Simple Capacitor Concept Question: A certain capacitor is fully charged by a battery, such that the positive plate holds a charge +q and the negative plate holds a charge -q. The plates of the capacitor are then pulled apart to twice their initial separation. Determine the new charge on the positive plate if, a). the battery is kept attached to the capacitor as the separation is increased b). the battery is disconnected BEFORE the plate separation is increased

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Simple Capacitor Concept Question: A certain capacitor is fully charged by a battery, such that the positive plate holds a charge +q and the negative plate holds a charge -q. The plates of the capacitor are then pulled apart to twice their initial separation. Determine the new charge on the positive plate if, a). the battery is kept attached to the capacitor as the separation is increased b). the battery is disconnected BEFORE the plate separation is increased

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Here is my line of thought: a). The charge is the same because the charges are still under the presence of a battery. Therefore the voltage at the point of the battery is the same at the point of the capacitor. b). Since the battery is disconnected, the charges even themselves to a state of equilibrium across the whole circuit. Therefore, the charge on both plates will be zero, since no work is being done by the battery to separate the negative charges to one side and the positive to the other. I am not sure if I am correct, so I was wondering if someone could check my logic. Any help would be greatly appreciated! :)
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Actually, wouldn't a). have different charges then? Since voltage is constant, but distance is changing, then we know that the capacitance changes. Which would mean that charge HAS to change
Given that \[C=\frac{ q }{ V }\]
that means that i am interested in the thread :-) another question was asked recently where the charges on the plates were different and i didn't have the time to go into it but addressing the question: for b), once the battery is disconnected, the charges are "stuck" on the plates. pulling them apart will then require work to be done, of course which is why the potential between the plates should increase. for a), well you're on to it now!!
Ah okay, I wasn't sure lol So for a).\[C=\epsilon*\frac{ A }{ d }\rightarrow C=\epsilon*\frac{ A }{ 2d }\] which means the C is halved. Since C is halved and is directly proportional to charge, then Q would be halved as well... right? b). Right, so since there is no battery and the charges have nowhere to go, then the charge would be the same for the whole system. But would they start to even out? Or stay in the same place leaving one plate positive and the other negative?
that makes sense to me too, FWIW!! \[V = const = \dfrac{Q_1}{C_1} = \dfrac{Q_2}{C_2} = \dfrac{Q_2}{\frac{1}{2}C_1} \implies Q_2 = \frac{1}{2}Q_1\]
Yay :) May I ask you another question? I'm pretty stuck on it..
if you think i can help....
stuff it in a new thread.

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