HW: Find the E field inside and outside a conductor of radius R for a nonconducting sphere of nonuniform density with

= B r 2 where B is a constant and #65 on page 641 (the solution is on the internet)
HW Question 1: Find the E field inside and outside a conductor of radius R for a nonconducting sphere of nonuniform density with

= B r 2 where B is a constant.
E inside =
#65
We did the inside yesterday
Q(inside) = Q(total)*r
3
/a
3
E *4 pi r 2 = Q(total) r 3 /(a 3 *epsilon naught)
E = Q(total) r/ (4 pi epsilon naught * a
3
)
Now since ρ(density) = total charge divided by total volume = 3Q/4 pi a 3
SINCE IT IS
UNIFORM!
An equivalent statement for E = (3/3) (Q(total) r/ (4 pi epsilon naught * a
3
)) and then factoring the expression for ρ out of it we get ρr /(3 epsilon naught) WHICH HAPPILY
IS THE EXACT ANSWER THEY HAVE THE IN BOOK! So see you can do this either using ρ or not and get the same answer.
Between a and b the answer is just Coulomb’s Law so E = 1/(4 pi epsilon naught) * Q/r 2
Between b and c the answer is zero since the E field in a conductor is zero. How can this happen if there is a net charge enclosed in a Gaussian surface due to the non-conducting sphere? That is because the conductor is a CONDUCTOR and so there is an induced charge opposite the non-conducting sphere on the inside of the conductor and an induced charge of the same sign as the non-conducting sphere on the outside of the conductor.
The conductor’s net charge is ZERO. But on the very inside surface and very outside surfaces there is charge. There is a great picture of this in the Princeton Review. Ask me and I will show you this. This leads us to the answer of the second part of this question.
(b) Charge density on the inside = -Q/(4 pi b 2 )
Charge density on the outside = +Q/(4 pi c
2
)