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Positive charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positive x-axis at x=a+r, a distance r to the right of the end of Q.

(a) Calculate the x- and y- components of the electric field produced by the charge distribution Q at points on the positive x-axis where x>a.

(b) Calculate the force (magnitude and direction) that the charge distribution Q exerts on q.

(c) Show if r>>a, the magnitude of the force in part (b) is approximately Qq/4pi(epsilon)r^2. Explain why this result is obtained.

Alright, I'm having trouble getting started. First off, I understand that the y components is 0, no need to go there.

I know that since the charge is distributed, I have to use an integral. the Limits of the integral are from (0,a)

But what how exactly is the integral? (I'm going to use k instead of 1/4piepsilon

To my understanding....

E=kq/r^2

So, dE=kdq/r^2

How would I go from there? (if that were correct...)

Thanks!