Electricity: Ifinite Line Charge

exiztone

Hi there, I hope you don't mind helping me out a bit because I'm really stuck. I'm doing computer science in university and the electrotechnology part can be a bit tricky for us.

For an infinite line charge Pl = (10^-9)/2 C/m on the z axis, find the potential difference points a and b at distances 2m and 4m respectively along the x axis.

I have a basic understanding of physics, Coloumb's Law, Voltage etc... but I don't know about this infinite line charge stuff. I'm not looking for a solution, I actually have the answer here (it's 6.24v in case you're interested) but I'd like to know how to calculate it. If this wasn't a line charge, i.e. if this was a point charge at the origin, I would be able to do it.

My attempt:
So if I were to go about this, I would get the potential of each point using this formula:

q/(4)(pi)(eps)(r). I derived that from integrating Coloumb's Law with respect to r. I'd then subtract point b from point a to get the voltage.

Coloumb's law wouldn't appear to work here though because it's a line charge and not a point charge. This is my attempt, I hope it's not too pathetic.

Thanks in advance

Civilian_Target

You're right, Coulomb's law won't help you.

Think of it like this - the line is infinite so you're going to get the same voltage using it at any point if you use Coulombs law as it's not length dependant.

I'm no EM expert, I gave it up a while back, but you'll probably need to know some sort of resistance or resistivity term for the line so you can tell at what pace the line loses voltage. This could be a given constant in an equation you have somewhere. Take a look at the question again...

mikelepore

exiztone:

You use Coulomb's Law but not the simplest form. And superposition. Select a point in space above the line, select a differential charge element on the line (not directly below the first point), write the distance formula between the two points (hypotenuse of triangle), calculate the potential (or electric field, whichever you want) at point 1 due to the charge element at point 2, integrate as the charge element goes from 0 to infinity, now double your answer because the charge is on two sides. (Rectangular coordinates.)

Civilian_Target:

Resistance has nothing to do with it. No current is flowing.

Mike Lepore in New York USA

(ancient education, BS EE '76, MS EE '85)
(and the brain damage doesn't help either)

lepore at bestweb dot net
http://www.crimsonbird.com/

Son Goku

It's an application of Maxwell's First Equation:


However to use it in this example I'll write it in integral form:

Where the first term, phi is the flux of the electric field across a surface.

Basically consider a finite cylinder around the line, at what ever distance you want.


Let us say that this cylinder has length L and radius R.

We don't actually need to work out the integral for this example. We just need to know the basic result that the integral equals the total charge enclosed by the surface divided by epsilon.

Since the Electric field flows out from the line, the flux through the ends of the cylinder is zero, because the field is always parallel to those ends.

The flux through the rest of the cylinder is simply the magnitude of the electric field times the area of the cylinder:

Flux = E*(area of cylinder)
Since the area of any cylinder is the same:
Flux = E*(2*Pi*R*L)

From the first equation above:
Flux = Charge enclosed/epsilon.

The charge enclosed by the cylinder is simply:
{(10^-9)/2 C/m}*L

So:
E*(2*Pi*R*L) = [{(10^-9)/2 C/m}*L]/epsilon.

So:
E = {(10^-9)/2 C/m}/(2*Pi*R*epsilon)

Now that you have the electric field, just use the usual method to find the potential at both points and get the difference.

Sev

Civilian_Target wrote:

You're right, Coulomb's law won't help you.

You can of course just integrate Coulomb's law along the length of the wire. You are given a charge density. This is the natural way I would approach the question. Gauss' law just never occurs to me.

Son Goku

Sev wrote:

You can of course just integrate Coulomb's law along the length of the wire. You are given a charge density. This is the natural way I would approach the question. Gauss' law just never occurs to me.

Coulomb's Law wouldn't work here because you can't surround the charge with a sphere and label that sphere the point charge.
Coulomb's Law is one solution of Maxwell's equations for the case of a spherically symmetric time invariant field and no Magnetic Field.
Since this isn't that case it won't work.

EDIT: There is an "infinitesimal length" way of doing using coulomb’s law, but it's "fake", for lake of a better word.

Sev

Son Goku wrote:

EDIT: There is an "infinitesimal length" way of doing using coulomb’s law, but it's "fake", for lake of a better word.

Yeah, thats what I was alluding to. Intuitively it makes perfect sense, is fairly straightforward and very clearly gives the correct answer.

Son Goku

Sev wrote:

Yeah, thats what I was alluding to. Intuitively it makes perfect sense, is fairly straightforward and very clearly gives the correct answer.

It only works here out shear luck and contradicts itself. If it were any other shape but an infinite straight line it wouldn't work.

Sev

Son Goku wrote:

It only works here out shear luck and contradicts itself. If it were any other shape but an infinite straight line it wouldn't work.

I must really have no grasp of electrostatics.. because I cant see any situation in which integrating coulombs law wont give the correct answer for a static distribution of charge.

Sev

Ok.. it occurred to me.. that maybe you think im suggesting something entirely different... and if im right.. something which is very silly indeed.

What I am suggesting.. is very simple.. so I dont quite understand how you dont follow. Its called integration.. you do it ALL THE TIME in physics, to for example calculate the gravitational force from a body of non uniform density or the magnetic field of a current in a wire by biot-savart... its nothing new.. its no surprise.

