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

  • #1

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/r2]


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/r2]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/r2]). 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?
 

Answers and Replies

  • #2
SammyS
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It looks like a good start.
 
  • #3
Alright. I was able to check the answer with a friend and I got it right. Thanks for your words.

Would an iterated integral be an easier way to approach this? Wouldn't I just keep my variables as is and just append their differentials to the end of the function?
 
  • #4
SammyS
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I don't see why you would use an iterated integral for this problem.
 

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