A uniformly distributed charge of density is contained within a...

Consider a coaxial cylindrical Gaussian surface of radius $x$ and length $L$ passing through point $P$.

By Gauss's law, the total electric flux through this surface is:

$\oint \vec{E} \cdot d\vec{A} = \dfrac{q_{\text{enclosed}}}{\varepsilon_0}$

Due to the symmetry of the long cylinder, the electric field $\vec{E}$ is radial and its magnitude is constant over the curved surface of the Gaussian cylinder. The flux through the flat circular ends is zero since the electric field is parallel to them.

$E(2\pi x L) = \dfrac{q_{\text{enclosed}}}{\varepsilon_0}$

The charge enclosed within the Gaussian surface is:

$q_{\text{enclosed}} = \rho \times \text{Volume} = \rho (\pi x^2 L)$

Substituting the enclosed charge into Gauss's law:

$E(2\pi x L) = \dfrac{\rho \pi x^2 L}{\varepsilon_0}$

$E = \dfrac{\rho x}{2\varepsilon_0}$