Video: Analyzing the Electric Flux between the Plates of a Capacitor

A rectangular area with side lengths of 5.4 cm and 2.4 cm is between two parallel plates where there is a constant electric field of 35 N/C. Find the electric flux parallel to the plates in this area. Find the electric flux perpendicular to the plates in this area. Find the electric flux normal to the area and making a 42° angle with the direction of the electric field.

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Video Transcript

A rectangular area with side length of 5.4 centimeters and 2.4 centimeters is between two parallel plates where there is a constant electric field of 35 newtons per coulomb. Find the electric flux parallel to the plates in this area. Find the electric flux perpendicular to the plates in this area. Find the electric flux normal to the area and making a 42-degree angle with the direction of the electric field.

In this situation, we have two parallel plates with a constant electric field between them. And positioned in between the plates is a rectangular area with given side lengths. And we arrange this area in different orientations and then want to solve for the flux through it. Let’s start out considering the first case: finding the electric flux between these plates that’s parallel to them. If we draw our two parallel plates a bit bigger, let’s say the electric field moves from the left plate to the right plate so the left plate relative to the right is positively charged. This electric field, which we’ll call 𝐸, has a given strength of 35 newtons per coulomb. We’re told of a rectangular area with given side lengths, which we can arrange in anyway we want between these two parallel plates.

In this first part, we want to solve for the electric flux that is parallel to our two parallel plates. To catch this electric flux so to speak, we will arrange our rectangular area perpendicular to the plates so that any electric flux lines which pass parallel to them would pass our rectangular area. Even though we’ve just drawn a line here, understand that it represents a downward view on what is actually a rectangular area of these given dimensions in the problem statement. We want to figure out just how much electric flux there is through this area when it’s arranged as shown, perpendicular to our two parallel plates.

To help us figure this out, let’s recall that electric flux 𝛷 sub 𝐸 is equal to electric field strength multiplied by the area that it moves through all multiplied by the cosine of the angle between the electric field and the normal to that area. This means that if the electric field pointed like this and we had an area whose area vector was represented by capital 𝐴 pointing like this, then 𝜃 is equal to the angle between these two vectors. That’s the 𝜃 in the cosine of 𝜃 in this equation. Applying this relationship to our scenario, it looks like we’ll want to find the electric field 𝐸, the area 𝐴, as well as the cosine of 𝜃. Let’s start with the cosine of 𝜃 and solve for it using our sketch.

Looking at our sketch, we know that the electric field is moving left to right; that is, it’s parallel to our area. And if we sketched in an area vector, it would point perpendicular to that field. This tells us that 𝜃 in this case is 90 degrees. And what’s the cosine of 90 degrees? It’s zero. So in this case, it doesn’t actually matter what our electric field and area are because we’re multiplying them by zero so the overall result is zero. What this means is that no electric field lines move through our area when the field and the area are arranged relative to one another this way. So the electric flux parallel to our two parallel plates is zero. That means that all the field moves left to right.

Next, rather than solving for electric flux parallel to the plates, let’s solve for it perpendicular to the plates. In this case, our rectangular area will be rotated 90 degrees so that it’s now parallel with our two parallel plates. In this case, we see the electric field lines do now pass through this rectangular area and that the angle between the area vector of this area and the electric field lines is zero degrees. That’s interesting because as we go back to our equation for electric flux, since 𝜃 is zero degrees, we know that the cosine of zero degrees is one, which means that the only things we need to calculate this electric flux are the electric field magnitude and the area through which these electric field lines are moving.

Taking what we know from the problem statement, we can write that 𝐸 is 35 newtons per coulomb and the area of this rectangular area is 5.4 centimeters times 2.4 centimeters. We’re almost ready to calculate flux. But before we do, we’ll need to make a units change. We want to change the length units from centimeters into meters, the SI based unit of length. So how do we do that? Well we recall that 100 centimeters is equal to one meter. In other words, converting from centimeters to meters involves moving the decimal place two spots to the left. That means that 5.4 centimeters turns into 0.054 meters and that 2.4 centimeters become 0.024 meters. Now that everything’s in standard units, we’re ready to calculate the electric flux through this rectangular area oriented this way. We find a result of 0.045 newton meters squared per coulomb. As expected, the electric flux through our rectangular area oriented perpendicular to the electric field lines is not zero; this is our answer for it.

Now let’s make one last change to the orientation of our rectangular area between these parallel plates. Now we’ll tilt or we’ll angle this rectangular area so that if we draw in the normal vector from it, the angle between that normal vector and the electric field lines is 42 degrees. In other words, that’s the value for 𝜃 that we’ll use in our equation for electric flux. And indeed that’s exactly what we want to know. Given this orientation of a rectangular area, what is the electric flux through it? To calculate this, our factors of electric field magnitude and rectangular area stay the same as before. Now we have another factor, the cosine of 𝜃, to multiply by. So as before, we have our electric field and our area, and now we’re multiplying those by the cosine of 42 degrees.

The outcome of this is an electric flux which is greater than our original calculation of zero, but less than the maximum electric flux we calculated at our last step. This is because our rectangular area is now angled away from being either parallel or perpendicular with the parallel plate. It’s at an in-between angle. And as a result, we get an in-between magnitude of 0.034 newton meters squared per coulomb for the electric flux.

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