Thin rod w/linear charge density, dipole moment

  • #1

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!
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
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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:
  • #3
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
BvU
Science Advisor
Homework Helper
14,134
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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 ...
 
  • #5
Well, I'm not going to give you a hard time about it. Thank you very much for your help!
 

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