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Electric Field over a charged cylinder using only Coulomb's Law

  • Thread starter PeteyCoco
  • Start date
38
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1. Homework Statement

What is the electric field at a point on the central axis of a solid, uniformly charged cylinder of radis R and length h?

2. Homework Equations

Well, I've set up the triple integral and have gotten to this point:

[itex]\bar{E}_{z}=\frac{2\pi\rho}{4\pi\epsilon_{0}}\int^{h/2}_{-h/2}[1-\frac{z'-z}{\sqrt{(z'-z)^{2}+R^{2}}}]dz[/itex] , z' is the location of the test point

When I integrate this I get a constant term and a mess of logarithms. I know that the field should be 0 when z' = 0, but it doesn't check out. What's wrong?
 

TSny

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Well, I've set up the triple integral and have gotten to this point:

[itex]\bar{E}_{z}=\frac{2\pi\rho}{4\pi\epsilon_{0}}\int^{h/2}_{-h/2}[1-\frac{z'-z}{\sqrt{(z'-z)^{2}+R^{2}}}]dz[/itex]
Re-evaluate how you got the 1 inside the brackets in the expression above. I suspect that you got this from assuming that ##\frac{z'-z}{\sqrt{(z'-z)^2}} = 1##. But, recall that ##\sqrt{x^2} = |x|##.
 
38
1
Alright. Now that I've changed that term, it vanishes when integrated, but the second term in the integrand still turns into

[itex]\frac{1}{2}ln| (z'-z) + \sqrt{(z'-z)^{2}+R^{2}} |^{h/2}_{-h/2}[/itex]

Which, as far as I can tell, can't be reduced to 0 when z' = 0
 
38
1
Nevermind, integrating this absolute value function is a bit more work than I thought it was.
 

TSny

Homework Helper
Gold Member
12,042
2,619
Also, I don't see how you are getting a logarithm expression in the result of the integration.
 

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