Problem 19.29

19.29.
We consider a Gaussian surface that is a cylinder of radius r away from the axis of the cylindrical charge distribution. Let's consider a length $\ell$ of the cylinder. Gauss' Law tells us that
$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{inc}}{\epsilon_{0}}$
$(E)(A) = \frac{\int \rho dV}{\epsilon_{0}}$
$(E)(2\pi r \ell) = \frac{\rho \pi r^{2} \ell}{\epsilon_{0}}$
$E = \frac{\rho r}{2 \epsilon_{0}}$