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Why potential of an infinite line of charge not equals to zero at r = infinite 
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#1
Feb104, 10:12 PM

P: 21

the potential of a point charge is defined to be zero at an infinite distance. why can we not define the potential of an infinite line of charge to be zero at r = infinite?



#2
Feb204, 12:10 PM

P: 45

For finite configurations like a point charge or a uniform spherical distribution, if you get far enough away the potential 'levels off,' and has a slope of zero. Since we have the freedom to set the zero level anywhere we want, we'll typically put that zero at infinity. This does NOT mean that being 'at infinity' automatically means having zero potential.
For an infinitely long wire, no point is special. In fact, if you find yourself at some point on an infinitely long wire, you have no way of distinguishing that point from any other on the wire. From this argument, it seems reasonable (and is true) that every point on that wire has the same potential. So if the potential wherever you put your origin is Vnaught, then the potential along the wire 'at infinity' (or any other location) is Vnaught as well. It can't be zero. Travel radially out from the wire, and the potential will decrease. At an infinite distance from the wire, it will level off to zero slope; you can call that level zero volts if you wishthat's the convention. So every twodimensional slice taken perpindicular to the wire looks similar to a point charge in 3D space. P 


#3
Feb204, 01:03 PM

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P: 2,327

No matter how far away you are from an infinite line charge, you still see an infinite line charge. In fact, the angular density of charge that you observe increases with increasing distance from the line charge. As far as I know, infinite line charges are much more unphysical than point particles. If you try to approximate infinite distance from the line charge in the lab, then either you will find that your lab is inadequately sized, or the line will start to look like a point.



#4
Feb204, 05:48 PM

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Why potential of an infinite line of charge not equals to zero at r = infinite
This makes for an interesting calculation. Assuming an infinitely long line of charge, of density ρ, the force on a unit charge at distance l from the line works out to be ρ/l. That means that the potential function, which is that "antiderivative" of the force is
ρ log(l)+ C. That is unbounded as l goes to infinity and so C cannot be chosen to make the potential 0 at infinity. Simplest thing to do is take C=0 so that the potential is 0 at l= 1, negative for l< 1, positive for l> 1. 


#5
Feb404, 03:48 PM

P: 8

By my reckoning (and that of Gauss's law), the field strength from a infinite line of charge is inversely proportional to the distance from the line of charge. Therefore at infinite distance the field is 0.



#6
Feb404, 04:46 PM

P: 922

Actually gauss' law is tricky to use here. Actually HallsofIvy has it right. You will have a none zero potential (or electric field) at infinity. The way I reconcile this to myself is that we are talking about a nonphysical system, something that is infinitely long, so the usual conventions (taking the potential to be zero at infinity or making sure the electric field vanishes there) do not apply.
Cheers, Norm 


#7
Feb404, 04:52 PM

P: 8

To use gauss's law assume a cylinder that is infinately long with charge density u. E=u/2PIe0r. I though gauss's law was very simple to use as you consider the flux per unit length of the cylinder.



#8
Feb404, 05:01 PM

P: 8

Besides which i got the answer from a textbook as well.



#9
Feb404, 05:29 PM

P: 45

Why can't we all be right?
The electric fieldstrength, big E with a vector sign on top, DOES fall as 1/r  so it goes to zero at infinity. HallsofIvy said this, before going on to state that The potential which gives rise to such a fieldstrength is a natural logwhich diverges to negative infinity at infinity (and, like that of a point charge, is undefined at the origin.) Bearing this in mind, I should probably revise some of the things I said, like my comment about the potential 'levelling off' if you got far enough away. Sure enough, an upsidedown natural log doesn't level off, at least not for a very long time. So I'll join with turin in saying that, while modelling a wire as 'infinitely long' makes a lot of sense if you're right next to it (i.e., the length of the wire is much greater than the radial distance from the wire to the test charge), any wire looks like a point if you get far enough away from it. Infinitely long wires don't exist, and unlike frictionless pendulums and massless springs, their lack of existence makes life easier, not harder. P 


#10
Jun1109, 08:11 AM

P: 19

I have a question along the same lines as this thread.
The expression for the potential is supposedly V = rho[ln(r1)  ln(r2)]/(2 pi epsilon). This is what you get when you integrate the electric field and evaluate at the "limits". Mathematically, If you try to set one of the limits to zero, that makes the distance from the line of charge equal to one, since ln(1)=0. So, is the distance one or infinity? The function changes sign at 1. Does the potiential change sign at one? If you use a ratio of the distance to 1, can you scale the potential and make it fit whatever you want the boundary conditions to be? I am so confused!!!!!!!!!!!!! 


