Thin rod w/linear charge density, dipole moment

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Homework Help Overview

The problem involves a thin rod with a linear charge density that varies along its length, ultimately resulting in a net charge of zero. The focus is on understanding the electric field expression for a point located at a distance from the rod and how it simplifies in the limit where the distance is much greater than the length of the rod.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the simplification of the electric field expression in the limit of large distances, questioning the relationship between the dipole moment and the electric field. There is an exploration of how to express the electric field in terms of the dipole moment and other variables.

Discussion Status

Some participants have verified calculations related to the dipole moment and are considering how to express the electric field in a familiar dipole form. There is an ongoing exploration of the implications of the limit d >> L and how to correctly incorporate the dipole moment into the electric field expression.

Contextual Notes

Participants note the importance of maintaining the definitions of variables and the implications of the limiting case on the electric field expression. There is an acknowledgment of the need to avoid overwriting established definitions with new interpretations.

Slightly Odd Guy
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Homework Statement


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.

Homework 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)

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!
 
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Slightly Odd Guy said:

Homework Equations


p=q*d ?
E = −kλ0/L(d/(L−d)−ln(d−L)+d/(L+d)+ln(L+d))
p=2λ0L2/3
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:
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?
 
BvU said:
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.
I should be flogged for giving all that away ...
 
Well, I'm not going to give you a hard time about it. Thank you very much for your help!
 

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