00:01
In this question, we have a conducting spherical shell.
00:05
We're told that it has a positive charge q located at its centre, and the total charge on the cell is equal to minus 3q.
00:12
The shell has an inner radius of a and an outer radius of b.
00:16
We want to derive expressions for the magnitude of the electric field in terms of the distance r, in the first case where r is less than a, the second case, when r is between a and b, so within the thickness of the shell, and then the third case when r is greater than b, so outside of the shell.
00:35
So what we know is the induced charge on the inner surface of the shell is equal to minus q as a result of the positive charge of the centre equal to q.
00:44
Therefore, on the outer surface, it is equal to plus q.
00:48
So as a result of this, we can say that the total inner charge on the inner surface is equal to negative q, but in order for it to be equal to negative 3q in total for the shell, we know that the outer charge must be equal to negative 2q.
01:07
So what we can do now is use gauss's law.
01:10
We know that gauss's law states that e .da is equal to the enclosed charge over epsilon 0.
01:28
And what we can do is we can say that the area element da is equal to 4 pi.
01:35
Squared, as that is the surface area of a sphere.
01:44
So substituting this into this expression, we get the e multiplied by 4 pi r squared is equal to the enclosed charge q over epsilon naught.
02:06
What we can do now is we can rearrange and say the e is equal to q over x0.
02:17
4 pi epsilon nr squared.
02:24
So this is our first expression here in the case where r is less than a.
02:32
Next we want to look at the case where r is between a and b, so it is within the thickness of the shell.
02:40
We know that as it's a conducting surface, there is no charge within the shell as it's all concentrated on the surfaces of the shell.
02:48
So as a result, in this case, we can say that e is equal to zero.
02:54
In our final case, we have that r is greater than b.
03:00
So again, what we can do is use gautus law and say that e .da, which is again 4 pi r squared, is equal to the enclosed charge...