Capacitors in a circuit with switches

[horsedeg]

Homework Statement

Consider the circuit shown in the figure below, where C1 = 4.00 µF, C2 = 7.00 µF, and ΔV = 22.0 V. Capacitor C1 is first charged by closing switch S1. Switch S1 is then opened, and the charged capacitor is connected to the uncharged capacitor by closing S2.

(a) Calculate the initial charge acquired by C₁.
(b) Calculate the final charge on each capacitor.

Homework Equations

Q=CV

The Attempt at a Solution

The answer to (a) was easy, though I'm not 100% sure my reasoning is accurate. My reasoning was that voltage across parallel devices is equal, so I just plugged in the voltage on the left device into Q=CV.

I don't really understand (b) though. The solution says "When S₁ is opened and S₂ is closed, the total charge will remain constant and be shared by the two capacitors." I'm assuming if something opens then it cuts off the connection and if it closes then it creates the connection. I still don't get it though. There must be some concept I'm missing.

 

[BvU]

Your reasoning in case (a) is impeccable. You agree that then opening S1 does not change anything ?
The same reasoning for case (b) "voltage across parallel devices is equal" can be expressed twice using your relevant equation (for C1 and for C2, with the same V -- which is not the battery voltage, though).

Can you describe what you think is happening when C1 is charged, S1 is open and then S2 is closed ?

 

[CWatters]

The solution says "When S1 is opened and S2 is closed, the total charge will remain constant and be shared by the two capacitors." I'm assuming if something opens then it cuts off the connection and if it closes then it creates the connection.

If you connect your phone to it's charger the battery will eventually become charged. When you disconnect it does all the charge you put into the battery instantly disappear or...

 

[horsedeg]

Can you describe what you think is happening when C1 is charged, S1 is open and then S2 is closed ?

Now that I think about it, of S2 is closed then C1 and C2 would be in parallel, meaning they add to make a single capacitor right?

 

[BvU]

Correct. And is the situation that C1 is charged while C2 is uncharged a stable situation, or will there be something happening ?

 

[epenguin]

I think your reasoning for answer a is not really right, though the answer is. There is no 'voltage across' C₂ when S₂ is open. There is no across S₂.

There is initially when S₁ is closed a voltage across only S₁, so that's the only one you need to consider, making your result right.

When you ask about missing something, maybe it is that opening S₁ isolates the acquired charge; more precisely it isolates the negative charge on the lower plate, and that holds in position the positive one on the upper. After S₂ Is closed, charge from both C₁ plates can redistribute itself onto C₂. It is unnecessary to calculate a voltage, but whatever it is, because of the conductive connections it is the same for both capacitors, and so this charge will redistribute in proportion to the capacitances.

(I think your instinct may be right if you feel uncomfortable and there is something missing. Just breaking - a circuit with no resistance and isolating a charge gave me a feeling of idealisations that could break down , maybe it shouldn't.)