# Is Griffiths simply wrong here? (Electrostatic Boundary Conditions)

Two ways of seeing that it’s right:

1. Consider the case of no charge. Then nothing interesting is happening at the sheet, so the fields should be equal: both sides of the equation are zero.
2. Gaus’s law: the sum of the fields going away, which is the outward flux, is given by the charge. Since $$E_{below}$$ is defined as towards the charge, it enters that calculation with a minus sign.

He’s picked a sign convention where the field upward is positive everywhere. That means that $$E_{below}$$ is defined such that a positive value means "the E vector points up" and a negative value means "the E vector points down". Which way the E vector points is given by the physics:

• if there's positive charge on the surface and no external field, the E field below it will point down, hence $$E_{below}$$ will have a negative value.
• If there's a large upward going external field, then in that case the E field points upward everywhere, and $$E_{below}$$ will be positive.

To put it another way, the length of the $$E_{below}$$ arrow in the picture isn't showing you the absolute magnitude; that has to come from somewhere else and is written as "$$E_{below}$$", a number. And it's not even showing you the direction, because if $$E_{below}$$ is less than zero, the actual E vector is pointing the other way. That arrow there is just defining a direction, like $$\hat{x}$$, $$\hat{y}$$ and $$\hat{z}$$.

Griffiths in correct. The flux through the top portion of the box is not just $$E_{\small{top}}^\top$$ but actually $$\vec E\cdot (A\vec n)$$ where $$\hat n$$ is perpendicular to the surface and points out: in your diagram, $$\hat n$$ would be along $$+{\hat z}$$ for the top portion of the flux calculation.

For the bottom of the box the $$\hat n$$ vector points along $$-\hat z$$ so the flux through through those sides of the box work out to $$(E_{\small{top}}^{\top} - E_{\small{bottom}}^\top)A \tag{1}$$ when $$\vec E_{\small{bottom}}$$ points along $$+\hat z$$: the minus sign in (1) comes from the $$-\hat z$$ for the direction of the surface element $$d\vec A$$ at the bottom of the box.

Thus, ignoring the thin sides of the box because the flux through those is arbitrarily small, you have $$(E_{\small{top}}^{\top} - E_{\small{bottom}}^\top)A=\frac{1}{\epsilon_0} \sigma A$$ and the area cancels out.