# Video: Determining the Electric Flux through a Variably Oriented Surface

A square surface of area 2.0 cm² is within a uniform electric field of magnitude 1.0 × 10³ N/C. The amount of flux through the surface depends on how it is oriented relative to the direction of the electric field. Find the electric flux through the surface when a line normal to the surface makes an angle of 30° with the electric field. Find the electric flux through the surface when a line normal to the surface is perpendicular to the electric field. Find the electric flux through the surface when a line normal to the surface is parallel to the electric field.

05:40

### Video Transcript

A square surface of area 2.0 centimeters squared is within a uniform electric field of magnitude 1.0 times 10 to the third newtons per coulomb. The amount of flux through the surface depends on how it is oriented relative to the direction of the electric field. Find the electric flux through the surface when a line normal to the surface makes an angle of 30 degrees with the electric field. Find the electric flux through the surface when a line normal to the surface is perpendicular to the electric field. Find the electric flux through the surface when a line normal to the surface is parallel to the electric field.

In this exercise, we have two elements: an electric field of given magnitude and a square surface of given area. We want to solve for the electric flux through the square surface in three different orientations of the surface to the electric field. Let’s start out by considering the first one which is when the normal to the surface makes an angle of 30 degrees to the direction of the electric field.

If we sketch out what’s happening in this first case, we can imagine our electric field is moving left to right in lines parallel to one another; it’s a uniform field. Within this uniform field, there is a square area which is arranged so that if we were to draw the normal vector that is the vector perpendicular to that area whose length represents the magnitude of the area, then we’re told that the angle between that area vector and the vector of the electric field is 30 degrees. Calling that angle 𝜃, we’re given that 𝜃 is 30 degrees.

The question is “given this orientation of the square surface and its size on the electric field magnitude, what is the electric flux through the surface?” To figure this out, let’s recall an equation that brings these three elements — electric field magnitude, surface area, and orientation — altogether.

In general, the electric flux Φ sub 𝐸 is equal to the magnitude of the electric field involved multiplied by the area through which that field may move multiplied by the cosine of 𝜃, where 𝜃 is the angle we’ve sketched in in our diagram. That is, in general, 𝜃 is the angle between the electric field 𝐸 and the normal to the area 𝐴.

So for this case, that is for this electric field, this area, and this angle, we know that 𝜃 is 30 degrees and that 𝐸 is 1.0 times 10 to the third newtons per coulomb. With these values plugged in, we also know that the area of our square surface 𝐴 is 2.0 square centimeters.

But before we go ahead and calculate electric flux, we’ll want to make sure that all of the units involved are in SI standard form. The units of newtons as well as coulombs passed that test. But the SI standard unit of length is not centimeters, but meters. So we want to convert this area.

Knowing that 100 centimeters is equal to one meter, that means to convert an area, we’ll need to square that equation so that 100 centimeters all squared is equal to one meter all squared. This works out to 10000 square centimeters is equal to one square meter. In other words, 2.0 square centimeters is equal to 0.0002 square meters.

Now, we’re all set with our units and we’re ready to solve for Φ sub 𝐸, the electric flux through our square. To two significant figures, this works out to 0.17 newton meters squared per coulomb. That’s the electric flux through our square surface oriented this way.

But now, what if we change the orientation of this surface? What if instead of the original angle, we rotate it so that it’s now parallel with the direction of the electric field lines.

In this case, if we draw the area vector that is the vector normal to the area of our square surface, we see it’s oriented at 90 degrees with our uniform electric field. This is an interesting angle because as we consider our electric flux through this newly oriented square surface, we know it will be multiplied by the cosine of 𝜃, where 𝜃 is now 90 degrees.

But wait a second! What’s the cosine of 90 degrees? It’s zero. So that means regardless of the values of the electric field in the area, when they’re arranged this way so that’s no electric field lines pass through that area, the flux through the area is zero. At this particular angle then, our electric flux is zero newton meters squared per coulomb.

Finally, in part three of the problem, we want to change the orientation of the square surface one more time. In this case, instead of being parallel with the electric field lines, the square is now perpendicular to them. Once more, we want to calculate electric flux through this square surface.

And we now see that a vector normal or perpendicular to this area points in the same direction as the uniform electric field. Based on that, we know that 𝜃, the angle between these two vectors, is zero degrees and the cosine of zero degrees we know to be one.

This means that the cosine of 𝜃 term effectively drops out of our calculation for flux. And all we need to do to find it is to multiply 𝐸 by 𝐴, the magnitude of the electric field by the magnitude of our area. We know that the electric field magnitude is 1.0 times 10 to the third newtons per coulomb. And we saw earlier that the overall area of our square in units of square meters is 0.0002.

Multiplying these values together, we find a result of 0.20 newton meters squared per coulomb. This is the electric flux through our square surface. And note that this is the maximum possible electric flux through this surface for any orientation it may have in this uniform field.