Thin rod w/linear charge density, dipole moment

In summary, the homework statement is that 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.
  • #1
Slightly Odd Guy
11
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!
 
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  • #2
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:
  • #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 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 ...
 
  • #5
Well, I'm not going to give you a hard time about it. Thank you very much for your help!
 

1. What is a thin rod with linear charge density and dipole moment?

A thin rod with linear charge density refers to a one-dimensional object with a uniform distribution of electric charge along its length. Dipole moment, on the other hand, is a measure of the separation of positive and negative charges within the rod. It is a vector quantity and is calculated by multiplying the charge of the rod by the distance between the positive and negative charges.

2. How is the electric field calculated for a thin rod with linear charge density and dipole moment?

The electric field at a point due to a thin rod with linear charge density and dipole moment can be calculated using the formula E = kλ / r^2, where k is the Coulomb's constant, λ is the linear charge density, and r is the distance from the point to the center of the rod. This formula takes into account the contribution of both the rod's charge and its dipole moment to the electric field.

3. Can a thin rod with linear charge density and dipole moment produce a non-uniform electric field?

Yes, a thin rod with linear charge density and dipole moment can produce a non-uniform electric field if the distance from the point to the rod is small compared to the length of the rod. In this case, the electric field will be stronger near the ends of the rod and weaker in the middle.

4. How does the dipole moment affect the behavior of a thin rod with linear charge density?

The dipole moment of a thin rod with linear charge density affects its behavior by determining the strength and direction of the electric field it produces. A larger dipole moment will result in a stronger electric field, while a dipole moment pointing in a certain direction will cause the electric field to be stronger in that direction.

5. Is the electric potential created by a thin rod with linear charge density and dipole moment zero at all points?

No, the electric potential created by a thin rod with linear charge density and dipole moment is not zero at all points. It is only zero at points along the axis of the rod, where the distance from the point to both ends of the rod is equal. At all other points, the electric potential will have a non-zero value due to the contribution of both the rod's charge and its dipole moment.

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