# Homework Help: Potential Difference of a Capacitor System

1. Mar 3, 2012

### forestmine

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

The capacitors in the figure are initially uncharged and are connected, as in the diagram, with switch S open. The applied potential difference is V_ab = 210V.

a) What is the potential difference V_cd?
b) What is the potential difference across each capacitor after switch S is closed?
c) How much charge flowed through the switch when it was closed?

2. Relevant equations

1/C_eq = 1/C1 + 1/C2

C_eq = Q_total/V

3. The attempt at a solution

So, as a side note, I'm calling the top left capacitor C_1. Top right is C_2. Bottom left is C_3, and bottom right is C_4.

Here's what I've got so far...

With the switch open, taking C_1 and C_2 to be a series of capacitors, and C_3 and C_4 to be a separate series, I calculated C_eq for both (getting the same value, of course) and adding them to get C_total, 4*10^-6 F.

From there, using Q_total = C_eq*V, I solved for Q, getting 8.4 *10^-4. So if I'm understanding correctly, this Q is as a result of all 4 capacitors and the given potential of 210 V. That being said, I would think my next step would be to divide that value by 4 (for each capacitor) to obtain the charge of each individual capacitor, but this turns out to be wrong, and I don't understand why.

Once I have the charge for each capacitor, I should be able to calculate V_cd by considering C_1 and C_3 to be part of a parallel system, and use V=Q/C, where Q is the charge of either capacitor (they should be the same in this case, I think?) and C is either C_1 or C_3.

At this point, I'm stuck on the the charge Q of each capacitor. If anyone could help clear that up for me, I would really appreciate it.

Thank you!

2. Mar 4, 2012

### Staff: Mentor

Instead of calculating the full C_eq for the circuit, calculate the individual C_eq's for the two series strings. Each string sees the 210V potential. What charge gets pushed onto each string?

3. Mar 4, 2012

### forestmine

Ok, so C_eq for either of the two series sets would be 2*10^-6. Using Q=CV, the charge for each set would be 4.2*10^-4, and since in a series combination, that Q is the same for all capacitors, that means each of the four capacitors has a charge of 4.2*10^-4 C on one conductor, and -4.2*10^-4 C on the other conductor?

Thank you! This definitely cleared that bit up.

Now for finding the potential for cd, I would think to use V=Q/C_1 for V_ad and V=Q/C_3 for V_ac and then find the difference which works out to be 70V.

For part b, once the switch is closed, we're dealing with two parallel combinations (C_1 and C_3, and C_2 and C_4). This means that the potential V_dc is the same. But at that point, I'm not really sure what to do next? Now that they're parallel, the charges shouldn't be the same across each capacitor, right?

4. Mar 4, 2012

### Staff: Mentor

Looks reasonable.
Make your life easy; By symmetry, what should be the potential of points c and d when the switch is closed and the dust settles?

5. Mar 4, 2012

### forestmine

Hm, I'm not sure. I know they should be the same, right? 70V at point c and at point d? But that doesn't really take into account the switch being closed...not really sure how to go about this.

6. Mar 4, 2012

### Staff: Mentor

When the switch is closed, conceptually the node cd is between two equal components...

7. Mar 4, 2012

### forestmine

That much makes sense, but I'm still not really sure what to do next.

When the switch closes, the total charge of the system should be greater, right? And therefore the total capacitance should also change?

If so, we have C_eq = C_1 + C_3 = 9*10^-6 and C_eq = C_2 + C_4 = 9*10^-6.

We know the potential along V_cd is 70V, but if the capacitance is changing, the ratio of Q/V should change accordingly, so there ought to be a new charge AND a new potential, both of which are unknown variables at this point...ah, I don't know...

8. Mar 4, 2012

### Staff: Mentor

If CD is halfway, shouldn't it be at half the potential between a and b?