You can ALWAYS use a brute force coloumbs law calculation to determine the electric field from a static distribution of charge. This is extremely TRIVIAL. Just like how you would get the field from two point charges by addition... in the continuous regime this becomes integration.

Using such an integral method the original poster's question could have been done using basic knowledge of coloumbs law and integration from the leaving cert without even considering Gauss' law.

Break up the line into a sum of infininitessimal charges dQ = ρdx

The field at any arbitrary point a distance R from the line will be made up from the integral of the electric field contributions from the charge element corresponding to each line element dx.

This is an integral of vector forces, but because of the symmetry of the situation, we need only consider the component of force directed radially outwards/inwards from the wire. (Other components cancel). We get this radial component by just multiplying the magnitude of this vector force by (R/r) where r is the distance from the line element to the point at which we want to calculate the electric field. r = Sqrt[R^2 + x^2]

so

dE = (R/r)(1/4πε)(ρdx)/r^2
= (ρR/4πε){dx/(R^2 + x^2)^3/2}

Integrating this gives

E = (ρ/4πεR)(x/Sqrt[R^2 + x^2]) + c

Putting in the limits x = Infinity,-Infinity gives.

E = ρ/2πεR

Which is the same result you get by using Gauss' law.

Son Goku

You can ALWAYS use a brute force coloumbs law calculation to determine the electric field from a static distribution of charge.

Yes, exactly. However it separates out a method from electromagnetic theory in general and that's the reason you're told not to do it in most texts on theoretical electrodynamics.
It is a good calculation method, however it is self-contradictory. I can show you the proof if you want. Basically it encounters difficulties in certain scenarios, which forces the use of distributions (i.e. like Dirac Deltas, not distributions of charges).

What I meant by the "any other shape, but an infinite straight line", I mean that's the only shape where it isn't "dippy", to use a phrase from Feynman.

You're not wrong. It does work.
It gives the right answer, but for the wrong reasons if you get me. Basically it has what is called an existence problem mathematically, if you've heard of the phrase.

I mightn't have made myself clear, so just point out any vagaries.

Sev

Yeah.. I wouldnt mind seeing this proof.. because I dont quite make sense of what youre saying.. this is all new to me. Existence Problem? Dippy process? Is this something im yet to be taught? Is this like how the laplacian of a coulomb potential is a delta function?

I don't take kindly to anything that goes against what I inuitively know to be true.

Son Goku

Sev wrote:

I don't take kindly to anything that goes against what I inuitively know to be true.

What do you mean?
I'm not saying it doesn't work in the sense of giving you the wrong answer, for a static field, I'm saying it has deeper problems that are best to avoid, for electromagnetic theory in general. Particularly for awkward coordinates.

Is this like how the laplacian of a coulomb potential is a delta function?

That's an example, but it's not a terrible problem.

Son Goku

F. Rohrlich, Syracuse University, New York talks about it, but give me a while to get some decent papers.

Sev

My point I guess.. that the integration approach to the infinite line of charge.. is an incredibly simple problem. You can clearly see.. by inspection that integrating the problem is always going to work.. youre just adding up all the charge elements from a length of wire, you know to have uniform charge density, and extending it to infinity. Numerically, by inspection... you can see that nothing is ever going to go wrong, the electric field goes to zero at infinity, the integral converges, and theres no singularites or anything crazy going on. So to even begin to believe that coulombs law might not work in that case is just naive.

I dont quite understand the need to consider even bringing in these far fetched seemingly purely academic complications. But, then again I havnt even ever contemplated such complications.. so I cant claim to know anything about this.. but it has piqued my curiosity so I'd be very interested to see this self contradictory proof thing, because I cant dig up anything on google about it anyway.

Also what is an example in which naively integrating coulombs law is not going to provide the correct result? But where Gauss' law does?

Son Goku

Sev wrote:

My point I guess.. that the integration approach to the infinite line of charge.. is an incredibly simple problem. You can clearly see.. by inspection that integrating the problem is always going to work.. youre just adding up all the charge elements from a length of wire, you know to have uniform charge density, and extending it to infinity. Numerically, by inspection... you can see that nothing is ever going to go wrong, the electric field goes to zero at infinity, the integral converges, and theres no singularites or anything crazy going on. So to even begin to believe that coulombs law might not work in that case is just naive.

It does work. My problem is foundational issues.

I dont quite understand the need to consider even bringing in these far fetched seemingly purely academic complications.

It's due to the fact that you'll eventually run into a contradiction. It'll cause a problem further down the line.

Also what is an example in which naively integrating coulombs law is not going to provide the correct result? But where Gauss' law does?

I think when there is a Vector Potential with a non-zero Time derivaitve.

Son Goku

I can see why this might seem a bit over the top, but stuff like this has actually held back certain areas, so I'm always over careful of it.

Sev

Ok.. well thanks for the responses.. this is something I'll look into. Id be grateful if you can let me know if you find any more details about where I can see this self-contradictory proof, becase im finding nothing with Rohrlich, hopefully its not to off the wall for an undergrad.

Also I dont quite get how a time varying vector potential is anything to do with a static distribution of charge.

Son Goku

Sev wrote:

Also I dont quite get how a time varying vector potential is anything to do with a static distribution of charge.

I know it seems wierd, but there is an quite a lot of second order non-constant Vector Potentials, which don't give rise to a magnetic field or dynamic charge.

The simplest example would be where the vector potential is of the form A(t).