A cube whose sides are of length 𝑑 is placed in a uniform electric field of magnitude 𝐸 equals four 4.0 times 10 to the third newtons per coulomb, and the field is perpendicular to two opposite faces of the cube. What is the net electric flux through the cube?
Let’s start off by drawing in our cube and then this field that moves through it. Here’s our cube, and we’re told the length of each side of our cube it is equal to 𝑑. We’re also told that a uniform electric field is present and the field is oriented in such a way that it’s perpendicular to two opposite faces of the cube. We could pick any opposite pair of faces in the cube. Let’s just pick the ones on the left and on the right so the electric field moves perpendicular to these two faces.
If we draw that in, that means the field looks something like this, where there are right angles between the electric field and the two faces of the cube that it passes directly through. Given all this, we want to solve for the net electric flux through the cube. And here, “net” is an important word. When we recall that the electric flux through a certain area 𝐴 is equal to the electric field magnitude through that area multiplied by the cosine of the angle between the normal to the area and the electric field, then we can see that for four of the six sides of this cube the electric flux through those sides is zero.
That’s true for the face on the front, the face in the back, the face on top, and the face on bottom. For all four of those faces, there is no electric field that passes through them. So the only two faces that can contribute to the electric flux are the left and the right faces, the ones with the field moving through them. Here’s how we’ll write that. We’ll say that the total or net electric flux through the cube is equal to the flux through the left face plus the flux through the right face as we’ve draw them.
Since our equation for electric flux relies on the normal vector to the areas that that flux is through, let’s go over to our cube to the left and the right faces and draw in the normal vectors to those two areas, the two faces that the electric field passes through. The area vector of the left face of our cube may look like this. We’ll call that vector 𝐴 sub 𝐿. And the area vector for the right face of our cube may look like this. We’ll label that 𝐴 sub 𝑅. Since these are two area vectors representing opposite faces of a cube, we can say that their magnitudes are equal but their signs are opposite; that is, 𝐴 sub 𝐿 is equal to negative 𝐴 sub 𝑅.
This realization is important because it tells us that when we find the cosine of the angle between 𝐴 sub 𝐿 and the electric field 𝐸, since that angle is 180 degrees, the cosine of 180 degrees is negative one. That tells us that the flux through the left face of our cube is negative 𝐸 times 𝑑 squared, the area of that face. On the other hand, the electric flux through the right face of the cube, where the angle between the normal vector and the electric field vector is zero degrees, must be equal to positive 𝐸 times 𝑑 squared.
When we add these two results together, we find zero. And that’s accurate! The net electric flux is 0.0 newtons meters squared per coulomb. That’s because the change in electric flux as the field passes through one face is perfectly offset by the change in flux as it passes through the opposite face. So the sum overall is zero.