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

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!

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