Question

A smooth spherical shell of electricity insulating material with outer radius a and inner radius a/2. Inside of this sphere, also with a radius of a/2, is a conducting solid sphere. The conducting sphere has an excess amount of charge q. The density of the insulating sphere is p.

A) What must be the value of p so that the total charge of this setup is 0?

B) Using the value of p from part (A), what are the magnitude and direction of the electric fields in the parts a<r, a>r> a/2, and a/2 > r > 0?

C) Where is there a discontinuity in the electrical field, and why is that possible?

6. A smooth spherical shell of electrically insulating material with outer radius a and inner radius a/2. Inside of this sphere, also with a radius of a/2, is a conducting solid sphere. The conducting sphere has an excess amount of charge q. The density of the insulating sphere is p. (a) what must be the value of ρ so that the total charge of this setup is 0? (b) Using the value of ρ frorn part (a), what are the magnitude and direction of the electric fields in the parts a < r, a>r > a/2, and a/2 > r > 0? (c) Where is there a discontinuity in the electric field, and why is that possible?

Answer 1

Charge density of involving sphere is p. then total charge inside the insulting sphere = integral_v p dT = integral_v p r^2 sin theta dr d theta d phi = p integral_a/2^a r^2 dr integral_0^pi sin theta d theta integral_0^2 pi d phi = p r^3/3 Vertical-bar _a/2^a 2 2 pi = 4 pi p[a^3/3 - a^3/24] = 4 pi p 7/24 a^3 = 7/6 pi p a^3 Now to make total charge 0 if the set up, total charge of insulting sphere must be -q. then, -q = 7/6 pi p a^2 implies p = -6/ q/pi a^3 ... (i) (I) For the region r > a, total charge is zero. As mentioned in the question, that net charge of The set up is zero. So, electric field must be zero for r > a. (II) Region: a/2 < r < a To calculate electric field in this region we have to consider both free charge of conducting sphere and bound charge