# Infinitely charged wire and Differential form of Gauss' Law

I have tried calculating the potential of a charged wire the direct way. If lambda is the charge density of the wire, then I get

$$\phi(r) = \frac{\lambda}{4 \pi \epsilon_0 r} \int_{-\infty}^\infty \frac{1}{(1+z^2/r^2)^{1/2}} dz.$$ But this comes to $+\infty$ unless I am doing the calculation wrong. Why doesn't this work the direct way?

Also, is it possible to calculate the potential of a charged wire using Gauss' differential law? What about in the case of an infinite charged sheet? Or does Gauss' differential law only apply to charged volumes?

• The potential of an infinitely long charged wire with uniform (line) charge density at a point in space is equal to the sum of the potentials of an infintesimal part of the wire (at that point in space). Sum up these potentials, and that's the potential of the wire at a point in space. That's what I am doing above. – Adam Rubinson Apr 21 '12 at 13:43
• Your calculation is right. You obtain the same result if you calculate potential from the electric field of infinite wire. The question is why is that so, that is, why is potential in space infinite compared to infinity for infinite wires. I am wondering too... – Pygmalion Apr 21 '12 at 13:50
• What? The potential is infinity? How does that match with the fact that the electric field is proportional to 1/r? – Adam Rubinson Apr 21 '12 at 13:56
• Potential is infinity compared to potential at infinity. But you can still find the difference in potentials between two points, that are let's say 1m and 2m away from the wire. And that is what matters and what proves that electric field non-infinitive. – Pygmalion Apr 21 '12 at 14:05
• @Adam Rubinson, As was mentioned in one of the answers, you are implicitly assuming that the potential at $\infty$ is zero. This is valid only for localized charge distributions. In your case the charge distribution extends to $\infty$ and so you should NOT take your zero of potential at $\infty$. Choose any other point $\mathcal{O}$. The choice of $\mathcal{O}$ does not matter since the electric field comes out the same irrespective of $\mathcal{O}$. See Griffiths section 2.3.2. – Vijay Murthy Apr 21 '12 at 18:35

## 4 Answers

1. The infinitely long wire has an infinite charge $$Q~=~\lambda \int_{-\infty}^{\infty} \! dz ~=~ \infty,\tag{1}$$ and EM has an infinite range, so one shouldn't be surprised to learn that the result $$\phi(r)~=~ \frac{\lambda}{4 \pi \epsilon_0} \int_{-\infty}^{\infty} \frac{dz}{\sqrt{z^2+r^2}} ~=~ \frac{\lambda}{4 \pi \epsilon_0} \left[ {\rm arsinh} \left(\frac{z}{r}\right)\right]_{z=-\infty}^{z=\infty} ~=~\infty \tag{2}$$ is infinite. (From a mathematical point of view, the integrand fails to be integrable wrt. the $z$ variable.) A similar situation happens often in Newtonian gravity if the total mass is infinite, see e.g. this question.

2. However, as Pygmalion mentions in his answer, the electric field $\vec{E}$ is well-defined for $r\neq 0$, and the corresponding integrand is integrable wrt. the $z$ variable. E.g. the radial component (in cylindrical coordinates) reads $$E_r(r)~=~ \frac{\lambda r}{4 \pi \epsilon_0} \int_{-\infty}^{\infty} \frac{dz}{(z^2+r^2)^{3/2}} ~=~ \frac{\lambda r}{4 \pi \epsilon_0} \left[ \frac{z}{r^2\sqrt{z^2+r^2}}\right]_{z=-\infty}^{z=\infty} ~=~\frac{\lambda}{2\pi\epsilon_0 r}\tag{3}$$ for $r\neq 0$.

3. Alternatively, apply Gauss' law $$d\Phi_{E} ~=~\frac{dQ}{\epsilon_0},\tag{4}$$ using an infinitesimally thin disk perpendicular to the wire. The disk has radius $r$ and thickness $dz$. The total electric flux $d\Phi_{E}$ out of the disk is $$E_r \cdot 2\pi r dz ~=~ \frac{\lambda dz}{\epsilon_0},\tag{5}$$ which leads to the same electric field $E_r$.

