An infinitely long conducting cylindrical rod with a positive charge lambda per
ID: 1655278 • Letter: A
Question
An infinitely long conducting cylindrical rod with a positive charge lambda per unit length is surrounded by a conducting cylindrical shell (which is also infinitely long) with a charge per unit length of -2 lambda and radius r_1, as shown in the figure. What is E(r), the radial component of the electric field between the rod and cylindrical shell as a function of the distance from the axis of the cylindrical rod? Express your answer in terms of lambda, r, and elementof_0, the permittivity of free space. What is sigma_inner, the surface charge density (charge per unit area) on the inner surface of the conducting shell? What is sigma_outer, the surface charge density on the outside of the conducting shell? Recall from the problem statement that the conducting shell has a total charge per unit length given by -2 lambda.)Explanation / Answer
for an infinitely long cylinderical rod with +ve charge lambda per unit length ( conductor)
surrounded by anotrher conducting cylindrical shell, charge density -2lambda, radisu 2R
a. between the rod and the cylinder, let electric fioeld be E
then E*2*pi*rl = lambda*l/epsilon [ from gauss' law, assume l length of the gaussean surface, then chargfe inside is lambda*l]
so E = lambda/2*pi*r*epsilon [ where epsilon is permittivity of free space]
b. as conductors have 0 electric field inside them
for a gaussean surface just inside the shell, q in = 0
so, let charge density on the inner surface of shell be lambda'
then lambda + lambda' = 0 for qin = 0
so sigma(inner) = - lambda
c. charge density on outer shell = -2*lambda
charge densioty on outer surface + chargfe density on inner surface = -2*lambda
sigma(outer) = -2*lambda + lambda = -lambda