anonymous
  • anonymous
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
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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! :)
IrishBoy123
  • IrishBoy123
.
anonymous
  • anonymous
@IrishBoy123 ??

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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
anonymous
  • anonymous
Given that \[C=\frac{ q }{ V }\]
IrishBoy123
  • IrishBoy123
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!!
anonymous
  • anonymous
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?
IrishBoy123
  • IrishBoy123
that makes sense to me too, FWIW!! \[V = const = \dfrac{Q_1}{C_1} = \dfrac{Q_2}{C_2} = \dfrac{Q_2}{\frac{1}{2}C_1} \implies Q_2 = \frac{1}{2}Q_1\]
anonymous
  • anonymous
Yay :) May I ask you another question? I'm pretty stuck on it..
IrishBoy123
  • IrishBoy123
if you think i can help....
IrishBoy123
  • IrishBoy123
stuff it in a new thread.

Looking for something else?

Not the answer you are looking for? Search for more explanations.