# Electric potential (rod)

#### misa

[SOLVED] Electric potential

1. 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)

2. Homework Equations

dQ = λdx
dV = dQ/(4*pi*ε0*r)

3. 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!

Last edited:
Related Introductory Physics Homework Help News on Phys.org

#### LowlyPion

Homework Helper
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

Homework Helper
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!

Last edited:

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving