Field of a finite line of charge

[ShayanJ]

I tried to find the potential of a finite line of charge with length 2l and constant charge density $\lambda$.So I set up the coordinates
somehow that the line is on the x-axis and the origin is at the center of the line.Then I did the following:

$\phi=\int_{-l}^l \frac{\lambda dx'}{4 \pi \varepsilon_0 \sqrt{(x'-x)^2+y^2}}=\frac{\lambda}{4 \pi \varepsilon_0}\int_{-l}^l \frac{dx'}{\sqrt{(x'-x)^2+y^2}}=sinh^{-1}(\frac{x'-x}{y})]_{x'=-l}^l=\ln{\frac{l-x+\sqrt{(l-x)^2+y^2}}{-(l+x)+\sqrt{(l+x)^2+y^2}}}$

Everything seems correct but there are two problem with this answer:

1-I expected the potential to be symmetric about x axis,But as you can see in the plot,its not!(Also I don't know what's the meaning of those empty places.)

2-when I take the limit of the potential as l goes to infinity,I get infinity instead of a formula!

I can't think of another way for doing it.What's your suggestion?

Thanks

 

[tiny-tim]

Hi Shyan!

When there's a square-root you must be careful …

that should be sinh^-1 of |(x'-x)/y| ​

 

[ShayanJ]

Thanks

Well,yeah.I really need to be more carefull.

But that still doesn't solve the problem.Don't know about the plot but now the the limit equals zero!

 

[Hassan2]

Thanks

Well,yeah.I really need to be more carefull.

But that still doesn't solve the problem.Don't know about the plot but now the the limit equals zero!

As for the second problem, you are calculating the potential difference between the point and a point at infinity. This difference is infinity. However you can set another point rather than infinity as your reference ( zero potential).

 

[ShayanJ]

I did some calculations and understood that things are not as easy as you said.

$\int \frac{dx}{\sqrt{x^2+a^2}}=\int \frac{a \ cosh{z} \ dz}{\mid a \mid cosh{z}}= \int \frac{a}{\mid a \mid} dz=sgn(a) \int dz=sgn(a) z$

From above,I can't reach what you said!
And the problems remain unsolved.

Also I should say that the plot of the function I derived with considering your advise,wasn't symmetric about x-axis but at least it had no empty places inside it.

Thanks hassan,But could you explain more?

 

[haruspex]

Shyan, I plotted the 'ln' version of your formula from your first post for the case of y = l = 1. It was nicely symmetrical, peaking at x = 0. So I'd check the plotting.

It will be misbehaved around y = 0, of course, since it tends to infinity there. And I would expect the potential at a given x, y to tend to infinity as l goes to infinity. Why do you think that's wrong?

 

[Hassan2]

Shyan,

I didn't mean to change the origin of the coordinate system. What I mean is that , for the infinite line, due to cylindrical symmetry we can use Gauss's law and calculate the electric field as
$E_{y}=\frac{λ}{2\pi ε_{0}y}$

thus $ΔV =-\int E_{y}dy'$ ( the limit of tegral from y'=a to y'=y)

so $ΔV =\frac{λ}{2\pi ε_{0}}ln(a/y)$

But I know you want to get it directly integrating the potentials due to the infinitesimal elements on the line. What you used as partial potential difference is with reference to infinity. subtract the the potential of your reference (0,a) from the integrand and I you'll obtain the above-mentioned result.

 

[ShayanJ]

But you're calculating the potential of an infinite line of charge hassan.
I want the potential of a finite line of charge!

And haruspex,you can calculate the equation of potential for an infinite line of charge using gauss's law.It is logical to think that the limit of the equation for a finite line of charge as the length goes to infinity,equals the above result.

Thanks both

 

[Hassan2]

Yes I know and I left the calculation of the finite line for you. If you calculate the potential difference between point B(x,y) and A(0,a) , you'll get:


$\phi_{BA}=\frac{\lambda}{4\pi\epsilon_{0}}(\ln{ \frac{l-x+\sqrt{(l-x)^2+y^2}}{-(l+x)+\sqrt{(l+x)^2+y^2}}}-\ln{\frac{l+\sqrt{l^2+a^2}}{-l+\sqrt{l^2+a^2}}})$

Now the limit becomes
$\frac{\lambda}{2\pi\epsilon_{0}}\ln\frac{a}{y}$

 

[ShayanJ]

Sorry,I attached the wrong plot in my first post.
The right one is this one.
It is symmetric about x-axis except for those empty places where it becomes infinite.
Its ok but I think the fact that it is just for x<l breaks the symmetry and makes it wrong.
Your function has the same plot hassan.
But at least it solves the limit problem.

Thanks

 

[Hassan2]

Your function has the same plot hassan.
But at least it solves the limit problem.

Thanks

Sorry my function must be corrected as:

$\frac{\lambda}{2\pi\epsilon_{0}}\ln\left|\frac{a}{y}\right|$

The function is not even so the graph is not symmetric because the potential is with reference to a point in one side of the line.I just subtracted the potential of the point from the result. Both ways are equivalent.

