Question

The charge is distributed uniformly throughout the volume of an infinitely long solid cylinder of radius R. (a) Show that, at a distance r from the cylinder axis,
$E=\frac{\rho r}{2{ϵ}_0}$
where $\rho$ is the volume charge density. (b) Write an expression for E when $r>R$.

Answer 1

(a) The diagram shows a cross-section (or, perhaps more appropriately, “end view”) of the charged cylinder (solid circle).
Consider a Gaussian surface in the form of a cylinder with radius r and length A, coaxial with the charged cylinder. An “end view” of the Gaussian surface is shown as a dashed circle. The charge enclosed by it is $q=\rho V=\pi r^{2}lp$, where $V=\pi r^{2}l$ is the volume of the cylinder.
If ρ is positive, the electric field lines are radially outward, normal to the Gaussian surface, and distributed uniformly along with it. Thus, the total flux through the Gaussian cylinder is $\Phi =E{A}_{cylinder}=E\left(2\pi rl\right)$. Now, Gauss’ law leads to :
$2\pi {ϵ}_{0}rlE=\pi r^{2}lp\Rightarrow E=\frac{\rho r}{2{ϵ}_0}$
(b) Next, we consider a cylindrical Gaussian surface of radius r > R. If the external field $E_{ext}$ then the flux is $\Phi =2\pi {ϵ}_{0}rl{E}_{ext}$ The charge enclosed is the total charge in a section of the charged cylinder with length A. That is, $q=\pi R^{2}l\rho$. In this case, Gauss’ law yields :
$2\pi {ϵ}_{0}rl{E}_{ext}=\pi R^{2}l\rho \Rightarrow E_{ext}=\frac{R^2\rho }{2{ϵ}_{0}r}$