Question about uniform charge distribution

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Chemmjr18
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1. The problem statement, all variables, and given/known data
Suppose you have a wire of length l and a uniform line charge density λ. Find the electric field at the midpoint that is height r above the x-axis

Homework Equations


(see attached)

The Attempt at a Solution


To solve, I used the following method.
20170622_170215.jpg


Could you help me find the flaw in my method?
 
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Sorry about that. Here's my attempt. I think it has something to do with how I'm integrating.
 

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Chemmjr18 said:
Sorry about that. Here's my attempt. I think it has something to do with how I'm integrating.
You may be confusing yourself by using l as the length of the wire and as the position of an element within it. Better to consider an element length dx at distance x from the midpoint of the wire, and integrate from x=-l/2 to x=+l/2.
 
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Thanks. I just want to make sure I understand this. When finding the electric field caused by some charge spread out uniformly over a line, circle, surface, or volume, what I need to do is find the electric field caused by some infinitesimally small piece of charge, dq, which is dE. The limits of integration will depend on the surface. The infinitesimally small charge, dq, is equal to the density of the charge times the length, area, or volume. Lastly, and perhaps the step I find most difficult, I need to find how the function I'm integrating, dE, varies with what I'm integrating over. For example, in this case, I'm integrating over the x-axis. Therefore, I have to determine how dE varies across the x-axis. In this case,

dE=(dq)/(x2+r2)3/2

And one last question, what if the field wasn't uniform? I can't think of any particular cases, but I still think it's something that could come up quite often. Again, thanks for the help, I appreciate it!
 
Chemmjr18 said:
Thanks. I just want to make sure I understand this. When finding the electric field caused by some charge spread out uniformly over a line, circle, surface, or volume, what I need to do is find the electric field caused by some infinitesimally small piece of charge, dq, which is dE. The limits of integration will depend on the surface. The infinitesimally small charge, dq, is equal to the density of the charge times the length, area, or volume. Lastly, and perhaps the step I find most difficult, I need to find how the function I'm integrating, dE, varies with what I'm integrating over. For example, in this case, I'm integrating over the x-axis. Therefore, I have to determine how dE varies across the x-axis. In this case,

dE=(dq)/(x2+r2)3/2
That is a correct description if you qualify the field references to be just the component in the normal direction. (That's where the square root comes from in the expression.)
Chemmjr18 said:
what if the field wasn't uniform?
I assume you meant if the charge is not uniform. No problem. In the uniform distribution dq=λdx for some constant λ. If the charge density at x is f(x) then dq=f(x).dx.
 
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Thanks for the help!