What Are the Units of μ in a Variable Charge Density on a Rod?

[TheLegace]

Homework Statement

I was just wondering if I had done this problem correctly

Take the rod axis to be the x-axis and the lefthand end of the rod to be x = 0. Assume
lambda is not constant but that lambda = μx where μ is a constant. What are the units of μ?
Find the force Vector F on the point charge q located a distance d from the righthand end of
the rod.

Homework Equations

dq= μxdx
dF = 1/4(pi)e_0 * (dq*q/d^2)

The Attempt at a Solution

Ok well, to start I want to find what the constant μ units will be. I figure since lambda needs to be C/m, and x would be a function of displacement, then the units for μ would be C/m^2; when you take product μx = C/m.

Now to find the force, I figured I should the find the dq for the rod, dq=lambda*dx=μxdx
Which I am hoping is correct.

Now dF = 1/4(pi)e_0 * (dq*q/d^2). Now integrating from 0 to x yields
F=(μq/(d^2)8(pi)e_0) * x^2, now I work out the units they I get a Newton unit, so that helps me confirm that maybe I did it right. What I may have issues with is are my bounds for integration correct and is my statement about the radius squared in Coulombs law correct.

Any help would be appreciated, sorry if it is a bit difficult to read.

Thank You.
TheLegace

 

[kuruman]

How long is the rod? You don't say, so let's assume it is L. Your limits of integration should be from 0 to L. Also, if your charge element dq is at distance x from the origin, then its distance to the point charge will be (L - x + d). The square of that is what should be in your denominator.

 

[TheLegace]

How long is the rod? You don't say, so let's assume it is L. Your limits of integration should be from 0 to L. Also, if your charge element dq is at distance x from the origin, then its distance to the point charge will be (L - x + d). The square of that is what should be in your denominator.

Thank You, I finally realized that since dq is being taken it will be subtracting the x, looking at geometry I was able to figure it out, but thank you very much.

Now is the integral easy to take? How would one take it?

 

[kuruman]

One would find a substitution that works. For example, let u = L - x +d.