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Quick conceptual question about disconnecting battery!

by xchococatx
Tags: battery, conceptual, disconnecting
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xchococatx
#1
Mar14-12, 09:06 PM
P: 7
If there is a parallel plate capacitor, connected to a battery, which is then disconnected, there is NO voltage now, but "Q" (charge) on the capacitor remains the same. YES OR NO? I think yes.
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gneill
#2
Mar14-12, 09:37 PM
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Quote Quote by xchococatx View Post
If there is a parallel plate capacitor, connected to a battery, which is then disconnected, there is NO voltage now, but "Q" (charge) on the capacitor remains the same. YES OR NO? I think yes.
What is your reasoning?
xchococatx
#3
Mar14-12, 09:58 PM
P: 7
Well basically I'm thinking that a capacitor stores charges (or energy) so, onces its all charged up by a battery, I don't see how it could lose those charges...even when the battery is disconnected. (immediately anyways, it wouldn't be all gone I don't think) Why I was thinking about this was because if the battery is now disconnected, and you wanted to calculate the work needed to insert a dielectric, fitting perfectly between the 2 plates, since there is no voltage, I don't see how you could calculate the work unless you used the original charge of the capacitor (without dielectric). (By means of E=Q^2/2C) So in order to use the original charge of the capacitor, I figured charge has to stay in the capacitor even if the battery is disconnected.

gneill
#4
Mar14-12, 10:17 PM
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Quick conceptual question about disconnecting battery!

Quote Quote by xchococatx View Post
Well basically I'm thinking that a capacitor stores charges (or energy) so, onces its all charged up by a battery, I don't see how it could lose those charges...even when the battery is disconnected. (immediately anyways, it wouldn't be all gone I don't think) Why I was thinking about this was because if the battery is now disconnected, and you wanted to calculate the work needed to insert a dielectric, fitting perfectly between the 2 plates, since there is no voltage, I don't see how you could calculate the work unless you used the original charge of the capacitor (without dielectric). (By means of E=Q^2/2C) So in order to use the original charge of the capacitor, I figured charge has to stay in the capacitor even if the battery is disconnected.
Okay, your reasoning is pretty good. Note that the charged capacitor will maintain a voltage (potential difference) according to Q = C*V, even without an external voltage source applied. Of course, if the capacitance changes (due, say, to the insertion of a dielectric) the voltage will change accordingly if the charges have nowhere to go.


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