Circuit with capacitors and switch

[Chetlin]

Homework Statement

This is the circuit diagram I am working with:

There is actually a switch under C₃ that I didn't draw. Initially, the switch was set so that the red dashed line was part of the circuit and the green dashed line was not (so C₁, C₂, and C₃ charged but C₄ did not). Then after the capacitors were fully-charged, the switch was flipped to the pencil-drawn circuit there.

I have to determine the charge on C₄ in this new configuration after the switch was flipped.

Here are the values given:
ε = 100 V
C₁ = 3 μF
C₂ = 6 μF
C₃ = 4 μF
C₄ = 5 μF

Homework Equations

[tex]C = \frac{Q}{V}[/tex]
All capacitors in series have the same charge on them, if the voltage is held fixed. (That is the condition, right?)
Charge on conductors will spread so that the surface of the conductor is an equipotential.

The Attempt at a Solution

If I did it correctly, then before the switch was flipped, C₁ and C₂ each had 200 μC of charge and C₃ had 400 μC of charge. At first I just tried distributing that 400 μC across both C₃ and C₄ (200 μC each) but that doesn't work. Looking at it later, I see that those two capacitors are still connected to the rest of the circuit, which includes two more capacitors as well as a battery. So it appears to me that a lot of complicated stuff is going on here—charges can still flow throughout the entire thing since it's all still connected (unless, maybe, since capacitors aren't actually touching, there are a few segments of conductor in this circuit and each one can have its own charge, but I don't think that's possible because the two sides of a capacitor must have equal charges), and two of the capacitors are still in a circuit with a battery.. But it really can't be this bad, because this was a former exam question so they can't make it too complicated.

What parts of the circuit should I really be looking at? Where can charges flow and where can't they? Will all capacitors have the same charge? I don't think they will.

 

[BvU]

Nah, charge on the bottom plates of C3 and C4 can't go anywhere, so the 'being connected to the outside world' has no consequences.

[edit] 200 μC each is not good as a starting point. Same voltage over C3 as over C4 would be a lot more sensible...

 

[rude man]

So to simplify, C3 is charged to E and then C3 is connected to C4 only. So C1 and C2 play what part in all this?
So C3 is charged, then how does charge distribute between C3 and C4 when V is the same for both?

 

[Chetlin]

[edit] 200 μC each is not good as a starting point. Same voltage over C3 as over C4 would be a lot more sensible...

Thanks, that helped a lot! :D I got the right answer.
I set the voltages of the two capacitors equal to each other: V₃ = V₄ so [itex]\frac{Q_3}{C_3} = \frac{Q_4}{C_4}[/itex].
Then, I also saw that the initial charge on C₃ was 400 μC and C₄ was uncharged. This would distribute over the two capacitors, so in the end, Q₃ + Q₄ = 400 μC.
Using these two equations you can solve to get Q₃ = 178 μC and Q₄ = 222 μC which is the correct answer.

So C1 and C2 play what part in all this?

There were multiple questions about this setup, one of which was to find the ratios of charges between the charges on capacitors C₁, C₂, and C₃ when the switch was set up so that C3 was being charged. I kept them in this drawing since I didn't know if they still would have an effect after the switch was flipped.

 

[rude man]

Thanks, that helped a lot! :D I got the right answer.
I set the voltages of the two capacitors equal to each other: V₃ = V₄ so [itex]\frac{Q_3}{C_3} = \frac{Q_4}{C_4}[/itex].
Then, I also saw that the initial charge on C₃ was 400 μC and C₄ was uncharged. This would distribute over the two capacitors, so in the end, Q₃ + Q₄ = 400 μC.
Using these two equations you can solve to get Q₃ = 178 μC and Q₄ = 222 μC which is the correct answer.


There were multiple questions about this setup, one of which was to find the ratios of charges between the charges on capacitors C₁, C₂, and C₃ when the switch was set up so that C3 was being charged. I kept them in this drawing since I didn't know if they still would have an effect after the switch was flipped.

All's well that ends well!

 

[BvU]

Good work Chet!