Homework Help: Capacitance, and a series

1. Feb 2, 2010

EV33

1. The problem statement, all variables and given/known data

My question is mainly just conceptual not an actual homework problem.

So lets say there are two capacitors in a series. Their C equiv= (1/c1 + 1/c2)

So to find the charge on either c1 or c2, you would do C equiv*V=Q.

Why is it that Q is the charge for each of them seperately,and not as a whole.

2. Relevant equations

CV=Q

3. The attempt at a solution

This doesn't make sense to me because When I think of C equiv I think of imagining the two capacitors as one, which is what I thought I was supposed to do. But if you imagine they are one capacitor, and wanted to find the charge on this whole imaginary capacitor wouldn't you use the same equation with the same values, and therefore get the same answer which should actually be double what it is?

2. Feb 2, 2010

EV33

Do two Capacitors in a series have the same capacity?

3. Feb 2, 2010

Staff: Mentor

First, you wrote: C equiv= (1/c1 + 1/c2)

But it should be: C equiv= 1/(1/c1 + 1/c2)

And Q=CV works for the series combination.

1 cap value C: Q=CV

2 caps in series, C-->C/2, and there is V/2 across each C, so Q/2 on each cap. Yes, the total charge separation seems like it should be 2*C/2, but remember that the two plates that are connected together between the caps would normally have had opposite charge, which cancels out to zero when you connect them, so that leaves you with a net of Q/2 across the series pair.

Does that make sense?

4. Feb 2, 2010

EV33

I'm not too sure what you are saying.

I see that V/2 is the V on each one, because v1+v2=V total.

I am not too sure what you mean by "C-->C/2". Which C's are you talking about?

5. Feb 2, 2010

Staff: Mentor

I was giving an example where each of the two series caps have the same value C. Then the total series combination is C/2.

6. Feb 2, 2010

EV33

I figured out what you were saying right after I posted that. My confusion comes a homework problem.

A 11.2 µF capacitor and a 17.6 µF capacitor are connected in series across the terminals of a 6.00-V battery.

Then it asks me to find C equiv which is 6.844 uF.

Then it asks,"Find the charge on each capacitor", all I have to do to get the correct answer (which is 41.064 uC) for each one is multiply Cequiv and 6V. But if the v is not equal for both Capacitors then shouldn't one capacitor be multiplied by a different voltage than the other in order to get the charge for one of them?

That is the impression I am getting at least. So would that mean that homework problem is wrong.

7. Feb 2, 2010

Staff: Mentor

Not necessarily. There can be another way to get the right answer.

You are correct, that the voltage across unequal capacitors in series will be unequal. It's kind of artificial to talk about a DC situation, though, because caps are opens for DC, so there is no way for current to flow to cause a DC voltage across caps.

So the right way to think about it is that they were initially at 0V across them, and were charged up to some DC value. During the charging, the voltage across them increased from 0V to V, and there was a current flowing through the serues combination to cause charge separattions. To get an intuitive feel for which cap gets a bigger voltage across it in the end, consider the case when the two values are C and 10C. The impedance of the C cap 10x the impedance of the 10C cap, right? So which cap develops a larger voltage across it -- the low impedance one, or the higher impedance one?

8. Feb 2, 2010

EV33

I haven't learned about impedance yet, but my guess would be that high impedience one because the higher your voltage the lower you capacitance but the equation C=Q/V?

How does this show that the charge and the voltage between c1=c2=c equiv? I am not making the connection.

9. Feb 2, 2010

vela

Staff Emeritus
You can do that, and you'll get the same answer either way. The voltages will work out so that Q=CeqVtotal=C1V1=C2V2.

Because of the gap between the plates of a capacitor, there's no way for charge to physically get from one side to the other. Now consider two uncharged capacitors in series. One section of the circuit consists of the plate of one capacitor connected by a wire to the plate of the second capacitor, and this section is cutoff from the rest of the circuit by the gaps in the two capacitors, which means it can't lose or gain charge as a whole. That means if one capacitor has -Q on one plate, the other capacitor has to have +Q on its plate so that the total charge remains zero. So if one capacitor is charged to Q, the other one will be as well.

So how does the first capacitor in series acquire charge Q? That comes from the current flowing into it from the rest of the circuit. Now the rest of the circuit doesn't know that there's actually two capacitors in series; it just sees a capacitance Ceq. So the amount of charge delivered Ceq is the same amount of charge dumped into the first capacitor, which is the same amount of charge on all the capacitors in series.