# Homework Help: Electric field undergoing discontinuity

1. May 14, 2012

### zezima1

Hi, I think I need some help understanding exactly what my book means when it says "the electric field undergoes a discontinuity passing a surface charge." In fact, using Gauss' law my book directly calculates by how much the field is discontinous, so definately I must be missing something.

What exactly is meant by the field being discontinous? I thought it meant that the strength of the field jumps but it can't be so from what I know about electric fields. For instance, the book considers a metal plate of positive charge and uses that as an example of something where the field is discontinous.
But the field, from Coulombs law must be extremly big, approaching infinity the closer you get to the surface of the plate, from either side. Thus infinitesimally close to either side of the plate the electric field is infinite point out perpendicular to the surface, correct? How does discontinuity come into the picture?

2. May 15, 2012

### tiny-tim

hi zezima1!

what we call the electric field is really only the "macroscopic" electric field

(i don't know what the official name is )

we pretend that charge is continuous

in reality, if you're exactly "on" the surface, you'll be between charges, a finite distance from them, and the "microscopic" electric field will be quite complicated, and nothing like the "macroscopic" field

3. May 17, 2012

### rude man

Think of a charged metal slab with surface charge density σ C/m^2. The E field just outside the surface is E = α/ε0. And just inside the surface it's zero! Going from a finite number to zero in zero distance, that's a discontinuity. E.g. the first deivative wrt distance is infinite.