Electrostatics

[Temperature101]

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A 1 mF capacitor is fully charged with a 12 V battery. The battery is then removed and the capacitor is connected to an uncharged 2 mF capacitor. After the system reaches equilibrium, what will be the charge on the 1 mF capacitor?
A. 3 mC
B. 4
C. 6
D.12

Edit...Never mind...I got it

 

[syoung]

I don't! plz explain!

 

[Temperature101]

I don't! plz explain!

I took the PR explanation since it's very comprehensive. These PR questions are rough.
Since Q = CV, we see that the battery delivers (1 mF)(12 V) = 12 mC to the capacitor. After the battery is removed and the second capacitor is connected, some of this charge will flow to the new capacitor. Choice D can therefore be eliminated. Equilibrium is established when the potential difference, or voltage, is the same across each capacitor. (Since electrons move toward regions of high potential and equilibrium implies there is no net motion, then that must mean that any two plates that are connected must have the same potential. An alternative way of thinking about this is that the two capacitors are in parallel, and voltages in parallel must be the same). The two capacitors share 12 mC of charge and, since Q ∝ C, the 2 mF capacitor should get twice as much charge as the 1 mF capacitor: 8 mC vs. 4 mC. The correct answer is choice B.