We talked already about how to calculate the electric flux through a flat surface and through an enclosed cube for a constant electric field. But what happens if the field is not constant? Or what if the surface is no longer flat? These notes will show how we modify the electric flux equation to account for varying fields and curved surfaces.

Before we said that for a flat surface, the area vector is given by the magnitude of the area times the vector that points perpendicular to the area. This makes sense for a flat surface, where the area vector will point in the same direction for all points on the surface. However, for a curved surface, this no longer makes sense. It becomes impossible to use a single vector to describe the surface.

But if we zoom in on a very small piece of the area (call this dA), this extremely small area will appear to be flat. Thus, we can write the area vector for this very small piece of area using our normal rule. Namely, that the vector for the very small piece of area ($\vec{dA}$) will be equal to the magnitude of the small chunk of area ($dA$) times the vector that is perpendicular to that small chunk ($\hat{n}$): $$\vec{dA}=dA \hat{n}$$

For example, take a hemisphere shell as a surface area. On the very top of the hemisphere, the little piece of area is horizontally flat, so the $\vec{dA}$ at this location would point straight up. If you move to either side of the top of the hemisphere, the dA would be locally flat (tangent to the hemisphere surface), so at that location, the $\vec{dA}$ would point perpendicular to the surface. No matter where you pick as your dA, the $\vec{dA}$ should point perpendicular to that small piece of area. In the case of the curved part of a hemisphere (or a whole sphere for that matter), the $\vec{dA}$'s all point radially away from the center of the sphere. So we could write any general $\vec{dA}$ for a sphere as: $$\vec{dA}=dA \hat{r}$$

To find the electric flux then, we must add up the electric flux through each little bit of area on the surface. In terms of calculus, this would mean we first would write the little bit of flux ($d\Phi_e$) as the cross product of the electric field through the little bit of area ($\vec{E}$) and the little area vector ($\vec{dA}$): $$d\Phi_e = \vec{E} \cdot \vec{dA}$$ Then to add up all of the little bits of electric flux, we would take the integral of both sides (over the whole area that you have): $$\Phi_e=\int d\Phi_e=\int \vec{E} \cdot \vec{dA}$$ This equation for electric flux is the most general equation that is always true - we have not made any assumptions about the kind of electric field or area shape. This means that this equation will always work to calculate the electric flux; however, the calculus can become very complicated very quickly if you are not careful. The next section of notes will explain how we make use of symmetry to make these calculations easier.

* Flux = integral of E dot dA
* What is dA? Why do you need it?
* Direction of dA
* Symmetry arguments
  * Dot product
  * E const