00:01
What we know from the problem is that there are two conductive spheres of radii a and b, and the region between them are filled with some conductive material of resistivity row.
00:13
And in part a, we're asked to show that the resistance between the two spheres is given by row over 4 pi, 1 over a minus 1 over b.
00:23
And the idea is to basically start with a thin, spherical shell with a thickness dr so that you can have a resistance of this spherical shell that's equal to dr, row dr over a.
00:51
So this is kind of like using r is equal to row l over a, where dr is playing the role of l and the cross -section area is still the cross -sectional area but what's happening here is that we're building the region between this sphere by these thin spherical shells by varying the radius from a to b so we're going to sum up all of the so we're going to pretend like we have little spheres here that fills the region starting with the radius of a and changing it by a very very small increment dr which is infinitesimal all the way up to b, and we add all of those contributions, all of those little resistance up to get the full resistance between the two spheres.
01:42
And the way to do this, the way you add infinitesimals is by integrating.
01:46
So we'll have to eventually do an integral, but let's now write the cross -sectional area, and that's nothing more than the surface area of the sphere.
02:04
So dr then is row d little r over 4 pi r squared, or i can rewrite that as row over 4 pi, r to the minus 2, dr.
02:27
And then when i take the integral, i have the resistance is equal to starting at a and going to b, row over 4 pi, r to minus 2, dr.
02:44
Now, row over 4 pi is a constant so i can pull out of the integration, and then i'm left with integrating r to minus 2 from a to b, which is nothing more than r to the minus 1 over minus 1 from a to b.
03:04
And that's row over 4 pi times minus 1 over b minus 1 over a.
03:19
4 pi times 1 over a minus 1 over b or if i want to put so that that's actually the result that's desired so now we'll solve part b of the problem and in part b we're asked to write the current density in terms of the terminal in terms of the potential difference between the two the two spheres so we know that that the current density is the current over the cross -sectional area.
04:00
And we also know that the potential difference is the current times the resistance.
04:09
And we found this in part a.
04:12
So we can solve this equation for a current...