Thin rod w/linear charge density, dipole moment

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1. Apr 18, 2016

Slightly Odd Guy

1. The problem statement, all variables and given/known data
A thin rod of length 2L has a linear charge density that isλ0 at the left end but decreases linearly with distance going from left to right in such a way that the charge on the entire rod is zero.

Given

E = −kλ0/L(d/(L−d)−ln(d−L)+d/(L+d)+ln(L+d))

for a point P that is distance d>l to the right of the center of the rod:

How does this expression simplify to a familiar result for the limiting case d≫L? (Hint: The dipole moment of the rod has magnitude p=2λ0L2/3.) Express your answer in terms of some or all of the variables p, d, and Coulomb's constant k.

2. Relevant equations
p=q*d ?
E = −kλ0/L(d/(L−d)−ln(d−L)+d/(L+d)+ln(L+d))
p=2λ0L2/3
λ=λ0(1-x/L)

3. The attempt at a solution
I don't have a clear idea of where I'm going in this problem. I started with the limiting case, and narrowed down the E equation to

E= (−kλ0/L)(1-ln(d)+1+ln(d))
E= (−kλ0/L)(1+1)
E= (−kλ0/L)(2)
E= -2kλ0/L

I don't think this is correct, because the answer has to be defined by no other variable than p, d, and k. Because of that, I know I have to use their equation for p, but I don't know how to use it. Frankly, I don't even know if my first step was correct.

I could definitely use some guidance in the right direction right now.

Thank you very much!

2. Apr 18, 2016

BvU

If you are given p, then it's not wise to overwrite that with something like p = q d. That d means something else than your d.
$$p = \int x dq = \int_0^{2L} x\ \lambda\, dx = \lambda_0 \int_0^{2L} x \left (1-{x\over L}\right ) \, dx$$which gives you p (check !)​
Where you are going in this exercise should be obvious: if they give you the dipole moment and mention d>>L, you are going towards an expression for the dipole field -- which you should write down for the given situation so you know where you are going.

In that context you can't simply ignore L/d (and write d/(L-d) = 1 ), but have to write out the E expression in terms of L/d (which is a small number...)

Last edited: Apr 18, 2016
3. Apr 18, 2016

Slightly Odd Guy

Okay, I've checked the integral for p, and it works, and I can appreciate how this condenses to a dipole field, since the rod and dipole would look the same given enough distance.

So that means I'm trying to condense Erod to the form E=c*k*p/d3, c being a numerical multiplier, correct?

4. Apr 19, 2016

BvU

I should be flogged for giving all that away ...

5. Apr 19, 2016

Slightly Odd Guy

Well, I'm not going to give you a hard time about it. Thank you very much for your help!