The field at the surface of a conductor is always $\frac{\sigma}{\epsilon_0}$.

An infinite conducting plane has two faces, each with a surface charge density $\sigma$.

  • The field at the surface is $\frac{\sigma}{\epsilon_0}\hat{y}$, then zero inside the plate and then $\frac{-\sigma}{\epsilon_0}\hat{y}$.

  • This satisfies boundary conditions as the value of the field goes down by $\frac{\sigma}{\epsilon_0}$ at every interface with a free charge density $\sigma$.

Fair enough.

But when we have an infinite plane sheet of charge, the field is $\frac{\sigma}{2\epsilon_0}$. Obviously, this cannot be a conducting surface, which means it must be a dielectric and that this charge is bound. In this case, the electrostatic boundary conditions say that the field below the sheet must be the same. But it goes down by $\frac{\sigma}{\epsilon_0}$

Now my question, why is this so?

I think that there might be another field inside the infinite plate that satisfies boundary conditions, but I can't think of what kind of field this might be or what its magnitude could be. Some help?


1 Answer 1


A sheet of charge is neither a conductor nor a dielectric. The charges are imagined to be fixed, and there is nothing to polarize. Sometimes a sheet of charge is just a sheet of charge.

The boundary conditions are found from Gauss's Law. The electric field on crossing a sheet of charge changes by $\sigma/\epsilon_0$. This condition is satisfied in both of your examples, the conducting slab and the charged sheet.

  • $\begingroup$ Ah! That explains it. So it's just a theoretical construct...well as much more more theoretical as something can be than an infinite plate? It's a layer of free charge that's not on a conductor? $\endgroup$
    – Parabola
    Commented Mar 29, 2017 at 13:36
  • $\begingroup$ Yes, an infinite sheet of charge is an idealization that can approximate real situations, or be used in theoretical developments. $\endgroup$
    – garyp
    Commented Mar 29, 2017 at 13:48

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.