# Question about uniform charge distribution

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1. Jun 22, 2017

### Chemmjr18

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

2. Relevant equations
(see attached)

3. The attempt at a solution
To solve, I used the following method.

Could you help me find the flaw in my method?

2. Jun 22, 2017

### Staff: Mentor

3. Jun 22, 2017

### Chemmjr18

Sorry about that. Here's my attempt. I think it has something to do with how I'm integrating.

#### Attached Files:

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4. Jun 23, 2017

### haruspex

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.

5. Jun 23, 2017

### Chemmjr18

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!

6. Jun 23, 2017

### haruspex

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.)
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.

7. Jun 23, 2017

### Chemmjr18

Thanks for the help!