9. Mar 4, 2012

### forestmine

Ok, in that case, then V_ac = Q/C_3 = 70V and V_cb= Q/C_4 = 140V, so V_ab = 70V, and V_cd = 35 V??

I feel like I'm missing some concept entirely. I don't completely understand what's going on at the point at which the switch is closed, other than I think the charges should change, and therefore the potentials.

10. Mar 4, 2012

### Staff: Mentor

With the switch closed the circuit is symmetrical. The total potential across it is 210 V. Therefore the halfway potential is 210 V/2 = 105 V. The charges on the capacitors must adjust themselves to reflect this new reality. But life is good, because you've now got the potential difference for each of them.

11. Mar 4, 2012

### forestmine

Ah jeez, I really was making this much harder than it ought to be. Ok, so, to say that the potential across each capacitor is 105 V, does that essentially mean that if we examine each capacitor by dividing the image into four squares, then the potential V_ad involving capacitor 1 is 105V, and the potential V_db involving capacitor 2 is 105 V, etc?

12. Mar 4, 2012

### Staff: Mentor

Yup.

13. Mar 4, 2012

### forestmine

Whew, ok, almost there. Thanks for all the help, by the way!

So, for calculating how much has charge has flowed through the system when the switch is closed, is it correct to think of C_1 and C_2 as a series, C_3 and C_4 as another series, C_1 and C_3 as parallel, and C_2 and C_4 as parallel? In which case, C_1 and C_2 should have the same charge, and C_3 and C_4 should have the same charge? If so, we can find C_eq for C_1 and C_2. 1/C_eq = 1/C_1 + 1/C_2, solving for C_eq = 2*10^-6, and then using the fact that V_ab = 210, the charge comes out to be 4.2*10^-4, but that puts me right back at the beginning, which can't be right, since we determined the charge should be greater now.

14. Mar 4, 2012

### Staff: Mentor

C1 and C2 cannot be considered to be in series when the switch is closed, since there's other stuff connected at their shared node. Same goes for C3 and C4.

I'll have to think about whether there's an elegant method to determine the total charge that flows through the switch after it closes.

Last edited: Mar 4, 2012
15. Mar 4, 2012

### Staff: Mentor

Okay, here's an approach. Find the Thevenin equivalent of the circuit looking into the open terminals of the switch. You already know the open circuit voltage there from a previous step. The 210V supply gets suppressed when you look for the impedance (which will be an equivalent capacitance) which should greatly simplify the procedure.

16. Mar 4, 2012

### forestmine

I'm not yet familiar with the Thevenin equivalent, and so something tells me I don't need to use it for this scenario. I think it's simply a matter of applying Q=VC correctly, but I can't seem to use it in a way such that I get the correct answer.

The solution manual suggests finding C_eq (like you said) when the switch is closed, but they show it as being equal to (1/C_1 + C_2)^-1 + (1/C_2 + C_3)^-1. If we're dealing with two parallel combinations, why is that equation applicable? I thought C_eq for parallel combinations is simply C_1 + C_2...etc. My C_eq in that case turns out to be exactly 2 times that of the one the solution suggests.

I would have thought it'd be simply a matter of finding C_eq for each half (from a to b), and then given that the voltage at V_cd is 105, use Q=CV, which gives me 4.725*10^-4, but this isn't correct at all...

17. Mar 4, 2012

### Staff: Mentor

Are you sure the formula isn't (1/(C_1 + C_2) + 1/(C_3 + C_4))^-1 ?

That would be the Thevenin equivalent capacitance as seen from the switch terminals.
The problem is, you want to find the current that flows through the switch from the state that the circuit was in while the switch was open, to the state some long time after the switch has been closed. So you won't have just one circuit arrangement to deal with. The effective circuit arrangement changes when the switch closes.

This issue can be eliminated as an issue by replacing the circuit with an equivalent circuit model. This is the Thevenin equivalent. You end up with a single voltage source in series with a single capacitance (and the switch). This equivalent circuit will behave in all ways like the more complex circuit it models.