Question

Gauss Law

Consider an infinitely long conductive cylindrical shell whose thickness can be neglected, basically a very long metal tube. The shell has a radius R and positive uniform charge distribution. Calculate the electric field inside and outside the cylinder. Now consider another cylindrical shell with radius R 'where R> R, basically another metal tube larger than in the previous exercise. The metal tube of radius R is placed within the other larger tube of radius R 'in a concentric manner. The big tube has the same uniform charge distribution as the small but negative one. Calculate the electric field between the two shells.

Answer 1

We have Gauss law as ,

   E da = Q_in/_o

a) Electric Field inside the cylinder (metal tube) is zero because cylindrical Gaussian surface inside the metal tube enclosed net charge is zero.i.e Q_in =0. so E=0 as per Gauss law

b) Electric filed outside the metal tube is E= 2k_e / R where is charge per unit lengt,. k_e is coulombs constant.,R is distance from metal tube.here cylindrical gaussian surface (radius R and length l) is constructed outside the metal tube and metal tube is considerd as infinite long wire with as charge per unit length.

E da = Q_in/_o

E da = l / _o

  E (2Rl) = l / _o

E= /2R _o

E = 2k_e / R by substituting k_e = 1/ 4_o

c) Electric filed is same as part b.i.e E= 2k_e / R   between two metal tube. because outer metal tube does not contribute for electric field.and charge enclosed by cylindrical gaussian surface is due to inner metal tube.