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Disclaimer: I've done the calculation myself, but trying to verify this is hard. I couldn't find much online, and when I did, the sources disagreed in their finding. Also, searching reveals one question on here like this, but the math is far different than what we use in my class. (And a tutor at my school had a completely different response. Anway, on to the problem...)

For an infinitely long nonconducting cylinder of radius R, which carries a uniform volume charge density $\rho$, calculate the electric field at a distance $r<R$.

I did:

$\phi_e = \int \vec{E}d\vec{A} = \frac{Q_{in}}{\epsilon_0}$, where I'm measuring $A$ to be the area of the Gaussian surface (not the real cylinder).

$\phi_e = \vec{E}(2\pi rl) = \frac{Q_{in}}{\epsilon_0}$ ; however I can write $Q_{in} = \rho(\pi r^2 l)$

$\rightarrow \vec{E} = \frac{\rho\pi r^2 l}{(2\pi rl)\epsilon_0} = \frac{\rho r}{2\epsilon_0}$

Can someone please confirm/deny if this is correct? I feel like the equation should be taking into consideration the cylinder radius R, but perhaps it doesn't matter (and is only in the problem to make me question it?).

Thanks!

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Your solution is correct, except that you get the magnitude of the electric field only.

That $R$ does not occur, I would argue as follows:

  • The charge distribution for an infinite thin, hollow cylinder is the same as for a conducting one, that is because of symmetry the charge will spread evenly on the thin shell.
  • Inside the now conducting, hollow cylinder, the electric field is zero, otherwise the charges would adjust.
  • If you stack these hollow cylinders, you end up with the part of the original cylinder that can be neglected.
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