(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

"An infinitely long rod of radius R carries a uniform volume charge density [itex]\rho[/itex]. Show that the electric field strengths outside and inside the rod are given, respectively, by [itex]E=\rho R^2/2\epsilon_0 r[/itex] and [itex] E = \rho r/2\epsilon_0[/itex], where r is the distance from the rod axis."

radius of rod = R

distance from axis to gaussian surface = r

length of gaussian surface = L

charge per unit volume = [itex]\rho[/itex]

2. Relevant equations

surface area of a cylinder (not including bases)

[tex]A = 2\pi rL[/tex]

Volume of a cylinder

[tex]V = \pi r^2 L[/tex]

Gauss's Law

[tex]\Phi = \int \vec{E} \cdot \vec{dA} = \frac{q_{encl}}{\epsilon_0}[/tex]

Answer for inside the surface

[tex]E = \frac{\rho R^2}{2 \epsilon_0 r}[/tex]

Answer for outside the surface

[tex]E = \frac{\rho r}{2 \epsilon_0}[/tex]

3. The attempt at a solution

First off, the surface area doesn't include bases because they're not relevant. I drew electric field lines on the diagram, and the one that is a side view of the rod shows that because it extends off to infinity, the electric field lines won't angle off because there are no ends, making them appear parallel from the side. This means, i don't need the ends of my gaussian cylinder in my surface area - for all practical purposes they're still there (and i'll still need to consider area when dealing with volume charge density), but because they're parallel to the field there is no flux. when viewed from the end (cross section), the lines are radial, which means i dont have to worry about [itex]cos \theta[/itex].

the dotted lines in the cross section show the gaussian surfaces i chose, and the cylinder surrounding the rod shows the outer gaussian surface i chose. the length of that, L, is the same for both surfaces. The outer one i got to work out:

there's no need for an integral, so i have [itex]E \cdot A = q_{encl}/\epsilon_0[/itex]

[itex]q_{encl}[/itex] is just charge density times volume.

[tex]2\pi rLE = \frac{\rho \pi r^2 L}{\epsilon_0}[/tex]

which simplifies to:

[tex]E = \frac{\rho r}{2 \epsilon_0}[/tex]

when i try to calculate the inner one, i know there is some ratio dealing with object volume and gaussian surface volume, but i don't know where it comes from or how it works... or how to get it:( please help.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Electric field of a rod

**Physics Forums | Science Articles, Homework Help, Discussion**