Answer
$\sigma = 3.2\times 10^{-9}~C/m^2$
Work Step by Step
In part (b), we found that the charge on the inner surface over a length of 1.0 meter is $-2.0\times 10^{-9}~C$
Since the net charge on the shell is zero, the charge on the outer surface of the shell must be $~~2.0\times 10^{-9}~C$
We can find the surface charge density:
$\sigma = \frac{Q}{Area}$
$\sigma = \frac{2.0\times 10^{-9}~C}{(2\pi~r)(1.0~m)}$
$\sigma = \frac{2.0\times 10^{-9}~C}{(2\pi)~(0.10~m)(1.0~m)}$
$\sigma = 3.2\times 10^{-9}~C/m^2$
