I Discontinuity of Electric field

[mondo]

Hi,
While reading griffith introduction to electrodynamic I have stumbled upon this:

I don't understand why when the pillbox hight goes to zero we have a discontinuity of electric field. On the figure 2.36 we can see the electric field penetrates the surface from below the plane and exits from its top so the magnitude of it must be the same hence I would expect $$E_{below} - E_{above} = 0$$ instead. In the same way as if n people entered a store and then n left what is left is 0 people.

Another thing that I don't understand in this paragraph is, author calculated the surface integral of E and gets $$E = \frac{1}{\epsilon_0}\sigma A$$ why does he calculate the integral here if all we use in this example is plain electric field at a point?

 

[Doc Al]

I don't understand why when the pillbox hight goes to zero we have a discontinuity of electric field.

Realize that there is a non-zero surface charge contained within the pillbox.

 

[Orodruin]

Griffiths is applying the divergence theorem:
$$
\oint_S \vec E\cdot d\vec S = \int_V \nabla \cdot \vec E \, dV$$ where V is the volume enclosed by the closed surface S. Because $\nabla \cdot \vec E = \rho/\epsilon_0$, the flux integral over the pillbox surface is therefore equal to the charge contained in the pillbox divided by the permittivity in vacuum, which is $\sigma A/\epsilon_0$. Just computing the flux integral directly for a thin pillbox gives $\Delta E_\perp A$. Identifying the expressions gives the desired result.

 

[mondo]

So does the equation 2.31 mean - we expect this field to be zero (left hand side of the equation) but it needs to be $$\frac{1}{\epsilon_0}\sigma $$ (right side of the equation) because of the presence of the surface charge?

 

[Doc Al]

So does the equation 2.31 mean - we expect this field to be zero (left hand side of the equation) but it needs to be $$\frac{1}{\epsilon_0}\sigma $$ (right side of the equation) because of the presence of the surface charge?

Yes. If there were no surface charge, just a surface in space, you'd expect the field above and below the surface to be the same -- thus the difference would be zero.

But if there's a surface charge the field above and below would have to be different -- due to the field of the surface charge itself.

 

[mondo]

@Doc Al , it sort of makes sense however, isn't the electric field above and below the gaussian pillbox generated by the surface charge on the surface? Therefore it should be the same below, above and on the surface.
And even if there are other changers in the space above or below the surface of interest, then the electric field should be a vectorial sum of all of them hence I am not sure why do we magically end up with a discontinuity on the surface if everything around it is already affected by its presence.

 

[Doc Al]

@Doc Al , it sort of makes sense however, isn't the electric field above and below the gaussian pillbox generated by the surface charge on the surface? Therefore it should be the same below, above and on the surface.

Realize that the field from the surface charge points in different directions above and below the surface. Those fields might have the same magnitude, but not the same direction. Don't forget that the field is a vector!

 

[Meir Achuz]

Eq. 2..31 is just Gauss's theorem.