Question

A point charge Q = 1.03 Times 10^-5 C is at the center of a conducting spherical shell of radius R = 23.70 cm. The total charge of the shell is -1.03 Times 10^-5 C. What is the field in the region between the point charge and the shell? (at r= 11.85 cm) What is the field just outside the shell?

Answer 1

For the First part, you only need to worry about the point charge. Everything on the shell cancels per gauss and per Newton's Shell Theorem.

E = Q/(4 π ε r^2)
r = 11.85 cm
&epsilon = 8.85 x 10^-12 C^2/N-m^2
E = (1.03x10^-5)/(4*3.14*8.85x10^-12*0.1185^2) = 6.599x10^6 N/C.

Part Second:

The field just outside the shell will be:
E = (Q + Q) /4 π ε r^2)
r = 23.70 cm

Here, Q+Q = 0

So,
E = 0