Gauss's Law for inside a long solid cylinder of uniform charge density?

Question

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!

Answer 1

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.