# Video: Relating the Electric Field between Parallel Plates to the Charge on the Plates and the Plate Size and Separation

Two thin parallel conducting plates are placed 2.0 cm apart. Each plate has sides of length 2.0 cm. One plate carries a net charge of +8.0 𝜇C and the other plate carries a net charge of −8.0 𝜇C. What is the charge density on the inside surface of each plate? What is the magnitude of the electric field between the plates?

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### Video Transcript

Two thin parallel conducting plates are placed 2.0 centimeters apart. Each plate has sides of length 2.0 centimeters. One plate carries a net charge of positive 8.0 microcoulombs. And the other plate carries a net charge of negative 8.0 microcoulombs. What is the charge density on the inside surface of each plate? What is the magnitude of the electric field between the plates?

This statement tells us the dimensions of the plates as well as the distance they are from one another and also the charge on each plate. First, we wanna solve for the charge density on the inside surface of each plate, which we’ll call 𝜎. And then we want to solve for the magnitude of the electric field that exists between the two plates, what we’ll call 𝐸.

We can start off by sketching these parallel plates. If we call the distance between the plates 𝑑, it so happens that that distance is also equal to the side lengths of each plate. 𝑑 is 2.0 centimeters. These parallel plates are oppositely charged. We can say the top plate has a positive charge, 8.0 microcoulombs, and the bottom plate has the opposite of that.

These opposite charges create an electric field in between the plates. That’s the field, 𝐸, we want to solve for in part two. If we call 𝑄 the magnitude of the charge on either plate, that can help us solve for 𝜎, the charge density on either plate. Charge density, often written as 𝜎, is equal to total charge 𝑄 divided by the area over which that charge is spread.

In our case, we know the magnitude of the charge on each of the top and bottom plates. And we also know the dimensions of the plate, that each one is 𝑑 squared where 𝑑 is 2.0 centimeters. So we can plug in and solve for charge density. When we do, we’re careful to write the charge in units of coulombs and the distance in units of meters, rather than centimeters in which it was given.

When we calculate 𝜎, we find it is equal to 0.020 coulombs per meter squared. That’s the amount of charge per unit area on these two parallel plates. Now that we know 𝜎, we want to solve for the magnitude of the electric field that exists in between these plates.

One interesting fact about that field is that it’s uniform. It’s the same everywhere as long as we’re in between these two parallel plates. The electric field in between parallel plates, 𝐸, equals the charge density on the plates 𝜎 divided by the permittivity of free space, 𝜖 naught. 𝜖 naught is a constant value. We’ll treat it as exactly 8.85 times 10 to the negative 12th farads per meter.

Calculating the magnitude of the field in between the plates, we’ve solved for 𝜎. And 𝜖 naught is a constant we know. We’re now ready to plug in and solve for 𝐸. When we enter these values on our calculator, we find that the electric field, to two significant figures, is 2.3 times 10 to the ninth newtons per coulomb. That’s the magnitude of the electric field in between these two parallel plates.