Question

The very long ( L 00) insulating cylinder has a uniform charge. Find the magnitude of the electric field as a function of in terms of the volume charge density p and radius R for both <T R and RCT Show that the two expressions agree on the boundary (r = R). L

Answer 1

This problem can be solved by applying Gauss's Law on a gaussian cylindrical surface both inside and outside the cylinder. Detailed solution given below. Feedback will be helpful.





Hence we see that both these expressions agree on the boundary (r=R)

Solution: Lets take a gaussian cylinder of radius r. whin ocrcr, R=radies of cylinder, Volume of gaussian cylinder, V = nr²l id= height of cylinder. 6 Charge enclosed in the gaussian cylinder is: &= Ne = nazep 2 P= charge density. Applying Gauss's law inside this gaussian cylinder, © Ea Ei(2003) = *XP « E: 2 гр — - 2 GO - for ryR, charge inclosed is RR² lp B (since there is no charge - outside the cylinder) Applying gauss law, .. E. (nr = ARXP Eoz outside electric - W E. - Re When rar; Eof Ree are 286 26 When rir, E;= Rp. Hence Ei= Eo at r=R. to .. field. 2rto 2.60