Calculating Charge in a Circuit with Multiple Capacitors

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To determine the charge (q) at a certain point in a circuit with multiple capacitors, use the equation q = CV, where C is capacitance and V is voltage across the capacitor. The discussion highlights that while capacitors allow current to flow, they limit the rate of voltage change across their terminals. When measuring charge along a circuit, it’s essential to consider the electric potential and capacitance of each individual capacitor. The charge distribution on a capacitor remains uneven when disconnected from a battery, unlike a wire that quickly reaches equipotential. Understanding these principles is crucial for accurately calculating charge in circuits with capacitors.
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Homework Statement


So this is just a general question. If I have multiple capacitors, how can I determine the number of electrons at a certain point along the circuit. I know q=ne so I just really need q but that's the biggie. I also know q=CV. So to figure out q at any point, what do I need to do? Is it the path an electron would take from the negative side of the battery? Also say the electron goes through multiple capacitors, how do I find q along the circuit line?


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The Attempt at a Solution

 
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Your question seems kind of ambiguous to me. My understanding of a capacitor is that it vastly limits (ideally completely) the flow of current so that the charge builds up at the plates. Are you asking for any arbitrary point on a wire or do you mean charge on one of the individual capacitors in a series of capacitors?
In the case of the latter you should be able to look at the individual capacitance and electric potential of that capacitor to know the charge of that capacitor.
In the case of a wire and capacitor I know we usually neglect resistance of the wire but that is obviously not reality so you'd consider charge along the wire as well and could determine the amount based on the electric potential.
I'm not sure how complex of an answer you want because the truth is at a certain point I don't know.
 
Capacitors do not limit the flow of current at all. They flow current just fine. They limit the rate at which the voltage across their terminals can change.

To count electrons in a capacitor, you have all the equations above. Just put a voltmeter across each cap and use q=cv to get the charge on each one. The rest is counting and adding.

Edit: also, ask yourself if this could be a trick question.

Edit edit. Ok, it's not a trick question but I did give a trick answer. Remember this; in a circuit charge cannot build up anyplace unless the current is not continous. And in a circuit, the current is always continuous.
 
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I don't understand how you can say that a capacitor doesn't limit current but that they limit the rate that voltage can change.
Isn't that just another way of stating that they have a resistance to current flow?

You'll have to excuse me if I don't understand completely I'm still taking the second part of electrodynamics so we've just recently adjusted our outlook with maxwell's correction of ampere's law. So I get that there is current flow given a changing magnetic field but let's say you charge a capacitor with a battery then disconnect the leads leaving a charge. The charge is going to stay unevenly distributed over the capacitor or at least a much longer time than it would if you did the same with a piece of wire.

I don't agree that a capacitor doesn't limit current flow unless you have a convincing argument to enlighten me.
 
charge doesn't pass through a capacitor.
 
Antiphon said:
Capacitors do not limit the flow of current at all. They flow current just fine. They limit the rate at which the voltage across their terminals can change.

As the charge builds up on capacitor's plates, it repels additional charge from approaching the plates ... thus current is max at t=0 and becomes 0 at t=∞
q=Q (1 - e-t/RC) where Q is max charge ... Q = CV

Bhumble said:
So I get that there is current flow given a changing magnetic field but let's say you charge a capacitor with a battery then disconnect the leads leaving a charge. The charge is going to stay unevenly distributed over the capacitor or at least a much longer time than it would if you did the same with a piece of wire.

Wire do not get charge during current flow.
 
@ Cupid
I was saying that if you connected a piece of wire to a battery of some electric potential that then there would be current flow to the wire which is at a lower electric potential. Then removing the battery from the wire leaving the wire at the same potential as the battery it would be equipotential very quickly versus a capacitor that would not.

Maybe not the best example but I was trying to give an example of how a capacitor limits current flow.
 
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