- #1
flyingpig
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Homework Statement
Trying to find the E-field inside a conductor using rings even though my book tells me it is 0.
I haven't learned how to do surface integrals yet but I think I only need Calc II to do this.
The Attempt at a Solution
[PLAIN]http://img18.imageshack.us/img18/699/unledoj.jpg
So one ring produces a E-field of
[tex]E = \frac{kq}{x^2 + a^2}[/tex]
Now graphing [tex]x^2 + y^2 = a^2[/tex]
[PLAIN]http://img713.imageshack.us/img713/9007/unledbz.jpg
Now what I want to do is to take cross sections of my rings with the E-field up there and sum it up like a volume.
Now the problem is, I am not sure how to set it up properly
I tried doing
[tex]dE = \frac{-2xkq}{x^2 + a^2} dx[/tex]
[tex]dx = \frac{x^2 + a^2}{-2xkq} dE[/tex]
Then I tried doing
[tex]A(x) = \pi r^2 = \pi (x^2 + a^2)[/tex]
And hence
[tex]V = \int \frac{\pi (x^2 + a^2)^2}{-2xkq} dE[/tex]
Now the problem is, what aer the bounds? Did I even set up my integral right?
I watched a video and it says that the E-field will point in opposite directions but same magnitude, but I don't understand why is that so.
I know that if my bounds are +E and -E, I would get 0 for the integral because my function is odd. Now the problem is, how do I distinguish this from an insulator? I could've done the exact same thing for an insulator and they would both come out 0
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