Electric potential and charge on capacitors-

[Kalibasa]

Homework Statement

Consider the circuit in the figure below, in which C2 = 16 µF and V = 80 V. Initially, the switch in is in position A and capacitors C2 and C3 are uncharged. Then the switch is flipped to position B. Afterward, what are the charge on and the potential difference across each capacitor?

http://www.webassign.net/knight/p30-70alt.gif

Homework Equations

Q= C * delta V where C is the capacitance in farads and Q is the charge on each capacitor plate (the charge on one plate is equal and opposite the charge on the other)
Equivalent capacitance of capacitors in series: (1/C1 + 1/C2)^-1
Equivalent capacitance of capacitors in parallel: C1 + C2

The Attempt at a Solution

I figured that the capacitors in series should have the same charge, so Q2=Q3. I also figured that the first capacitor should have the same potential difference as the other two capacitors put together, or deltaV1= deltaV2+deltaV3. I finally figured that the sum of the final charges on all the capacitors should equal the initial charge, which I found to be Qi= C1deltaVi = 15 microfarads * 80 V= 0.0012 C.

Is this all right? If it is, even with these equations I haven't been able to solve this problem. I've tried endless rearrangements of the equations above and I always still have a variable I don't know.

I don't know if it will come into play, but I solved for the equivalent capacitance of C2 and C3 and got 10.438 microfarads.

Please give me a detailed answer if you can. I'd appreciate any help, but really short answers are probably not going to help me at this point, given that I've already spent over two hours on this problem!

 

[collinsmark]

I figured that the capacitors in series should have the same charge, so Q2=Q3.

Very good. That gives you one equation.

I also figured that the first capacitor should have the same potential difference as the other two capacitors put together, or deltaV1= deltaV2+deltaV3.

Very nice! That's important and it gives you a second equation, ΔV₁ = ΔV₂ + ΔV₃

Now, noting that the voltage across the terminals of a given capacitor is ΔV = Q/C, try reformulating your second equation to be in terms of C₁, C₂, C₃, Q₁, Q₂, and Q₃.

I finally figured that the sum of the final charges on all the capacitors should equal the initial charge, which I found to be Qi= C1deltaVi = 15 microfarads * 80 V= 0.0012 C.

Excellent. Now you have a third equation:

0.0012 [C] = Q₁ + Q₂ + Q₃. [Edit: Corrected mistake of using 'C' instead of 'Q'. Ooops. ]

So now you have 3 independent equations, and 3 unknowns (the unknowns are Q₁, Q₂, and Q₃). You should be able to solve for the unknowns.

 

[Kalibasa]

For the last equation, did you mean Q1 + Q2 +Q3= 0.0012? Or do you really mean the sum of the capacitances?

 

[collinsmark]

For the last equation, did you mean Q1 + Q2 +Q3= 0.0012? Or do you really mean the sum of the capacitances?

Ooops. Yes, Sorry about that. Serves me right for trying to reply too fast. 'Made correction in above post.

 

[Kalibasa]

I've mixed up C for capacitance and C for coulomb many times so far ;)

Thanks for the help, but I am still getting the wrong answers! And I did assume that you meant Q1+Q2+Q3= .0012. Since your post I tried two things. I tried reworking the second equation as you said to Q1/C1= Q2/C2 + Q3/C3. I set Q3 equal to Q2, and I set Q1 equal to 0.0012-2Q2. I tried solving for Q2, but it involved a lot of crazy algebra and then I got the wrong answer. I'm not sure if it's because of a math mistake, which is totally possible, or if that isn't the way to solve the problem.

The other thing I tried is reducing the two capacitors on the right to an equivalent capacitor. I used the equivalent capacitance I found earlier, I set deltaV of the equivalent capacitor equal to Q1/C1 (deltaVeq= deltaV2+deltaV3= deltaV1= Q1/C1), and I set the equivalent charge to what I'd found for Q2, (0.0012-Q1)/2. Still didn't work.

This problem shouldn't be that hard either, or at least not according to the difficulty level marked in the book. I can't help but think there's a much easier way to do this...

 

[collinsmark]

Thanks for the help, but I am still getting the wrong answers! And I did assume that you meant Q1+Q2+Q3= .0012. Since your post I tried two things. I tried reworking the second equation as you said to Q1/C1= Q2/C2 + Q3/C3. I set Q3 equal to Q2, and I set Q1 equal to 0.0012-2Q2. I tried solving for Q2, but it involved a lot of crazy algebra and then I got the wrong answer. I'm not sure if it's because of a math mistake, which is totally possible, or if that isn't the way to solve the problem.

Sounds like you are setting stuff up correctly. Based on what you've said above, your three original equations are:

...Q₂ = Q₃,

...Q₁/C₁ = Q₂/C₂ + Q₃/C₃,

...0.0012 [C] = Q₁ + Q₂ + Q₃.

Substituting the first equation into the other two gives,

...Q₁/C₁ = Q₂/C₂ + Q₂/C₃,

...0.0012 [C] = Q₁ + 2Q₂.

