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!

• If you look at the solution for sphere you get the same type of relationship. No dependence on the radius of the sphere. See also the shell theorem. The field depends only on the charge contained by the Gaussian.
– nasu
May 27, 2022 at 11:21

That $R$ does not occur, I would argue as follows:
• To obtain the magnitude you need to assume $\vec{E}$ is parrallel to $\hat \rho$ anyway May 27, 2022 at 11:26
The cylinder radius, $$R$$, can be added to the equation if you replace the charge density ($$\rho$$). $$\begin{equation*} \rho=\frac{Q}{\pi R^2 l} \end{equation*}$$ then your
$$\begin{equation*} E = \frac{Qr}{2 \pi l E_0 R^2} = \frac{2 K Q r}{l R^2} \end{equation*}$$