Electric Field of Line Charge with Displaced Origin

[Chansu]

Homework Statement

Positive charge Q is distributed uniformly along the x-axis from x = 0 to x = a. Determine
a. The electric field produced by the charge distribution Q at points on the positive x-axis where x > a.
b. A point charge q is then placed at x = a + r. Determine force on q due to Q
c. If r >> a, show that the force approaches that given by two point charges.

With the origin being to the left of the point, it is throwing me off quite a bit. I'm not sure if my answers are correct, but I feel like I'm headed in the right direction.

I've attached the diagram as well as mine to this thread. The video below also is extremely relevant.

Homework Equations

E = kq / r² = k dq / r²

λ = Q / L = dQ / dL

F = qE

x = a + r

The Attempt at a Solution

[/B]
a. dE = (k dQ) / L² i^

using λ and solving for dQ,

dE = ∫[(k λ dL) / L²] i^ (from r to x)

taking the integral,

E = [k λ (-L^-1)] i^ (from r to x)

E = [k λ(-x^-1 + r^-1)] i^

E = [k λ(-x^-1 + (x-a)^-1)] i^ N/C, as long as x > a

b. F = q [k λ(-x^-1 + [x-a]^-1)] i^

if x = a + r,

F = q [k λ(-[a+r]^-1 + [a+r-a]^-1)] i^

F = q [k λ(-[a+r]^-1 + r^-1)] i^ N

c. if r >> a,

F = q k λ [-(a+r)^-1 + r^-1]

F = q k λ [-r^-1 + r^-1]

F = q k λ

 

[haruspex]

F = q k λ [-r^-1+ r^-1]
F = q k λ

The second line does not follow from the first. Indeed, the first line is not helpful. Too much detail has been lost.
Go back to the preceding line and find a more accurate approximation.