I'm trying to understand why if I consider a cylindrical capacitor whose height is $h$, the radii are $R_1$ (external) and $R_1$ (internal) and I put myself between the two plates I can say that the electric field is radial. My teacher says

For symmetry reasons

To my ears "symmetry reasons" isn't at all a reasonable explanation, I'd expect my teacher to explain what those symmetry reasons are but it didn't happen and now I'm struggling to figure out why. I already encountered this "for symmetry reasons" in another case, which is the spherical capacitor but in that case it was much easier to figure by myself: to achieve null electric field inside on the surface of the sphere the charge had to redistribute uniformly, and given any point of the space I could imagine my sphere as made of coaxial rings where of course the axis would be the radius from the center of the sphere to the point I was interested in. But now charge has to redistribute in a non uniform way to make the electric field null at any point inside, and I'm lost.


1 Answer 1


The plates are cylinders with circular cross sections and they are also equipotential surfaces.

The electric field has to be at right angles to equipotential surfaces ie the electric field is radial.

Another way of looking at the problem.

A line of uniformly distributed charge produces radial field line and equipotential surfaces which are cylinders with axis of symmetry along the line of charge and at right angles to the field lines.

Adding conductors along equipotential surfaces does not change the field distribution.

Connecting the line of charge to the inner conducting cylinder with a conductor does not charge the field lines outside the inner cylinder.
So there you have your radial field lines between the two conducing cylinders.

  • $\begingroup$ I think that what you just said about the fact that the plates are equipotential surfaces is always true cause if I want to calculate the potential difference between two points on the surface of a conductor I can always take a line which is fully inside the conductor and then at any points of the line the electric field is $\vec{0}$ and therefore the potential difference is zero. In other words the surface of a conductor is always an equipotential surface. What am I missing? $\endgroup$ Jul 26, 2019 at 15:12
  • $\begingroup$ @Bafforasta you aren't missing anything. The electric field close to a conductor is always parallel to the normal to the surface. $\endgroup$
    – lr1985
    Jul 26, 2019 at 16:39
  • $\begingroup$ @Ir1985: Yes, I know that but the two plates are not supposed to be necessarily close to each other, so how do I prove that it's radial everywhere between the plates? $\endgroup$ Jul 26, 2019 at 17:24
  • $\begingroup$ @Bafforasta the system is invariant for rotations around the axis of the cylinder. This means that the physics must remain the same if you rotate the system (or the reference frame), which implies that the electric field must have the same radial symmetry. $\endgroup$
    – lr1985
    Jul 26, 2019 at 17:45
  • $\begingroup$ @lr1985 I don't think that's enough. Symmetry wrt rotations does not imply radial field. The fields could be "slanted" and still have rotational symmetry. One also needs symmetry wrt rotations of 180 deg about an axis perpendicular to the axis of the cylinder. $\endgroup$
    – garyp
    Jul 13, 2020 at 0:54

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