This is an old revision of the document!
Example: Flux through a Closed Cylinder
A constant electric field →E is directed along the x-axis. If we imagine a cylindrical surface with radius R and height h is situated in the field so that the bases of the cylinder are parallel to the xz-plane. What is the electric flux through the cylinder?
Facts
- The cylinder is closed.
- The cylinder has radius R and height h.
- The electric field is directed along the x-axis, which is parallel to the bases of the cylinder.
Lacking
- Φcylinder
- →A of cylinder, or d→A pieces.
Approximations & Assumptions
- The electric field is constant.
- The electric flux through the cylinder is due only to →E.
- We can approximate infinitesimally small pieces of the cylinder's area as flat.
Representations
- We represent the electric flux through a flat surface with:
Φ=→E∙→A
- We represent the situation with the following diagram:
Solution
Before we dive into the math, let's reason about the nature of the situation. It will be helpful to visualize how the area vector d→A would look for different parts of the cylinder's surface. Here is a visual from a couple perspectives, where each arrow represents the direction of d→A at its location. Remember that the convention for closed surfaces is that area-vectors are directed outside.
Notice that the area vectors on the bases of the cylinder are pointing along the y-axis. Since the electric field is aligned with the x axis, there will be no flux through the top and bottom of the cylinder. The math is here:
Φtop=→E∙→Atop=(Eˆx)∙(πR2ˆy)=0

I think it'd be worth showing the flux through A1 and the flux through A2 to emphasize that the parallel E bit will make a positive flux and an anti-parallel E bit will make a negative flux so they cancel. I think we're missing that step.
Within the sum, we can match up the little pieces of area with their opposites (like →A1 and →A2 in the figure). It shouldn't be too surprising at this point to see that everything cancels out. When we match up the terms, we get something like: Φmatching pieces=ΦA1+ΦA2=→E∙→A1+→E∙→A2=→E∙(→A1+→A2)=→E∙(0)=0