When deriving the equations for distinct charge distributions, either by the integration of the electric field of each charge $\text{d}Q$ or by Gauss' Law or any other method, we sometimes find some expressions for the electric field that are constant in magnitude. How can it be that some of these electric fields are constant?

Let me explain my confusion. For some charge distributions it is reasonable to say that the electric field is constant. For example, if we first derive the electric field for a disk using integration, then take the limit in which the size of the disk goes to infinity, the disk is then the same as an infinite sheet of charge and the expression for the magnitude can become $\sigma/2\epsilon_0$. In this case, it is reasonable to say that the magnitude is constant because no matter what the distance from the sheet is, the field will always be the same because the distance is negligible compared to the size of the sheet.

However, my confusion came when studying the electric field of a conductor. The fact that the electric field inside of the conductor is 0 is reasonable, but outside of the conductor it is found with Gauss' that it is $\sigma/\epsilon_0$. How can this be possible?! How can a non-infinite charge distribution produce a constant electric field? I'm struggling to get intuition on why this is possible without the assumption of saying that the conductor is infinite or at least large.

I would appreciate any response, thanks.


1 Answer 1


The electric field is not constant. The expression $\sigma/\epsilon_0$ that you quoted only applies to the electric field right at the surface of the conductor (in other words, infinitesimally close). At this distance, the surface will always look like a uniformly-charged plane.

This can be established from 3 reasons:

  1. All charge in an electrostatic conductor goes to the surface
  2. The electric field at the surface of a conductor is always normal to the surface
  3. The electric field is discontinuous at a surface charge by $\sigma/\epsilon_0$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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