00:01
For this problem on the topic of electromagnetic induction, we are given the area as well as the separation of the parallel plates in a capacitor.
00:11
We are also given the dielectric constant, the potential difference between the plates, the conduction current as well.
00:18
And we are asked to calculate the charge on each plate, the rate of change of the charge on the plates, as well as the displacement current in the dielectric.
00:28
Now, we know for a parallel plate capacitor, the charge, on the plates of the capacitor, q is equal to c times v, where c is the capacitance and b is the potential difference between the plates.
00:42
And we can write this as by replacing c as epsilon times a over d, where epsilon is a dialectic constant, a is the area and d is the separation of the plates.
00:59
And that's multiplied by the potential difference v.
01:02
Now, since we know these values, we can substitute them and we can hence calculate the charge on the plates.
01:10
So this is the dielectric constant of 4 .7 times the permittivity of free space epsilon not, which is a constant, times the area of 3 times 10 to the minus 4, and that's square meters, times the potential difference between the plates of 120 volts, all divided by the separation between the plates, which is 2 .5 times 10 to the minus 3 meters.
01:49
So calculating, we get the charge held on each plate to be 5 .99 times 10 to the minus 10 kuloms.
02:02
And that's our first solution...