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## Homework Statement

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.

## Homework Equations

E = [1/(4[itex]\pi[/itex][itex]\epsilon[/itex]

_{0})][q/r

^{2}]

## 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?