# Field of a finite line of charge

1. May 14, 2012

### 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

#### Attached Files:

• ###### Potential of finite line charge.jpg
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2. May 14, 2012

### tiny-tim

Hi Shyan!

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

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

3. May 14, 2012

### 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!

4. May 14, 2012

### Hassan2

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).

5. May 15, 2012

### 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?

6. May 15, 2012

### 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?

7. May 15, 2012

### 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.

Last edited: May 15, 2012
8. May 15, 2012

### 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

9. May 15, 2012

### 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
$\frac{\lambda}{2\pi\epsilon_{0}}\ln\frac{a}{y}$

10. May 16, 2012

### 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

#### Attached Files:

• ###### Potential 1.jpg
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Last edited: May 16, 2012
11. May 16, 2012

### Hassan2

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.

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.

Last edited: May 16, 2012
12. May 16, 2012

### haruspex

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.

13. May 17, 2012

### 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.

14. May 17, 2012

### 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.

15. May 17, 2012

### 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.

16. May 17, 2012

### haruspex

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.

17. May 17, 2012

### 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

18. May 17, 2012

### Hassan2

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.

19. May 17, 2012

### 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?

20. May 17, 2012

### ShayanJ

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.

Last edited: May 17, 2012