Answer to Question #206929 in Electric Circuits for Sarah

Question

A charge of uniform linear density /m is distributed along a Long, thin, non conducting rod. The with a long conducting cylindric shell (inner raclius =5.0 cm , outer rudius The nel charge on the shell is ) what is the magnitude of the electric held from the axis of the shell ? What is the curface charge density on the b) inner and c ) outer surface of the shell ?

Answer 1

Answer: full question must be

A charge of uniform linear density 2.0nC/m is distributed along a long, thin, nonconducting rod. The rod is coaxial with a long conducting cylindrical shell (inner radius =5.0 cm, outer radius =10 cm). The net charge on the shell is zero. (a) What is the magnitude of the electric field 15 cm from the axis of the shell? What is the surface charge density on the (b) inner and (c) the outer surface of the shell?

a) we know that : λ=$q\over L$  where q is the net charge enclosed by a cylindrical Gaussian surface of radius r. The field is being measured outside the system (the charged rod coaxial with the neutral cylinder) so that the net enclosed charge is only that which is on the rod.Consequently, 

$|\vec{E}|=\frac{2\lambda}{4\pi\epsilon_or}=\frac{2(2.0\times10^{-9}C/m)}{4\pi\epsilon_o(0.15m)}=2.4\times10^2N/C$

(b) Since the field is zero inside the conductor (in an electrostatic configuration), then there resides on the inner surface charge –q, and on the outer surface, charge +q (where q is the charge on the rod at the center). Therefore, with

 r_i= 0.05 m, the surface density of charge is :

$\sigma_{inner}=\frac{-q}{2\pi r_iL}=\frac{-\lambda}{2\pi r_i}=-\frac{2.0\times10^{-9}C/m}{2\pi (0.050m)}=-6.4\times10^{-9}C/m^2$ For the inner surface.

(c) With r_o​=0.10 m, the surface charge density of the outer surface is : 

$\sigma_{outer}=\frac{q}{2\pi r_oL}=\frac{-\lambda}{2\pi r_o}=3.2\times10^{-9}C/m^2$