Question

A spherical shell with a charge of 3.8 µC uniformly distributed throughout the shell has an inner surface with radius of 4.0 cm and outer surface with radius of 5.0 cm. Use Gauss’s Law to find the electric field
a. 1.0 cm from the inner surface, inside the shell.
b. halfway between the two surfaces.
c. 5.0 cm from the outer surface, outside the shell.

Answer 1

a) Inside the inner surface shell there is no enclosed charge. Hence the electric field there is 0 N/C

b) the charge density of the shell is

where a = 5 cm = 0.005 m is the outer surface

b = 4 cm = 0.004 m is the inner surface

The charge enclosed inside a gaussian sphere of radius r (b< r< a) inside the shell is

Therefore using Gauss's law the electric field at distance r is

where k is the electrostatic force constant

r = 0.5+4 cm,= 4.5 cm = 0.045 m  

Therefore

c)

The charge enclosed inside the sphere is Q

The electric field 5 cm from the outer surace of the shell is

(r = 5+5 = 10 cm = 0.1 m)