Electric Field at a Point


by Broem
Tags: electric, field, point
Broem
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#1
Feb24-13, 03:05 PM
P: 9
1. The problem statement, all variables and given/known data
A charged rod of length L = 1.00 m lies centered on the x axis as shown. The rod has a linear charge density which varies according to λ = ax where a = −90.0 μC/m.
What is the x component of the electric field at a point on the x axis a distance of D = 2.00 m from the end of the rod?

2. Relevant equations
E=kQ/r^2
charge density = Q/A

3. The attempt at a solution
I really do not understand...Where do I get the r from? Since its not uniformly distributed how can there be any field outside?
Would my r = (D-(half of L))?
I'm assuming lambda is just regular charge density at a location of x?
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File Type: bmp rod_symetric_on-end.bmp (271.0 KB, 7 views)
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tiny-tim
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#2
Feb24-13, 03:44 PM
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Hi Broem!
Quote Quote by Broem View Post
Would my r = (D-(half of L))?
I'm assuming lambda is just regular charge density at a location of x?
Yes.

(and of course it's a linear charge density, ie coulombs per metre, not per metre3 )
Since its not uniformly distributed how can there be any field outside?
Not following you.
Broem
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#3
Feb24-13, 04:11 PM
P: 9
Thanks for the quick response!!
Ok so here's where I am:

I now know that lambda = Q/L so Q = L * (lambda)

So I'm left with
9e9(-9e-6)/(2-.5)^2

This however is not the correct answer...Where can I go from here?

tiny-tim
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#4
Feb24-13, 04:18 PM
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Electric Field at a Point


oh, no, you'll have to integrate

use coulomb's law to find the electric field from a tiny section [x,x+dx] (whose charge will be λdx, and which you can assume is concentrated entirely at x), and integrate from -L/2 to L/2
Broem
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#5
Feb25-13, 09:26 AM
P: 9
I got it!!!
Thank you so much for your help!!!


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