WEBVTT
00:00:00.800 --> 00:00:09.840
Two parallel conducting plates, each of cross sectional area 580 square centimeters, are 3.2 centimeters apart and uncharged.
00:00:10.260 --> 00:00:15.630
2.0 times 10 to the 12 electrons are transferred from one plate to the other.
00:00:16.120 --> 00:00:18.500
What is the charge density on each plate?
00:00:18.830 --> 00:00:22.180
What is the magnitude of the electric field between the plates?
00:00:22.670 --> 00:00:25.780
Let’s consider first this question about charge density.
00:00:25.820 --> 00:00:27.900
And we’ll call our charge density σ.
00:00:28.450 --> 00:00:42.210
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.
00:00:42.660 --> 00:00:49.150
σ, the charge density of the plate, is equal simply to the total charge each plate has divided by its area.
00:00:49.490 --> 00:00:56.270
When we consider just how much charge 𝑄 is on each one of these parallel conducting plates, we’re not told that directly.
00:00:56.390 --> 00:01:01.970
But we are told that 2.0 times 10 to the 12 electrons are transferred from one plate to the other.
00:01:02.410 --> 00:01:14.340
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.
00:01:14.870 --> 00:01:21.240
And we can recall that that charge by itself is negative 1.6 times 10 to the negative nineteenth coulombs.
00:01:21.580 --> 00:01:28.250
If we multiply these two numbers together then, we can get a value for the total charge magnitude on one of the conducting plates.
00:01:28.530 --> 00:01:32.870
But we can go a step further by dividing 𝑄 by the area of the plate 𝐴.
00:01:33.300 --> 00:01:41.410
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.
00:01:41.880 --> 00:01:50.680
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.
00:01:51.220 --> 00:02:02.770
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.
00:02:03.280 --> 00:02:09.100
We see that our value of σ will now have units of coulombs per square meter, which is just what we want.
00:02:09.670 --> 00:02:15.700
To two significant figures, our answer is 5.5 times 10 to the negative sixth coulombs per square meter.
00:02:16.070 --> 00:02:18.450
That’s the surface charge density on each plate.
00:02:18.900 --> 00:02:26.300
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.
00:02:26.830 --> 00:02:34.530
Because the two parallel conducting plates have opposite charges, that means an electric field is set up between them pointing from positive to negative.
00:02:35.010 --> 00:02:39.470
It’s the magnitude of this electric field, we’ll call it a capital 𝐸, that we want to solve for.
00:02:39.910 --> 00:02:51.370
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.
00:02:51.860 --> 00:02:57.470
This constant value we can approximate as 8.85 times 10 to the negative twelfth farads per meter.
00:02:57.870 --> 00:03:03.500
Since we solved for σ in a first part, we’re ready now a plug in and solve for the electric field magnitude.
00:03:03.950 --> 00:03:15.100
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.
00:03:15.590 --> 00:03:21.360
After all the dust settles, this comes out as 6.2 times 10 to the fifth newtons per coulomb.
00:03:21.780 --> 00:03:25.660
That’s the magnitude of the electric field between these parallel plates.