Question

What does cylindrically symmetric charge distribution mean? Use Gauss’s law to determine the electric field of an infinite solid cylinder of radius R having volume charge density(rho) at a point outside it.

Answer 1

We call these types of charge densities cylindrically symmetric, as the charge density changes as a function of the distance. from the z-axis only (i.e., is independent of coordinates or z. φ 

Gauss law for an electrostatic field in the integral form is as follows

$\oint _s \epsilon_0 \bar{E}. nds =q$

Here $\bar{E}$ is the electric field intensity, $\epsilon_0$ is the permittivity of free space, n is the normal on the surface and q is the total charge inclosed by the surface

Consider the following diagram

Consider the charge on the cylinder is positive hence, the electric field and the normal vector to the surface 3 will point in the same direction. but for surface 1 and 2 electric fields and the normal vector will make 90⁰

Hence,

$\oint _1 \epsilon_0 \bar{E}. nds +\oint _2 \epsilon_0 \bar{E}. nds+\oint _3 \epsilon_0 \bar{E}. nds=q$

$\oint _1 \epsilon_0 (Ecos 90)ds +\oint _2 \epsilon_0 (Ecos 90)ds +\oint _3 \epsilon_0 (Ecos 90)ds =q$

$0+0+\oint _3 \epsilon_0 (Ecos 90)ds =q$

Since $\lambda$ is the linear charge density hence, total charge in length L will be $\lambda$L. Hence , charge inclosed by the cylinderical surface is $\lambda$L

Substitute $\lambda$L for q in the equation above

$\oint _3 \epsilon_0 Eds =\lambda$L

$E \epsilon_0\oint _3 ds =\lambda$L

$\oint _3 ds$ is the integral over the gaussian surface which is the total surface area of surface 3

$E \epsilon_0(2 \pi rL) =\lambda$L

$E = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r}$

Outside only

$E = \frac{1}{4 \pi \epsilon_0} \frac{\lambda}{R}$