# What is the electric field due to a 1D line of charge on the edges of a pillbox geometry?

The "Introduction to Plasma Physics C17 Lecture Notes", which I'm reading online, set out the following scenario in section 1.1.2:

Consider the electric field due to a 1-D line of charge (see Fig. 1.2). Applying Gauss' theorem to the pillbox shown, we find $$\int E.ds = 2AE = \rho A dx/\epsilon_0$$

Here is the diagram:

I'm struggling to reconcile this with my understanding of Gauss' law. Looking at the above scenario I would say that the total charge $$Q$$ is equal to the charge density times the height, $$A$$ (area in 1 dimension). The total flux is the $$E$$ times the area at each side, so $$2A$$. Equating these via Gauss' law gives:

$$\Phi_E = 2AE = Q/\epsilon_0 = \rho A/\epsilon_0$$

This is not the same as the result in the lecture notes ($$\rho A dx / \epsilon_0$$). Why is the $$dx$$ relevant here? I have a suspicion that it's to do with my (mis)understanding of the geometry of a "Pillbox", but I'm not sure.

• Note that your charge density appears to be per unit length, whilst the author's is per unit area. Jun 7, 2020 at 13:53
• @jacob1729 in the lecture notes the author says $\rho$ is the charge per unit length of the line. Jun 7, 2020 at 13:56
• If $\rho$ is charge per unit length then the author's expression $\rho A dx$ is not a charge. They must be interpreting $\rho$ as charge per unit volume (area). Jun 7, 2020 at 13:58