The problem is as follows: suppose we charge two capacitors whose capacitances $C_1$ and $C_2$ are $5\mu F$ and $10\mu F$ respectively until they have charges $q_1 = 2\mu C$ and $q_2 = 3\mu C$. We now put them in a circuit with a $10V$ battery as follows:

Given circuit

Suppose we then close the switch S1. Find the final charges for $C_1$ and $C_2$ after an infinite amount of time (that is, do not bother with calculating the times.) Now, the answers are $q_1 = 31.7, q_2 = 36.7 \mu C$. To get to these answers, I assumed the following: the capacitors are in series, therefore the equivalent capacitor $C_{eq} = \frac{10}{3}$.

Using that result, we use the $q = CV$ formula to find the charge that should be in both capacitors if they were wired positive-to-negative instead of positive-to-positive. The result is $q_{final} = 33.333...$

However, since the capacitors are wired positive-to-positive, this does not result in the usual balance between the capacitors. So what I assume is: the positive charge flowing through the positive end of the battery encounters the negatively charged plate of $C1$ and "loses" $q_1$ charges, thus resulting in $31.333...$, while the negative charge of the battery wired to the negative plate results in the opposite, which results in $36.333...$

I suspect I am wrong to assume this, and that I am failing to apply the charge conservation in this case. (I know that in negative-to-positive capacitor plates, the charge will be the same between them due to the implications of charge conservation, but can't find a way to apply it here other than the approach described above). I am having a lot of trouble finding information on this problem, and I would like someone to evaluate what my proposed answer is.


This is a fairly standard problem which illustrates the point that you should be very careful in solving the problem by using the formula for capacitors in series.

Diagram $1$ is the two charged capacitors before the battery was connected.

enter image description here

I have not labelled the equal magnitude negative charges which reside on the other capacitor plates.
The charge $q$ will be assumed yo be in $\mu \rm C$ and the capacitance in $ \mu \rm F$ for convenience.

In diagram $1$ the voltage across the $5 \mu \rm F$ capacitor is $\dfrac 2 5 \rm V$ and the voltage across the $10 \mu \rm F$ capacitor is $\dfrac {3}{1o} \rm V$ with node $C$ being at the higher potential relative to node $B$ by $\dfrac {1}{10} \rm V$.

So when a wire is connected between nodes $B$ and $C$ a current will flow and eventually the potentials of nodes $B$ and $C$ will be the same which the same as saying that the voltages across the two capacitors will be the same.

Conservation of charge means that $q_5 + q_{10} = 2 + 3 = 5$ and equality of voltages across the two capacitors that $\dfrac {q_5}{5}= \dfrac{q_{10}}{10}$

Solving these two simultaneous equations gives $q_5 = \dfrac {5}{3} \mu \rm C$ and $q_{10} = \dfrac {10}{3}\mu \rm C$ with the voltage across each of them $\dfrac 1 3 \rm V$.

So now to your circuit.
The conservation of charge equation is still the same.
Looking at diagram $3$ need to make the sum of the voltages across the $5 \mu \rm F$ capacitor and the $10 \rm V$ battery equal to the voltage across the $10 \mu \rm F$ capacitor ie $V_{\rm AB} + V_{\rm BC} = V_{\rm AC}$ but be a little careful with signs.
I think that it might be informative if you draw a final diagram showing the final charges and voltages?

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  • $\begingroup$ "The reason for that is that the capacitor in series formula was derived on the assumption that both capacitors were initially uncharged.": The capacitance of two capacitor in series is a parameter of the circuit, and it is independent on any assumption on the initial charges. $\endgroup$ – Massimo Ortolano Oct 18 '16 at 9:09
  • $\begingroup$ @MassimoOrtolano Thank you for your comment. Do you think that I should amend my answer as the use of the formula directly in this case can cause a lot of problems as the OP found? I have changed the first paragraph. $\endgroup$ – Farcher Oct 18 '16 at 9:13
  • $\begingroup$ Your edit if fine ;-) $\endgroup$ – Massimo Ortolano Oct 18 '16 at 9:35
  • $\begingroup$ Thank you very much. Got to the answer, but one of the final charges was of negative sign. I guess that has to do with the current direction, so adjusting for that we'd get both to be positive? $\endgroup$ – NeedForMath Oct 18 '16 at 11:41
  • $\begingroup$ That is why I made the comment about being careful with the signs. What you have found is that the voltage across the $10 \mu \rm F$ capacitor is less than $10 \rm V$ showing that voltage across the $5 \mu \rm F$ was reversed with a plate initially having positive charges on it now has negative charges. $\endgroup$ – Farcher Oct 18 '16 at 11:52

Intuitively, there is an unknown amount of charge transfer between capacitors and the source because battery basically can supply/take charge, however, charge conservation may not hold true always as the charges may cancel each other out in series(verified). In a parallel connection, charge conservation holds since like charges are in contact with each other always, hence they redistribute accordingly.

Thus, instead of equating charges, we equate voltage. V1+V2=Vsupply

Final charges are arranged as +c1- +c2-

Also, the current is the same everywhere in the loop - because the elements are in series. Capacitors in series build/lose the same amount of charge. So if they start out with different charges they can end up with different charges.

Initial Q across c1 = -2uC , Initial Q across c3 = +3uC (both Q are on the upper plate, lower has equal and opposite)

Assuming a common charge Q flows in the circuit loop, 10 = (-2+Q)/5 + (3+Q)/10 , we get Q= 33.7uC (which is the amount of charge flown out of the battery)

THERE IS NO SUCH REDISTRIBUTION OF CHARGES IN SERIES CONNECTION OF THE CAPACITORS(mutually). Only EMF drives the charges so that the system satisfies KVL law.

Now, final Q on c1= 33.7 - 2 = 31.7uC ,final Q on c2= 33.7+3 = 36.7uC

Verify KVL, 10 = 31.7/5 + 36.7/10 is true

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