Assume that the linear charge density is the same for the charged infinite cylinder and the infinite line. By Gauss' Law, I know the charge enclosed is the same given a Gaussian cylinder of a certain length and so the E fields must be the same at any given distance from the axis of symmetry for both objects, given that the distance is greater than the radius of the cylinder.

I just don't understand intuitively why this is the case. For a charged infinite cylinder, since the charges are distributed only on the surface of the cylinder, shouldn't the E field produced by the cylinder at a distance r > a (a being the radius of the cylinder) from the axis of symmetry of the cylinder be greater in magnitude than that produced by an infinite line of charge at the same radius r?


Maybe it helps to compare your cylinder with a cylinder with a tiny radius, but with the same linear charge density. To have the same linear charge density, the smaller cylinder has to have a large surface density, because its surface per unit length is smaller.

  • $\begingroup$ Out of all of the explanations on this thread, I think this one is the simplest and the clearest. $\endgroup$ – ABCDF Dec 25 '16 at 17:59

Contours of equal field strength are concentric rings around a single charge (in 2D, spheres in 3D).

So if you instead create a ring of equal charges it's as if you created one of those contours, from outside the ring and in the plane of the ring anyway.

Inside the ring it is v. interesting because there is an integral proof that the summed forces cancel, there is no field strength. This is a starting point for modelling the fairly well known Farady cage.

Fields propagating between the charges neighbouring each other on the ring are summed not lost, so the charges on the ring individually produce weaker fields than the contour.


Comparing the fields of the line and the cylinder, there are a couple of important differences. First, there are differences in distance. For the cylinder, some of the charges are closer to the point at which are measuring the field than the distance to the symmetry axis, and some of the charges are farther away. Because of the $1/r^{2}$ behavior of the field, you might be inclined to believe that the increase in the field due to the closer charge would be greater than the decrease due to the charges farther away. However, this is not the only effect.

The other difference between the two configurations, which weakens the physical electric field, is that the bits of charge "off to the sides" of the cylinder produce fields that are not pointed in the radial direction. Everything except the radial field cancels by symmetry, so the oblique components of the field are effectively "lost." This further weakens the field (relative to what I discussed in the previous paragraph) and the result is a field identical to that of the line charge.

  • $\begingroup$ "This is not the only effect" - indeed the other key effect is that there is a lot more charges on the further away side than there are on the near side. Just the right amount to cancel the field. $\endgroup$ – JMLCarter Dec 23 '16 at 21:02
  • $\begingroup$ Non radial fields sum. they are not "lost". (Only fields from opposite charges cancel.) $\endgroup$ – JMLCarter Dec 25 '16 at 2:48
  • $\begingroup$ The forces due to the fields do cancel radial, to be fair, though. $\endgroup$ – JMLCarter Dec 25 '16 at 3:04
  • $\begingroup$ Can you clairfy by what you mean when you say "the bits of charge off to the sides?" $\endgroup$ – ABCDF Dec 25 '16 at 18:01

Imagine that you have your line of charge with a certain charge per unit length. This will produce a radial electric field whose magnitude is inversely proportional to the distance from the line of charge.
The equipotential surfaces will be concentric cylinders centred on the line of charge.

Now add an uncharged conducting cylinder along an equipotential surface.
Nothing changes with the electric field distribution and the equipotential surfaces staying exactly the same.
The outside of the conducting cylinder will have the same charge per unit length as the line of charge and the inside of the cylinder the same charge per unit length but of opposite sign.
Those charges on the cylindrical conductor are induced charges.

Now connect the line of charge and the cylinder with a conductor.
Inside the cylinder there is no longer any charge and so there is no electric field and it is an equipotential volume.
Outside nothing changes with the electric field the same as that which was produced by the line of charge.

The key thing is that the conductor was added along an equipotential surface.

  • $\begingroup$ I still don't exactly understand why the line of charge+cylindrical conductor+conductor to connect the two would produce the same electric field as the original line of charge. $\endgroup$ – ABCDF Dec 25 '16 at 17:51
  • $\begingroup$ It is because the electric field outside the cylindrical conductor is unchanged by the neutralisation of the charges inside it. You are really just moving the lines charges onto the outer surface of the cylindrical conductor. $\endgroup$ – Farcher Dec 25 '16 at 19:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.