Problem 19.36

19.36. a)
On the inner surface of the conductor, some negative charges will accumulate to balance the charge of the wire. The magnitude of the charge per unit length will be identical, only the sign of the charge will be zero. This will imply a charge per unit length on the inner surface of the conductor of
$\frac{q}{\ell} = -\lambda$
This is reasonable since outside of the inner surface of the conductor and inside the outer surface of it, the total enclosed charge, and therefore the electric field, is zero. Now, on the outer surface of the conductor, a positive charge will build up and since the total charge on the conductor is $2\lambda$, a charge per unit length of $3\lambda$ will build up on the outer surface. Mathematically, $Q_{total} = -\lambda \ell + 3 \lambda \ell ~=~ 2\lambda \ell$
which is what we were given for the charge on the conductor.

b)
Using Gauss' Law,
$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{inc}}{\epsilon_{0}}$
$(E)(A) = \frac{3\lambda \ell}{\epsilon_{0}}$
$(E)(2\pi r \ell) = \frac{3\lambda \ell}{\epsilon_{0}}$
$E = \frac{2}{4\pi \epsilon_{0}} \frac{3\lambda}{r}$
using our definition for k,
$E = \frac{6\lambda k}{r}$