# Video: Finding the Electric Flux through the Surface of a Square Region of Charge

A uniform electric field of magnitude 2.3 × 10⁴ N/C is perpendicular to a square sheet with sides 4.0 cm long. What is the electric flux through the sheet?

02:21

### Video Transcript

A uniform electric field of magnitude 2.3 times 10 to the fourth newtons per coulomb is perpendicular to a square sheet with sides 4.0 centimeters long. What is the electric flux through the sheet?

In this situation then, we have a sheet with sides of equal length: 4.0 centimeters. We also have a uniform electric field at that’s oriented perpendicular to the sheet. We could draw that field like this. And since it’s perpendicular to the square sheet, we know that some of the field lines are moving through that sheet.

That electric field, which we’ll label 𝐸, has a magnitude of 2.3 times 10 to the fourth newtons per coulomb. And knowing all this, we want to solve for the electric flux through the sheet. What flux tells us is just how much of our electric field 𝐸 passes through the area of this square sheet.

When the direction, the electric field, points and the plane of the area that we’re considering and moving through are perpendicular to one another as they are in this case, then we can write the electric flux, Φ sub 𝐸, as the product of the electric field strength multiplied by that area.

In our case, we already know the electric field magnitude. That’s given to us in the problem statement, and we’ve called it 𝐸. The area 𝐴 is the area of our sheet that the field was moving through. Since the side lengths of our sheet are 4.0 centimeters, we could see the area is 4.0 centimeters squared. But we’ll want to convert those length units from centimeters to meters. We do this so that all the units in our expression are in SI base units.

To make this conversion, we recall that 100 centimeters is equal to one meter in length. This means the area of our sheet is 0.04 meters, the length of one side, multiplied by 0.04 meters, the length of a perpendicular side, or written in another way 0.04 meters all squared. That’s the area of our square sheet.

At this point, to solve for electric flux, all we have left to do is to multiply these numbers together. And notice the units we’ll end up with. The units will be newtons meters squared per coulomb. To two significant figures, we find that Φ sub 𝐸 is 37 newton meters squared per coulomb. That is the electric flux through this sheet.