00:01
Going to apply goss's law to this gossian cylinder here for part a we can say that a is going to be less than r which is less than b and we can say that the electric flux is then equaling the electric field times two pi r l and we know that the charge enclosed would be equalling the linear charge density lambda times l the charge this would be the charge this would be the charge charge on the length l of the inner conductor that is inside the gaussian surface.
00:39
And so we can say that then the electric flux equals the charge of the enclosed divided by the primitivity of free space.
00:58
This gives us that then the electric field multiplied by 2 pi r times l will be equalling pi l sorry lambda l divided by epsilon not of course this gives us then that the length is going to cancel out and we have that the electric field is going to be equaling the linear charge density lambda divided by two pi r times the electric permit the permittivity of free space or the permittivity of the vacuum.
01:38
We can say that the enclosed charge is positive, therefore the direction of the electric field is, we can say, radially outward.
01:53
For part b, this would be the, this would be essentially the diagram.
02:01
So we're going to play goss's law where for part b, r is greater than some length c.
02:09
And we can say that the electric flux would be equal to the electric field times 2 pi r times l once again and again the enclosed charge is equaling pi times l this is going to give us here as we can see the enclosed charged the magnetic flux all are giving us the exact same values so here the electric charge is again the linear charge density divided by 2 pi r multiplied by epsilon not.
02:49
And here the enclosed charge is again positive.
02:52
So the direction of e would be again radially outward.
03:00
So we can now, for part c, we have that within a conductor, the electric field is equaling zero.
03:12
So we can say electric field equals zero within a conductor...