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A homework problem asked me to find the voltage across a series of capacitors after they came to equilibrium.

Essentially a capacitor C1 with a capacitance of 6.0 F is charged until the potential difference (V) is 9.0V across it using a battery.

This battery then is removed and replaced with a second uncharged capacitor C2 with a capacitance of 3.0F. The goal of the question is to determine the voltage across C1 after the system has came to equilibrium.

The answer is that C1 has a voltage of 6V across it however in solving this problem a number of questions came up. And a number of different ways of solving it came up.

One way of getting to that solution hinges on the assumption that the voltage across capacitor 1 needs to be equal to the voltage across capacitor 2 and thus their q/C ratios should be the same. We are not sure whether this conclusion is correct.

This results in us having some confusion about the potential difference across capacitor 2 at the end. And then if the potential difference does not add up at the end can we even use Kirchhoff's laws since the potential difference in the system is 9 and thus there wouldn't be a point where the sum is 0?

To explain the confusion here are some diagrams of our thinking. Are we even right about our assumptions of the potential difference of the system?

enter image description here

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  • $\begingroup$ In the absence of resistance, any stray inductance will make the solution an infinite oscillation... which never really comes to equilibrium. $\endgroup$ – Whit3rd May 16 '19 at 21:34
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Kirchhoff's voltage law is still solid and forces $$Q_1/C_1 - Q_2/C_2 = 0,$$ where $Q_1,C_1$ are the final charge and capacitance of the first capacitor and $Q_2, C_2$ are the final charge and capacitance of the second capacitor.

Kirchhoff's current law needs to be time-integrated in such a case. It says that charge does not accumulate within a wire, so that the amount of current flowing into a wire is equal to the amount flowing out of a wire. In this context it then concludes that the charge on the top branch of the circuit cannot change because the charge has nowhere to go, in other words if $Q_0$ was the initial charge on the first capacitor and the second capacitor was uncharged, it states that $$Q_1 + Q_2 = Q_0.$$

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  • $\begingroup$ So you would say the V across the 2nd capacitor after a bit of time would be -6 since the 1st capacitors is 6? $\endgroup$ – CalebK May 16 '19 at 19:39
  • $\begingroup$ I mean it depends how you measure it; I would say $V_1 = Q_1/C_1$ and $V_2 = Q_2/C_2$ and that Kirchoff's voltage law forces $V_1 = V_2$, so I would conclude that $V_1 = V_2 = 6\text{ V}$. But yes, from the point of view of a loop that travels clockwise from the bottom, that convention for $V_1$ is encountered additively, increasing the voltage and $V_2$ is encountered subtractively, and so if you chose the clockwise convention and called those $\bar V_{1,2}$ you would say $\bar V_1 = \text{6 V},~~\bar V_2=-\text{6 V}.$ $\endgroup$ – CR Drost May 16 '19 at 20:34
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It's very often overlooked that charge obeys the conservation principal (along with energy and momentum). If you couch your problem in those terms, you may be better served. Previous answers refer.

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