How Do You Calculate Electric Potential of a Nonuniformly Charged Rod?

[misa]

[SOLVED] Electric potential

Homework Statement

The charge on the rod of the figure (length 2l, center at the origin) has a nonuniform linear charge distribution, λ = ax.
Determine the potential V at:

(a) points along the y-axis.
(b) points along the x-axis. (Assume x > l)

(express all answers in terms of a, x, l, ε₀ and appropriate constants)

Homework Equations

dQ = λdx
dV = dQ/(4*pi*ε₀*r)

The Attempt at a Solution

For part a, V = 0 because
dV = dQ/[4*pi*ε₀*(x² + y²)^1/2] dx with limits of integration from -l to l.

For part b, I'm having a really hard time determining the limits of integration and what "r" in dV is (ie, is it x or is it r, a segment of the rod?). I tried a lot of things, none of which produced the correct answer. Right now, I have

Let k = 1/(4*pi*ε₀)
I'm treating "x" as a fixed distance from the rod, while calling r a segment or distance along the rod starting at x - l.

dV = [k(ar)]/(x - l + r) dr with integration limits from x - l to x + l (one end of the rod to the other)

With change of variables,
u = x - l + r
dx = du
r = u - x + l
integration limits become 2x - 2l to 2x


dV = [ka(u - x + l)] /u du
= ka (1 + (l - x)/u) du

V = ka(u + (l - x)ln(u))
V = ka(2x - (2x - 2l) + (l - x)ln(2x) - (l - x)ln(2x - 2l))
V = ka(2l + (l - x)ln(2x / 2x - 2l))
V = ka(2l + (l - x)ln(x / (x - l)))

I have a feeling that this is wrong, especially because I still don't completely understand what I'm supposed to integrate along, etc. Can someone explain how to go about solving this problem and point out what I am doing incorrectly?

Thank you!

 

[LowlyPion]

Along x, aren't you integrating in r at any point x > l from the distance x - l the closest point to x + l the farthest distance?

 

[misa]

umm, I was until the change of variables part--unless I don't need a change of variables and integrated the wrong expression?

 

[LowlyPion]

umm, I was until the change of variables part--unless I don't need a change of variables and integrated the wrong expression?

I guess what I was trying to suggest was that you have an r for some point X > l such that r(x) = X - x. You would be taking the integral of this I was thinking from x = l to x = -l

dV = k*a*x/r(x)*dx = k*a*x/(X - x)*dx

 

[misa]

Oh! Okay, that makes sense (where X is treated like it's fixed but r(x) is still the distance from X to x, some point on the rod). Thank you!