Now we are down to 2 equations and 2 unknowns.

Solving for Q₁ in the last equation and substituting that into the one before it gives,

...(0.0012 [C] - 2Q₂)/C₁ = Q₂/C₂ + Q₂/C₃,

which is 1 equation and 1 unknown.

Is that how you set things up? If so, solve for Q₂. If you show your work, I might be able to help out if there is a math mistake.

 

[Kalibasa]

I did exactly what you listed up there. Starting with the last equation you lsited, I multiplied both sides of the equation by C2 and by C3 in order to get a like denominator. This gave me [C3*C2*(0.0012-2Q2)]/C1=C3Q2 + Q2C2. Then I plugged in numbers for all the capacitances:

[30 * 16 * (0.0012 - 2Q2)]/15= Q2 * (30+16)
32 microfarads * [.0012-2Q2]=46 microfarads * Q2.
0.0012 - 2Q2= 1.4375Q2
0.0012= 3.4375 Q2
Q2=0.0003491 C

And apparently that's wrong...?

 

[collinsmark]

I did exactly what you listed up there. Starting with the last equation you lsited, I multiplied both sides of the equation by C2 and by C3 in order to get a like denominator. This gave me [C3*C2*(0.0012-2Q2)]/C1=C3Q2 + Q2C2. Then I plugged in numbers for all the capacitances:

[30 * 16 * (0.0012 - 2Q2)]/15= Q2 * (30+16)
32 microfarads * [.0012-2Q2]=46 microfarads * Q2.
0.0012 - 2Q2= 1.4375Q2
0.0012= 3.4375 Q2
Q2=0.0003491 C

And apparently that's wrong...?

According to my calculations, your Q₂ value is right! Ya!

Now you should be able to calculate Q₁ and Q₃ easily enough. But the problem statement is asking for the 'ΔV's across each capacitor, not the 'Q's. So you have to use the ΔV = Q/C equation before you get to the final answers.

 

[Kalibasa]

It's asking for the charges and the deltaV's on each one, though- I'm doing an online homework and there are six boxes to fill in. It's telling me that that value for Q2 is wrong. So I don't know.

Maybe I'll try putting in a deltaV I've calculated off of that and see, but I don't know what could be wrong

 

[collinsmark]

It's asking for the charges and the deltaV's on each one, though- I'm doing an online homework and there are six boxes to fill in. It's telling me that that value for Q2 is wrong. So I don't know.

Maybe I'll try putting in a deltaV I've calculated off of that and see, but I don't know what could be wrong

Sometimes these online homework programs can be very touchy about significant digits. Also, sometimes when units are involved they might barf it you're not careful (e.g. entering in a value in Coulombs, but it wants you to enter the answer in units of μC). Could it be related to something like that?

If you wish to post your values here for the other 'Q's and 'ΔV's, I can compare them with mine if you would like. Alternately, substitute your final values back into your original equations, and check to see that they make sense.

 

[Kalibasa]

So I finally gave up and got the answer from someone else, and I'll post it here since you'll probably be curious now! The answer we got for Q2 was clearly wrong (these values all worked in my online homework, so they must be right). But I have no idea why our method didn't work- I don't see anything inherently wrong with it. I think I have a long date with the TA tomorrow... :)

Regardless, thanks for all your trouble! You didn't abandon me on this god-awful problem, I appreciate it a ton. :)

Answer:

This is really very simple. here are the keys:
1. charge is conserved
2. rules of caps in parallel and series
3. caps in series have the same charge
4. Q = CV

first of all C2 and C3 are equivalent to a 16*30/46 = 10.4µF

initial, Q = CV = 80*15 = 1200 µC

final, the capacitance is 15 + 10.4 = 25.4µF
charge is conserved, so Q = CV, V = Q/C = 1200/25.4 = 47.2 volts (voltage on C1)
Charge on C1 = CV = 47.2 x 15 = 709µC
Voltage on C2/C2 combo is 47.2 volts and charge is 47.2 x 10.4 = 491µC
That charge is on both C2 and C3 and the voltage on each is:
V3 = Q/C = 491/30 = 16.4 volts
V2 = 491 / 16 = 30.7 volts

 

[collinsmark]

I think I see where I went wrong. Earlier I had,

...Q₂ = Q₃,

...Q₁/C₁ = Q₂/C₂ + Q₃/C₃,

...0.0012 [C] = Q₁ + Q₂ + Q₃

But when capacitors are in series, the charge on each is the total charge that enters. In other words, if you charge up a single uncharged capacitor at 1 Amp for 1 second, that capacitor will store 1 Coulomb. But if 2 uncharged capacitors are in series, and you charge them at 1 Amp for 1 second, they each store 1 Coulomb.

My third equation should be modified to be:

0.0012 [C] = Q₁ + Q₂

(We could have picked either Q₂ or Q₃, but not the sum of both).

Sorry about that. We all make mistakes, including myself.