Gaussian Surface - General Physics - Solved Past Paper, Exams of Physics

Download Gaussian Surface - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 3. (25 pts) Consider a long cylindrical charge distribution of radius R with a uniform charge density ρ0 C/m 3, of which a segment is shown below. Find the electric field at a distance r from the axis of the cylinder in the case where r < R. Keep in mind that (a) this is a solid cylinder, not a shell, (b) it is an insulator, not a conductor, and (c) the charge density is given in terms of volume, not the just the length of the cylinder. R r L Use a cylindrical Gaussian Surface, as shown in blue above. There is only flux through the curved outer edge, not the circular ends. The net flux is therefore ΦE = 2πrL E Then we need the charge inside this surface. It is a cylinder, so the volume is the area of the base (a circle) times the length, V = πr2L So the enclosed charge is qenc = ρ0πr 2L Putting these back into Gauss’s Law gives 2πrL E = ρ0πr 2L ǫ0 E = ρ0r 2ǫ0