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Parallel-plate capacitor with flipping charge

by giladsof
Tags: capacitor, charge, flipping, parallelplate
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giladsof
#1
Jun19-11, 05:22 AM
P: 13
1. The problem statement, all variables and given/known data



2. Relevant equations


U = 0.5CV^2

3. The attempt at a solution

I'm really lost on this one...?
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zhermes
#2
Jun21-11, 03:17 PM
P: 1,261
Yes, you do seem to be fairly lost on this one.
Is that the only question you had?
giladsof
#3
Jun21-11, 06:52 PM
P: 13
well usually in these kind of questions where there is some kind of change in the capacitor (the plates were move closer for instance)- We need to calculate the energy in the system before the change, calculate the energy afterwards and see the change's size and sign.

But in this problem I don't seem to manage to do so.... more help and less sarcastic remarks will be most appreciative

berkeman
#4
Jun21-11, 07:12 PM
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Parallel-plate capacitor with flipping charge

Quote Quote by giladsof View Post
1. The problem statement, all variables and given/known data



2. Relevant equations


U = 0.5CV^2

3. The attempt at a solution

I'm really lost on this one...?
Do you see intuitively whether it will require energy to be put into the system, or give energy back, to switch the position of the plates? Why?

And to calculate it, I think I would just calculate the energy required to move the charges those distances in that Electric Field. What is the equation for the force on a charge in an Electric Field? What is the equation for the work done by a force through a distance? Don't forget to account for each plate's charge separately (so two energy calculations).

Quote Quote by zhermes View Post
Yes, you do seem to be fairly lost on this one.
Is that the only question you had?
I believe he is just gently reminding you of the PF Homework Help rules, which require you to show an attempt at a solution.
giladsof
#5
Jun25-11, 05:39 AM
P: 13
Thank you I solved it.

One small question remains- I don't understand why is it so intuitive that the capacitor will resist this change?
berkeman
#6
Jun25-11, 11:09 AM
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P: 41,085
Quote Quote by giladsof View Post
Thank you I solved it.

One small question remains- I don't understand why is it so intuitive that the capacitor will resist this change?
Because you're having to move charges against the forces from the E-field. You have to do work to move the charges against those forces. If you were going the other way, well, that's the way the charges would go anyway if they were not constrained to be held on the plates (if they were free charges).


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