Consider the uniform electric field 𝐄 equals quantity 4.0𝐣 plus 3.0𝐤 times 10 to the third newtons per coulomb. What is the electric flux through a circular area of radius 2.0 meters that lies in the 𝑥𝑦-plane?
We can call the radius of our circular area 2.0 meters 𝑟. We want to solve for the electric flux that goes through the circular area. We’ll call that flux 𝜑 sub 𝐄. If we draw a sketch of the circular area on the 𝑥𝑦-plane, we can see that this circular area which we’ve centered on the origin of the 𝑥𝑦-plane has a radius 𝑟 of 2.0 meters.
Looking at the equation for our uniform electric field, 𝐄, we see that it has a 𝐤 or 𝑧- direction component to it. Since our circular area lies in the 𝑥𝑦-plane, only that component, that 𝑧-component, of the uniform electric field will create electric flux through this area.
If we were to draw that component of 𝐄 on our diagram, if positive 𝑧 points out of the page towards us, then the 𝐤- or 𝑧-component of 𝐄 would point through our circular area and out of the page. It’s this flux which represents 𝜑 sub 𝐄, the electric flux passing through our circular area.
In general, electric flux, 𝜑 sub 𝐄, is equal to the electric field 𝐄 multiplied by the area through which the electric field moves times the cosine of the angle between the electric field vector, 𝐄, and the normal to the area, 𝐴.
In our case, we can write that electric field flux, 𝜑 sub 𝐄, equals the 𝑧-component of our electric field times the area of our circle, recalling that the area of a circle in general is equal to its radius squared times 𝜋. We can now plug in for the 𝑧-component of our electric field, 𝐄, as well as the area of our circle.
That 𝑧-component of 𝐄 is 3.0 times 10 to the third newtons per coulomb. And since 𝑟 is 2.0 meters, 𝐴 is 𝜋 times 𝑟 squared. When we multiply these numbers together on our calculator, we find that 𝜑 is 3.8 times 10 to the fourth newtons per coulomb meter squared. That’s the electric flux that passes through the circular area.