00:01
Part a of the given problem, the current passing through the material equals the potential difference between the place divided by resistance.
00:07
So i function of t, v function of t, divided by r.
00:12
The potential across the capacitor, vt, is a q of a t divided by c, let's call this equation 1, where the capacitance is a ck, not a divided by d.
00:26
We substitute into equation 1 then i of t will become v of a t that is r so which is 1 over r d divide by kappa epsilon d a sorry a times a q d f f f f f f f f f f f f f f f f f f f f f f f ft of t the resistance of a dielectric is given by oom's law that is r is equal to resistivity z d time area from here, we substitute, let's call this equation number one.
01:01
Substitute this into equation number two to simplify i of a t will be q of a t times d divide by k, epsilon nod, a times a row times d.
01:19
A will cancel out, so q of a t divide by k, epsilon nod, row.
01:25
Let's call this equation number three.
01:28
Then the current is the time derivative of the charge.
01:33
So we can write i of t is minus d q of t over d t, where the negative charge came from the fact that the charge on the capacity decreases...