# Electric field due to infinitely charged plane does not satisfy boundary conditions

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

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.

• 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? – Parabola Mar 29 '17 at 13:36
• Yes, an infinite sheet of charge is an idealization that can approximate real situations, or be used in theoretical developments. – garyp Mar 29 '17 at 13:48