About your plot ( of your function)
For x$\in$[-l,l], the function is undefined ( infinity) but for values of x outside that range you have finite values. The graph shows right for x<-l but wrong for x>l. It should be symmetric with respect to both X and Y axes. Try to plot with a discretisation that X axis is not included. You can get two values very close to y=0 , for example y=-0.001 and y=0.001. If still asymmetric, there is a problem in your code.

Edit: Due to the square root in your first integral, the function in post #9 must be modified duly for value of x outside the interval.

 

[haruspex]

you can calculate the equation of potential for an infinite line of charge using gauss's law.

Yes, but not directly. Gauss's law gives you the field, ~ 1/r. The potential will be the integral of this, ln(r), but what is the constant of integration? Maybe it becomes infinite.

Going back to your original formula, for the purposes of figuring out the potential as the length tends to infinity we can fix x = 0. Abbreviating (y/l)^2 to r we get
$\phi$(r) = ln(1 + 2/($\sqrt{(1+r)}$-1))
For a given y, as l tends to infinity, r tends to 0, $\phi$(r) tends to infinity.
If the problem is that the constant of integration is infinite, we can sidestep that by looking at potential differences:
exp($\phi$(r) - $\phi$(s)) = ($\sqrt{1+r}$+1)($\sqrt{1+s}$-1)/($\sqrt{1+r}$-1)($\sqrt{1+s}$+1)
For small r, s this approximates s/r.
That is, $\phi$(r) - $\phi$(s) ~ ln(s) - ln(r)
Subsituting back r = (y/l)^2, s = (z/l)^2 just applies a factor of two. The l's cancel, leaving still ~ ln(z) - ln(y).

So it seems to me that a finite charge density on an infinite wire does give an infinite potential everywhere. If you compensate for that with a suitable background charge (on a cylinder at infinity?) you can rescue a sensible field.

 

[ShayanJ]

Looks like you're saying that because $\phi(r) \rightarrow \infty$ as $r \rightarrow 0$ , $[ \phi(r)-\phi(s) ] \rightarrow \infty$ too.
But that's wrong.because we have $\infty - \infty$ which is not equal to $\infty$ but is an unknown number which should be known in the process of taking the limit.

 

[ShayanJ]

Looks like you're saying that because $\phi(r) \rightarrow \infty$ as $r \rightarrow 0$ , $[ \phi(r)-\phi(s) ] \rightarrow \infty$ too.
But that's wrong.because we have $\infty - \infty$ which is not equal to $\infty$ but is an unknown number which should be known in the process of taking the limit.

 

[Hassan2]

As as $r \rightarrow 0$ and $l$ is constant, $\phi(s)$ does NOT change. It's the potential of a fixed point. However as length $l \rightarrow ∞$ and $|r|>0$ both potentials become infinite but their difference is finite according to your argument.

When $r \rightarrow 0$ and $l \rightarrow ∞$, the difference is infinite again.

 

[haruspex]

Looks like you're saying that because $\phi(r) \rightarrow \infty$ as $r \rightarrow 0$ , $[ \phi(r)-\phi(s) ] \rightarrow \infty$ too.
But that's wrong.because we have $\infty - \infty$ which is not equal to $\infty$ but is an unknown number which should be known in the process of taking the limit.

No, I'm not saying that [ϕ(r)−ϕ(s)]→∞.
If you fix y and z but let l tend to infinity then r and s tend to zero in a related way. The potential difference converges to a finite value. If you allow r and s to tend to zero without knowing the relative rates then the potential difference becomes indeterminate.

 

[ShayanJ]

Well,I should confess I don't understand what your telling haruspex.
Could you start again and clarify more?
Thanks

And there was a problem with my substitution in the integral,I should have written$x-x'=y \ sinh \ z$ but written $x'-x=y \ sinh \ z$.The extra minus sign changed the result a little(but solved no problem)
Then by hassans advice for changing the potential origin,I got the following:
$\phi=\frac{\lambda}{4 \ \pi \ \varepsilon_0} \ln \left( {\frac {x+l+\sqrt {{x}^{2}+2\,xl+{l}^{2}+{y}^{2}}}{x- l+\sqrt {{x}^{2}-2\,xl+{l}^{2}+{y}^{2}}}} \right) +\ln \left( {\frac {-l+\sqrt {{l}^{2}+{a}^{2}}}{l+\sqrt {{l}^{2}+{a}^{2}}}} \right)$

It has the right limit.
and about the problem with the plot.In those empty places,we have $\ln{\frac{0}{0}}$
I use maple 13 for that.Looks like it finds a number for $\frac{0}{0}$ when plotting 2D and doesn't when doing it in 3D.
So looks like there is no problem with the function.
Thanks to all

 

[Hassan2]

and about the problem with the plot.In those empty places,we have $\ln{\frac{0}{0}}$
I use maple 13 for that.Looks like it finds a number for $\frac{0}{0}$ when plotting 2D and doesn't when doing it in 3D.
So looks like there is no problem with the function.
Thanks to all

I think in those empty places " on the line",we have $\ln{\frac{some\:positive\:number}{0}}$. But due to numerical error, the denominator may become -0 ( a very small negative number) so the argument of ln function can become negative which makes the ln function undefined. Empty places of x>l have no justification. They should show a finite positive potential.

