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

There is a non-conducting charged rod with length L=.0815(m) and linear charge density λ=-5.9x10^{-14}(C/m). The rod is placed parallel to and on the x-axis, and at a distance a=.12(m) from the right-most end of the rod is point P. Calculate the magnitude and direction of E (electric-field) at point P due to the charged rod.

2. Relevant equations

E = [1/(4[itex]\pi[/itex][itex]\epsilon[/itex]_{0})][q/r^{2}]

3. The attempt at a solution

First, we know we are dealing strictly with x-components of the E-field due to the charged rod because the x-axis runs through the length of the rod and point P is on the x-axis. The equation for E becomes E = [1/(4[itex]\pi[/itex][itex]\epsilon[/itex]_{0})][q/r^{2}]cos([itex]\theta[/itex]). At the same time we note that the angle that any element of the rod makes with point P is [itex]\theta[/itex]=0, so the cos([itex]\theta[/itex]) term is 1.

Because the charge is distributed throughout the rod uniformly (as opposed to being concentrated at a single point), we need to find the contribution for each charge element dq of the rod. I do this by changing q in my equation to dq.

At this point I have two variables, dq and r (E = [1/(4[itex]\pi[/itex][itex]\epsilon[/itex]_{0})][dq/r^{2}]). Here's where my trouble begins. Since I'm not utilizing iterated integrals (don't mind you guys giving me tips on how to do so) I want to get all variables in terms of just one, namely dq=(λ)(dx) and r=.12(m)+(.0815(m)-x). Am I on the right path?

**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: Magnitude and Direction of E-field at a Point Due to a Charged Rod

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