Energy density due to infinite uniform line charges

In summary: Yes, meaning the two rods move (while staying perpendicular to the ##x,y## plane) and thus the intersection points move. I have to find differential equations for ##E_x,E_y## and show that the component ##E_z## of the electric field satisfies:$$ \left(-\frac{\partial}{\partial x^2}-\frac{\partial}{\partial y^2}+\epsilon_0 \mu_0 \frac{\partial}{\partial t^2}\right)E_z = 0$$but ##E_z## should be zero, because the potential ##C\cdot \log(|r_2-r_1|
  • #36
Karl86 said:
The answer is given: it is ##-\frac{2\lambda^2}{4\pi\epsilon_0} \log(|r_2-r_1|)##. What I am asked is to prove that it can be written like that. How is it you got that expression naively?

You start with one "infinite" wire. You bring a unit of line charge in from "infinity" - and don't worry that the potential is infinite at infinity - and you have the energy to create the configuration, per unit length of the second wire.

But, now it seems very clear, if you released that unit line charge, it has infinite potential energy. So, there is no finite answer.

If, however, you consider the wire long but finite, then eventually the logarithmic potential will cease to hold. The PE of the unit length of charge is, therefore, finite for any finite wire, but tends to infinity for an infinite wire, so the problem is not well-defined.
 
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  • #37
PeroK said:
You start with one "infinite" wire. You bring a unit of line charge in from "infinity" - and don't worry that the potential is infinite at infinity - and you have the energy to create the configuration, per unit length of the second wire.

But, now it seems very clear, if you released that unit line charge, it has infinite potential energy. So, there is no finite answer.

If, however, you consider the wire long but finite, then eventually the logarithmic potential will cease to hold. The PE of the unit length of charge is, therefore, finite for any finite wire, but tends to infinity for an infinite wire, so the problem is not well-defined.
You are seeing it in terms of energy to get a certain configuration, but the wires are fixed in a constant position. There are no assumptions on how close or far they are, not infinitely far though, one intersects the xy plane in a point r_1 and the other in a point r_2. What the question seems to point to is they generate a field and this field carries some energy. Calculate it between the planes z=a and z=a+1. Is the way you are trying to see it equivalent to this?
 
  • #38
Karl86 said:
You are seeing it in terms of energy to get a certain configuration, but the wires are fixed in a constant position. There are no assumptions on how close or far they are, not infinitely far though, one intersects the xy plane in a point r_1 and the other in a point r_2.

Nevertheless, they store infinite energy per unit length.

The analysis I did is necessary to calulate the energy stored, either explicitly or through a previously derived formulas, that uses the "bring a charge in from infinity" approach. Note also:

PeroK said:
To get that equation - the integral of ##E^2## - you need to neglect a boundary surface integral, which is not valid if the charge distribution is not bounded.

Do you know another equation for electrostatic energy ?

I overlooked that the same objection applies to the other formula using the potential.

In short, you cannot define the energy stored in the field of an infinite wire, because the potential is not finite at infinity.

The moral is that you need to be careful with these hypothetical charge distributions that go to infinity.
 
  • #39
Ok so you stand by your analysis that it seems not well defined, but what did you mean when you said that ##C\log(|r_2-r_1|)## can be deduced from the configuration? How?
 
  • #40
Karl86 said:
Ok so you stand by your analysis that it seems not well defined

What happens if you release a positive point charge in the field of an infinite positively charged wire?

What is the final KE of the charge?
 
  • #41
PeroK said:
What happens if you release a positive point charge in the field of an infinite positively charged wire?

What is the final KE of the charge?
The field is constant and pushes it away from the wire, the charge never stops, so the KE is never 0. IS that right?
 
  • #42
Karl86 said:
The field is constant and pushes it away from the wire, the charge never stops

It's not a question of whether it stops or even whether the force is never zero - those apply to a point charge source as well - the issue is that the force drops off as ##1/r## and the potential is logarithmic, so that there is no final asymptotic KE "at infinity". The KE increases without bound.

Therefore, the PE of the point charge is not well defined.

The infinite wire configuration, although useful for calculating the potential difference between two finite points, as an approximation of a long finite wire, is ultimately a purely hypothetical configuration, with infinite energy density.
 
  • #43
Yes, I'm convinced. The discussion that arose from a seemingly innocent problem was remarkable though :D
 
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  • #44
PS The only question now is whether this question was a deliberate "sandbag" or a genuine misconception on the part of whoever gave it to you!
 
  • #45
It was not a deliberate sandbag, it is meant to be answered correctly, so as things stand I am failing to answer it properly, and yet I communicated it to you correctly, and I am persuaded the discussion is correct.
 
  • #46
Oh, ok I asked my instructor and I was told to set the reference point not at infinity but at a finite distance from the wire. Do you think that can help?
 
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