Gauss Law Problem with Cylinders

  • #1

Homework Statement


An infinitely long, non-conducting cylinder of radius R carries a uniform volume charge density ρ. Find the magnitude of the electric field for 0 < r < R


Homework Equations


EA=Qin/εo


The Attempt at a Solution


I am debating whether the answer should be ρr^2/2εoR or ρr/2εo.
I said E(2πRL)=(ρπr^2L)/εo and solved to get ρr^2/2εoR.

I was confident in my answer but then found the same problem in the book, but it asked to find E at distance r from axis, where r<R. The answer to this was ρr/2εo. Can someone explain the difference between these two problems, if there is one. They may be exactly the same and my calculation is wrong, but I am confused over the two
 

Answers and Replies

  • #2
ehild
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I said E(2πRL)=(ρπr^2L)/εo and solved to get ρr^2/2εoR.

If you want to find E at r<R your Gaussian surface is a cylinder of radius r.

ehild
 
  • #3
Since r<R, R will never appear in the equation since my dA vector relates to r and Qin also relates to r. I would simply replace R with r in my equation and solve, yielding ρr/2εo. Is this logic correct?

I think I just failed to remember that in the basic form of the equation, you are integrating over the Gaussian surface. But does Qin relate charge enclosed by the Gaussian volume drawn or by the entire volume in question?
 
  • #4
ehild
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Qin is the charge of volume enclosed by the Gaussian surface.

ehild
 
  • #6
ehild
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So I would get E(2πrL)=(ρπr^2L)/εo and E=ρr/2εo.
But this website says I am wrong so I am now very confused
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html#c3

It is saying for r<R, E should be completely different. Can someone please clear this up?


In that formula at hyperphysics, E is given in terms of lambda, the charge per unit length. In your problem, the charge per unit volume (rho)was given. Compare your new result with that you cited from your book in your first post.


ehild
 
  • #7
When I calculate it, I get ρr/2εo.
But the website above says something different, so I am clueless as to what to put if this problem were on my test tomorrow(today lol).
 
  • #8
We posted at the same time. I did not notice that. I'll change my calculation and see what I get
 
  • #9
ehild
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See my previous answer.

ehild
 
  • #10
I've concluded that the calculation is correct for my problem and it is ρr/2εo.
But, I don't understand where they get the R^2 in the bottom of their equation on hyperphysics. It has to come from the A part of EA=Qin/eps, but how? Shouldn't A always be in terms of the gaussian surface, which has nothing to do with R.
 
  • #11
nevermind, found the link that explains it. I would never come up with that on my own but at least I now understand the difference and why it is there
 

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