Answer to Question #178130 in Electricity and Magnetism for jay rajak

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

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