# What is the gravitational/electric/inverse-square field inside a cylinder?

I've read from the shell theorem that an inverse-square potential has zero field inside a spherical shell. What about the field inside a cylinder? Are objects inside a long cylinder attracted to the center, or to the sides? Is there a simple analytic form for the field in a (possibly infinite) cylinder?

(Edit: To be more precise I think I should have said that electric charge is evenly distributed over the surface of the cylinder - I think this allows us to use the same result for a gravitational field as well as an electric field. I guess that's not a very realistic assumption for electrical applications, since charge tends to redistribute itself to create a constant potential (at least if the surface is conducting). Also, my original intention was to ask about an open-ended cylinder, but I think it doesn't matter so much if the cylinder is long.)

• @Josh: It's awesome that you referred me to your hand-written notes from 1972 (and it's awesome that you saved them and put them online; cool handwriting too!). However, it seems that the boundary conditions you treat are slightly more or less general than the ones implied by my question. Perhaps I could do the computations myself and learn something about Bessel functions, but I can't very easily look at the formulas and figure out if objects are attracted to the center or the sides of the cylinder. Anyway, I guess that's why you commented instead of posting an answer :) – Metamorphic May 30 '16 at 4:52

The field inside an hollow infinite cylinder is $0$, just like the field inside an hollow sphere.

This is because of Gauss' law: the flux of the electric field $\vec E$ through any closed surface $S$ is

$$\Phi = \int_S \vec E \cdot d \vec S = \frac Q {\epsilon_0}$$

Where $Q$ is the charge inside the volume enclosed by the surface.

Let $R$ be the radius of our cylinder and let's take a cylindrical surface with radius $r<R$ coaxial to it. Since the charge is all on the surface we will have

$$\Phi = \int_S \vec E \cdot d \vec S = 0$$

Now since the cylinder is infinite the field must be directed in the radial direction for symmetry, so that $\vec E \cdot d \vec S = E d S$, hence

$$\Phi = E \int_S d S = E S = 0$$

Which means that $\vec E$ must be $0$. The same is true also for the gravitational field.

If the cylinder is not infinite, the last part of the argument is not valid and the field should be 0 only at the exact center of the cylinder if I'm correct.