00:01
For this problem on the topic of kauser's law, we are given the positive charge density of a very long insulating cylinder, which has a radius r cylinder.
00:10
A particle that has a charge minus q and mass m is orbiting the cylinder at a constant distance, which is r orbit.
00:17
This is greater than the radius of the cylinder.
00:20
And we want to use this information to find the linear charge density lambda of the tube in terms of r cylinder and row not.
00:26
And we want to determine the period of motion in terms of r orbit.
00:32
Now, we want the linear charge density of the cylinder lambda, and lambda is the charge q divided by the length of the cylinder l.
00:45
So we need to find q in terms of l.
00:48
Now, the charge q is equal to the integral of the charge density row times dv, which is the integral over the radius of the cylinder from 0 to r of row not, into 1 minus little r over big r times and we can replace dv by 2 pi r l d r and so if we solve this integral this becomes pi times l times r squared over 3 and so therefore the linear charge density lambda is the charge q over l, which is pi l ro -not r squared over 3, all divided by l.
01:54
And so the linear charge density is pi ro -not r squared over 3.
02:04
And so we have an expression for the linear charge density of the cylinder...