Problem 19.43

19.43. a)
Again, to determine the electric field, we turn to Gauss' Law.
$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{inc}}{\epsilon_{0}}$
$(E)(A) = \frac{q}{\epsilon_{0}}$
The Area of our Gaussian surface is given by
$A = 4\pi r^{2}$
For r<a, the charge q varies as a function of the radius. The enclosed charge is given by
$q = \rho (\frac{4}{3} \pi r^{3})$
So, the electric field is given by
$E = \frac{\rho r}{3 \epsilon_{0}}$
For a<r<b, and r>c, the electric field is just that of a point charge centered at the origin, or
$E= \frac{Q}{4\pi \epsilon_{0} r^{2}}$
or, more familiarly written,
$E = \frac{kQ}{r^{2}}$
Inside the conductor, the net enclosed charge is zero, and therefore the electric field is zero.

b)
The definition of surface charge density is charge per unit area. So, for the inner surface, a total charge of -Q is present to balance that of the point charge at the origin, so the surface charge density is given by
$\sigma_{1} = \frac{-Q}{4\pi b^{2}}$
and on the outer surface, a total charge Q is present to balance the -Q on the inner surface since the total charge is zero on the conductor. Therefore, the surface charge density on the outer surface is given by
$\sigma_{2} = \frac{Q}{4\pi c^{2}}$