Video: Determining the Electric Field due to a Charge Separation

Two parallel conducting plates, each of cross sectional area 580 cm², are 3.2 cm apart and uncharged. 2.0 × 10¹² electrons are transferred from one plate to the other. What is the charge density on each plate? What is the magnitude of the electric field between the plates?

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Video Transcript

Two parallel conducting plates, each of cross sectional area 580 square centimeters, are 3.2 centimeters apart and uncharged. 2.0 times 10 to the 12 electrons are transferred from one plate to the other. What is the charge density on each plate? What is the magnitude of the electric field between the plates?

Let’s consider first this question about charge density. And we’ll call our charge density σ. If we consider one of the two parallel plates in this example, we know that some amount of charge sits on that plate the same magnitude as is on the other one and that the plate has some total area we can call capital 𝐴. σ, the charge density of the plate, is equal simply to the total charge each plate has divided by its area.

When we consider just how much charge 𝑄 is on each one of these parallel conducting plates, we’re not told that directly. But we are told that 2.0 times 10 to the 12 electrons are transferred from one plate to the other. If we call that number of electrons capital 𝑁, then we can say the total charge magnitude on the plate 𝑄 is equal to 𝑁 times the magnitude of the charge on a single electron 𝑞 sub 𝑒. And we can recall that that charge by itself is negative 1.6 times 10 to the negative nineteenth coulombs.

If we multiply these two numbers together then, we can get a value for the total charge magnitude on one of the conducting plates. But we can go a step further by dividing 𝑄 by the area of the plate 𝐴. If we do this, then we see by a relationship for σ that by calculating this ratio we’ve also calculated σ, which is what we want. There’s just one last thing left to do before we can calculate this number, and that is to convert the units in our area from square centimeters to square meters.

If we recall that 10000 centimeters squared is equal to one meter squared, then that means 580 square centimeters is equal to 0.0580 square meters. We see that our value of σ will now have units of coulombs per square meter, which is just what we want. To two significant figures, our answer is 5.5 times 10 to the negative sixth coulombs per square meter. That’s the surface charge density on each plate. Now that we’ve solved for σ, we can move on the part two, where we wanna calculate the magnitude of the electric field between these parallel plates.

Because the two parallel conducting plates have opposite charges, that means an electric field is set up between them pointing from positive to negative. It’s the magnitude of this electric field, we’ll call it a capital 𝐸, that we want to solve for. And to do that, we can recall that for parallel plate capacitors like we have here, the electric field between them is equal to the charge density of the plates divided by 𝜖 naught, the permittivity of free space. This constant value we can approximate as 8.85 times 10 to the negative twelfth farads per meter.

Since we solved for σ in a first part, we’re ready now a plug in and solve for the electric field magnitude. That magnitude is σ over 𝜖 naught, which equals 5.5 times 10 to the negative sixth coulombs per square meter divided by 8.85 times 10 to the negative twelfth farads per meter. After all the dust settles, this comes out as 6.2 times 10 to the fifth newtons per coulomb. That’s the magnitude of the electric field between these parallel plates.

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