Thin rod w/linear charge density, dipole moment

AI Thread Summary
The discussion revolves around deriving the electric field expression for a thin rod with a linear charge density that decreases to zero, specifically in the limit where the distance from the rod is much greater than its length. The provided electric field equation simplifies to a form that relates to the dipole moment, which is given as p = 2λ0L²/3. Participants emphasize the importance of expressing the electric field in terms of the dipole moment, distance, and Coulomb's constant, aiming for a final form of E = c*k*p/d³. The conversation highlights the need to carefully handle the variables and integrate the charge density to find the dipole moment correctly. The goal is to connect the electric field of the rod to the familiar dipole field expression as the distance increases.
Slightly Odd Guy
Messages
11
Reaction score
0

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!
 
Physics news on Phys.org
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!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top