 

[haruspex]

Well,I should confess I don't understand what your telling haruspex.

When your integral gave a formula for the potential that tended to infinity as l did, you thought it must be wrong. It wasn't. An infinite wire with uniform charge will produce an infinite potential everywhere. One way to get a sensible answer is to balance this charge with an equal and opposite one "at infinity". That's not quite straightforward.

As often happens when two 'balancing' terms both tend to limits, you have to specify their relationship as they do so; otherwise the result is indeterminate. E.g. y/x as both tend to 0 is indeterminate, but if I tell you x and y are linked by y = sin(x) then you can get a meaningful answer.
In the present case, you could balance the charge in a finite wire length 2l with an equal and opposite total charge spread uniformly over an enclosing (open ended, say) cylinder, 2l in length and l in radius. The details are not important, just the asymptotic behaviour.
Now as you let l tend to infinity you can hope to get a sensible potential.

An alternative is to stop worrying about the potential as an absolute entity. The enclosing cylinder introduced above would, as l increases, produce an increasingly uniform background potential inside it. So we can get the same effect by only looking at relative potentials. When we do that, we find that letting l tend to infinity produces relative potentials which would give rise to exactly the same field as deduced from Gauss's law.

Is there anything else I said that needs clarifying?

 

[ShayanJ]

An infinite wire with uniform charge will produce an infinite potential everywhere.

Why do you say that?
I guess in the case of an infinite charged wire,because we have charge at infinity,we can't choose infinity as the potential origin and if we do,we get non sense answers,like I did.

Another point.
There is nothing called absolute potential.We're always dealing with potential with respect to a potential origin which is an arbitrary point.

 

[haruspex]

Why do you say that?

Because that's what your calculation showed, and I got the same answer. But I suppose I'm also saying it didn't surprise me. So while you looked for an error in your calculations, I looked for an explanation of how it could be consistent with Gauss's Law.

There is nothing called absolute potential.We're always dealing with potential with respect to a potential origin which is an arbitrary point.

Yes, and I regard that as another way of saying what I'm saying. But more specifically, you needed to apply that principle to your formula before taking the limit. The infinity would then have disappeared.

 

[ShayanJ]

Because that's what your calculation showed, and I got the same answer. But I suppose I'm also saying it didn't surprise me. So while you looked for an error in your calculations, I looked for an explanation of how it could be consistent with Gauss's Law.

The only things that a potential function should satisfy are not gauss's law and the boundary conditions.It should also make sense mathematically and physically.
From some physical backgrounds,we can make some rude predictions about the behavior of the potential in some points and because infinity isn't a well known number and no one knows the difference between two infinities,A function which is every where infinity can't satisfy us to be the potential.

 

[haruspex]

Shyan, it's a question of limits and doing things in the right order, always a tricky area mathematically.
You take the view that potentials are always relative. Fine, let's stick with that.
So when you performed the integral to find the potential induced by a wire of length 2l, you should have expressed it as the difference in potential between two points (omitting x, ϕ(y)-ϕ(z)). Fixing y and z and letting l tend to infinity this converges to ln(z/y). The infinity is gone.
In terms of a physical reality, an infinite straight wire with uniform charge density does not fit with reality even in principle. It would carry infinite charge. You should not be surprised to get infinities as a result. It turns out that in this case they can be tamed as above or by introducing a balancing charge which drifts off to infinity (in the right way) as l tends to infinity. This works because the difference between the relative potentials in that picture and the relative potentials in the original picture tends to 0. I introduced this alternative model to show how bringing it a little more into line with reality resolves the problem, but clearly it would be algebraically more difficult.

 

[vanhees71]

I do not understand, where the problem is. Just use the Green's function of the Laplacian in free space. In Heaviside Lorentz units the electrostatic potential of a linear homogeneous charge distribution is given by

$$\Phi(\vec{x})=\int_{-L}^{L} \frac{\lambda}{4 \pi \sqrt{(x-x')^2+y^2+z^2}}=\frac{\lambda}{4 \pi} \ln \left (\frac{x+l +\sqrt{(x+l)^2+y^2+z^2}}{x-l + \sqrt{(x-l)^2+y^2+z^2}} \right ).$$

The electric field is then given by the gradient, and in the limit $l \rightarrow \infty$ this goes to

$$\vec{E}(\vec{x})=\frac{\lambda}{2 \pi r_{\perp}} \vec{e}_{\perp},$$

where $r_{\perp}$ is the distance of the point from the line of charge and $\vec{e}_{\perp}$ the corresponding unit vector pointing radially away from the line of charge.

Of course, there's also a potential for this field, but you must chose an arbitrary reference point at a finite distance from the charge distribution:

$$\Phi(\vec{x})=-\frac{\lambda}{2 \pi} \ln \left (\frac{r_{\perp}}{r_0} \right ).$$

 

[haruspex]

vanhees71, the problem, as I understand it, is that Shyan was bothered by the fact that the potential function, as you quote it, tends to infinity as L tends to infinity. I was trying to explain why that is (a) not surprising and (b) not really a problem.