# Capacitor Circuit -- Violating conservation of charge?

Discussion in 'Homework Help' started by HunterDX77M, Aug 2, 2016.

1. ### HunterDX77M Thread Starter Active Member

Sep 28, 2011
104
2
So I was doing a practice problem from a workbook, and I feel like I'm missing a critical concept here.

In the attached circuit, the capacitors can be reduced to a single 20 uF cap. That gives it a charge of

$20 \mu F \times 400 V = 8 mC$

But then the solution starts breaking up the single 20 uF capacitor into its original component capacitors, applying the same charge across the 20 uF, 40 uF and 9 uF. That is, saying they each have 8 mC of charge on them. How is that possible? Where is the extra charge coming from?

2. ### WBahn Moderator

Mar 31, 2012
23,408
7,116
There's no extra charge -- in fact there's no net charge at all. Put the capacitor in a black box and ask how much charge is in the box. The answer is zero. All that has happened is that the charge has been moved around within the box, but the charge on the box itself has not changed.

With a single 20 uF capacitor charged to 400 V you have to move 8 mC of charge from one place to another such that there is 400 V between them. Doing so took 1.6 J of work.

Now let's say that this single capacitor is replaced with two 40 uF capacitors in series. Now we will create a charge separate of 8 mC on each capacitor, but that capacitor will only be charged to 200 V. The amount of work required to charge each capacitor is 0.8 J, for a total of 1.6 J of work.

So basically we created twice the charge separation but the charges were separated by a smaller amount keeping the total energy involved the same.

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