### Video Transcript

Consider the uniform electric field π equals quantity 4.0π£ plus 3.0π€ 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 π₯π¦-plane?

We can call the radius of our circular area 2.0 meters π. We want to solve for the electric flux that goes through the circular area. Weβll call that flux π sub π. If we draw a sketch of the circular area on the π₯π¦-plane, we can see that this circular area which weβve centered on the origin of the π₯π¦-plane has a radius π of 2.0 meters.

Looking at the equation for our uniform electric field, π, we see that it has a π€ or π§- direction component to it. Since our circular area lies in the π₯π¦-plane, only that component, that π§-component, of the uniform electric field will create electric flux through this area.

If we were to draw that component of π on our diagram, if positive π§ points out of the page towards us, then the π€- or π§-component of π would point through our circular area and out of the page. Itβs this flux which represents π sub π, the electric flux passing through our circular area.

In general, electric flux, π sub π, is equal to the electric field π multiplied by the area through which the electric field moves times the cosine of the angle between the electric field vector, π, and the normal to the area, π΄.

In our case, we can write that electric field flux, π sub π, equals the π§-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 π§-component of our electric field, π, as well as the area of our circle.

That π§-component of π is 3.0 times 10 to the third newtons per coulomb. And since π is 2.0 meters, π΄ is π times π squared. When we multiply these numbers together on our calculator, we find that π is 3.8 times 10 to the fourth newtons per coulomb meter squared. Thatβs the electric flux that passes through the circular area.