1 seems to have been permanently set to have a charge q0 greater than the charge on C2 and nothing appears able to change that, which seems to have no physical justification...

Who said that $$q_o$$ is permanent ?
To make things more clear in order to explain the physical interpretation of $$q_o$$ consider the following example.

Consider a capacitor, with capacity $$C$$ whih is connected through a resistance $$R$$ with an ideal battery of voltage $$V$$ and let it there until it is full charged. The final charge on the capacitor is $$q_o=C\,V$$.
Now disconnect the capacitor and set up a new circuit with the same elements, but now connect the negative plate (say A) with the positive pole of the battery, and the positive plate (say B)with the negative pole of the battery. Which is now the function $$Q_A(t)$$ describing the charge on the capacitor's plate A?
I think you agree that firstly the capacitor will discharged and then will be charged up to voltage $$V$$. Correct?
If you do agree, that means that the final charge on the plate A will be $$Q_A(\infty)=C\,V$$, while the final (=total) charge moved by the battery is $$q(\infty)=2\,C\,V$$. Correct?
Now if you don't distinguish between the charge $$q,\,q(0)=0$$ running through the circuit and the chagre $$Q_A(t),\,Q_A(0)=-q_o=-C\,V$$ you will run into problems.

The charge $$q$$ is given by

$$q(t)=2\,C\,V\,(1-e^{-t/\tau}), \tau=R\,C$$

while the charge $$Q_A$$ on the plate is

$$Q_A(t)=C\,V\,(1-2\,e^{-t/\tau}), \tau=R\,C$$

Does this clears out the situation?

 

[i_island0]

ok.. your explanation works well with what i have read