4. This electric field $\vec{E}=-\vec{\nabla}\phi$ is consistent with a potential of the form $$\phi(r) ~=~-\frac{\lambda}{2\pi\epsilon_0}\ln r \qquad \text{for}\qquad r\neq 0.\tag{6}$$

• +1 for actually doing the math. But you've missed out on the differential form of Gauss' law ($\nabla\vec E=\rho$). $\vec E=-\nabla\phi$ is NOT the diff form of Gauss' law. – Manishearth Apr 22 '12 at 14:53
• Well, one could argue that the Gauss' law used is in some sense differential, since the thickness is infinitesimal. – Qmechanic Apr 22 '12 at 16:04
• Ahh I see. Though when talking about Maxwell's equations, differential f orm' has a specific meaning. – Manishearth Apr 22 '12 at 17:25
• Yes I agree, cf. en.wikipedia.org/wiki/Gauss%27s_law#Differential_form – Qmechanic Apr 22 '12 at 17:30
• @Qmechanic What do you think about my discussion with Manishearth about self energy/potential? – Pygmalion Apr 22 '12 at 17:58

You're not calculating the potential here, you're calculating the self energy of the wire. The self energy is the energy required to bring charges from infinity to create the wire. And this is obviously infinite.

The potential of a wire is defined as the energy required to bring a point charge from infinity to a point on the wire. Note that, by this definition, the potential of the wite is infinite as well. Also note that not all objects have a defined "potential". A uniformly charged sphete has a varying potential at different points, so we cannot assign a "potential" to it (it has a self energy though).

No, the differential form of Gauss' law cannot be used simce $\rho$ is infinite at points. I guess you could use limits and do it via the differential form. Or you could convert it to the integral form via Stokes law and do it(sort of cheating). Usually its best to use the integral form when you have infinite volume charge densities.

• I do agree with your comment and it gives rational explanation (thank you to remind me to that!), but there is one thing I disagree. Every potential calculation is actually calculation of self energy of the system that exist in space. So he was calculating potential/self energy, but in this case it equals infinity. – Pygmalion Apr 21 '12 at 14:02
• @Pygmalion Potential=energy to get a test charge from infinity to specified point. Self energy=energy reqd to create system. Potential is for a point, self energy is for a system. this gets even morcomplicated when we have multiple objects, it can be split into self energy, interaction energy, and you will still have potentials at every point, independant of the other two – Manishearth Apr 21 '12 at 14:05
• You say "This is because different points have a different potential on a wire." However, by symmetry, all the points on a wire should have exactly the same potential. By the way, this wire is infinite (forgot to mention). – Adam Rubinson Apr 21 '12 at 14:06
• @Adam oh whoops. Alright, the wire is equipotential(though its potential taking infinity as a datum is infinite). For a general object, this does not hold though. – Manishearth Apr 21 '12 at 14:09
• OK then, but then Adam did not calculate self energy. Nevertheless I thing that self energy and potential are somehow related. – Pygmalion Apr 21 '12 at 14:44

First, you must be aware that you was calculating the potential in space, under assumption that potential at infinity is zero.

You would obtain the same result if you calculated potential using the expression for electric field around infinite wire

$$E = \frac{\lambda}{2\pi\epsilon_0 r}$$

I guess it is general property of all infinite distributions of charges that they give infinite potential in whole space, regarding to the potential at infinity. But there can exist even some more fancy theoretical explanation.

Try starting with a different problem: an infinitely long cylinder of non-zero radius, with a uniform surface charge density, nested inside an infinitely long, coaxial, conducting cylinder with non-infinite radius. Use the integral form of Gauss' Law + symmetry; the math is straightforward. (You can set the potential of the outer cylinder to zero for definiteness.) Once you have that solution, you can see how it blows up as the inner cylinder's radius goes to zero, or as the outer's goes to infinity.