00:01
For this problem on the topic of gauss's law, we are told that charge is uniformly distributed throughout an infinitely long solid cylinder that has a radius of big r.
00:12
We want to show that at a distance less than the radius from the cylinder axis, the electric field, strength is row r over 2 epsilon 0, where row is the volume charge density.
00:22
We then want to find an expression for the electric field outside the radius.
00:27
Now, from the diagram, we are shown a cross -section of the charged cylinder, which is a solid circle.
00:34
We'll consider a gaussian surface in the form of a cylinder with the radius r and length l, which is coaxial to the charged cylinder.
00:41
An end view of this gaussian surface is shown as a dashed circle in our diagram, and the charge enclosed by it is q, which is the charge density row times the volume v, which is the volume v, pi r squared, l times row.
00:59
Now if row is positive, the electric field lines are readily outward, normal to the gaussian surface, and distributed uniformly along it.
01:06
Thus, the total flux through the gaussian cylinder, phi, is equal to the electric field strength, e times the area of the cylinder, which is e times 2 pi r times l...