#11
Jun1109, 10:04 PM

P: 67

In simple and nonmathematical terms, the infinite line of charge would look EXACTLY the same at some ridiculously large distance away as it would if you were close to it. If you define the potential to be zero at infinity (large distances), then there would be no way to define the potential at less than infinity, since it would be the same mathematics on a different scale.
The reason for introducing "infinite lines of charges" into electrostatics is to allow an approximation for the situation in which a finite line of charge seems infinite (ie when you are very close to it). The closer parts of the wire make a much larger contribution to your measurements than do the sections of wire that are far away. 


#12
Jun1209, 12:52 AM

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P: 11,587

For a more extreme example of this situation, consider an infinite plane sheet of charge. The electric field on each side of the sheet is uniform (does not depend on position), directed perpendicularly away from the sheet for a positivelycharged sheet, and towards a negativelycharged sheet. Therefore a graph of potential versus distance from the sheet is a straight line whose slope equals the magnitude of the electric field. This obviously goes to either plus or minus infinity as the distance goes to infinity.
In reality, of course, infinite plane sheets of charge don't exist, so this analysis is useful only as an approximation near a finitesized sheet. 


#13
Jun1209, 04:54 AM

P: 19

Are you saying that the mathematical expression is only valid for a range of distances specified for a particular configuration? Does this mean that the potential is not zero at r = 1 meter? Does this mean that the exression is not valid for r < 1 meter? 1 meter is close to the line.



#14
Jun1209, 09:08 AM

P: 490

If you want to do it right for finite sheets and lines, just integrate. It's not so bad.



#15
Jun2609, 12:00 PM

P: 19

I can integrate. That's how I got the equation.



#16
Jun2609, 06:48 PM

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P: 7,030

Gauss's law is based on a finite charge per unit distance on an infinite line, or a finite charge per unit area on an infinite plane. In both cases the total charge of the objects generating the field is infinite. This could help explain why the potential doesn't go to zero as distance goes to infinity.
C = Coulomb, sometimes defined as the charge of 6.24150962915265 x 10^{18} electron charges r = distance of line perpendicular from infinite plate to the point m = meter N = newton q = charge of object (C) s = sec k = electrical constant = 8.9875517873681764 x 10^{9} N m^{2} / C^{2} σ = charge per unit area λ = charge per unit length E(r) = field intensity at distance r from source infinite plane case: E(r) = 2 π k σ [tex]\Delta V = 2 \pi k \sigma \int_a^b dr = 2 \pi k \sigma (b  a)[/tex] Neither a or b can be ∞ infinite line case: E(r) = 2 k λ / r [tex]\Delta V = 2 k \lambda \int_a^b \ dr / r = 2 k \lambda (ln(b)  ln(a)) =  2 k \lambda ln(b/a)[/tex] Neither a or b can be ∞ finite charge finite sphere or point case: E(r) = k q / r^{2} [tex]\Delta V = k q \int_a^b \ dr / r^2 =  k q (1/b  1/a)[/tex] a or b can be ∞, leading to V = k q / r 


#17
Jun2609, 07:45 PM

P: 1,133

Think of it like this. Let us say that the line of charge was of finite length. Then, there would be some distance r that would approximate the line of charge as a point charge (if you go infinitely far from a finite line of charge, the line of charge will look like a point). However, if the line of charge was infinite (as stated in your question), there would not be an effective distance for which the the line of charge would look like a point (even if you were infinitely far from the line of charge, you would still see a line of charge and not a point).



#18
Jun2609, 07:59 PM

P: 164

In physics an "infinite line of charge" is a line of charge with length L considered by such an observer a radial distance r away from the line (which is taken to lie on the zaxis) that [itex]\frac{r}{L} << 1 [/itex]. Not only is this the way that the expressions like [itex]E = \frac{\lambda}{2 \pi \epsilon_0 r}[/itex] are applied in practice, but it is also true that if you do a power expansion of the exact expression for the finite rod and drop (r/L)^2 and higher powers, then you obtain the same expression for the electric field which I gave earlier in this sentence.



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