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

AI Thread Summary
To calculate the electric field (E) at point P due to a non-conducting charged rod, the focus is on the x-components of the E-field since both the rod and point P are aligned along the x-axis. The electric field equation is modified to account for the continuous charge distribution, leading to E = [1/(4πε0)][dq/r²], where dq is the differential charge element. The variables dq and r are expressed in terms of linear charge density (λ) and the distance from the charge element to point P. The discussion raises the question of whether using iterated integrals would simplify the calculation, though the initial approach seems valid. Ultimately, the problem was successfully solved, confirming the correctness of the method used.
MurdocJensen
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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\pi\epsilon0)][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\pi\epsilon0)][q/r2]cos(\theta). At the same time we note that the angle that any element of the rod makes with point P is \theta=0, so the cos(\theta) 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\pi\epsilon0)][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?
 
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It looks like a good start.
 
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?
 
I don't see why you would use an iterated integral for this problem.
 
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