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Electromagnetics  parallel plate capacitor energy storage 
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#1
Jun2112, 05:18 PM

P: 17

1. The problem statement, all variables and given/known data
A parallelplate capacitor 0.5m by 1.0m has a separation distance of 2cm and a voltage difference of 10V. Find the stored energy, assuming that ε=εo. If a 200volt potential is applied in the capacitor, (a) determine the energy stored; (b) hold d1 at 2cm and the voltage difference at 200V, while increasing d2 to 2.2cm, determine the final stored energy. 2. Relevant equations U=(1/2)(CV^{2}) C=εS/d 3. The attempt at a solution (a) C=((8.854x10^{12})(0.5x1.0))/(0.02) C=2.2135x10^{10} F U=(1/2)(2.2135x10^{10})((10+200)^{2}) U=4.8808x10^{6} J (b) C=((8.854x10^{12})(0.5x1.0))/(0.022) C=2.0123x10^{10} F U=(1/2)(2.0123x10^{10})((10+200)^{2}) U=4.4371x10^{6} J I'm not sure what to do with the voltage difference of the capacitor, so I added it with the voltage potential applied. 


#2
Jun2212, 05:25 AM

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P: 3,440

Yeah, I'm not sure what it means by the voltage difference of the capacitor and the applied potential. I would guess that they are two separate questions, i.e. work out the answer with 10V, then also work out the answer with 200V (in other words, don't add them together).
And for part b, what does d1 and d2 mean? Are they the positions of the two plates, or are they the positions of one of the plates in two different situations? 


#4
Jun2212, 08:38 AM

P: 17

Electromagnetics  parallel plate capacitor energy storage
So I saw my classmate's answer for this problem and for (b), d in C=ε_{o}S/d is d=(d_{1}+d_{2})/2? Why is this so?



#5
Jun2212, 08:40 AM

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#6
Jun2212, 08:43 AM

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#7
Jun2212, 08:48 AM

P: 17

The thing is, our professor didn't give us an illustration or anything. One of my classmates just drew the setup  two plates separated by d and that's it.
Sir, can I post another question here and my solution to it? It's about electromagnetics too. 


#8
Jun2212, 09:00 AM

HW Helper
P: 3,440

If it's another question, you will probably get more replies by making a new thread for it. But you can post it here too if you'd like. About this question, I guess you are meant to use (d1+d2)/2 for the separation, and I would guess to use 200V, but it might be different depending on what your classmate had in mind. Anyway, It looks like you get the idea of how to do the calculation, but you were just unsure on the specific setup in this case.



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