Answer
The magnitude of $E$ is $~~0.214~N/C$
Work Step by Step
The charge on the shell is $-6.80\times 10^{-12}~C$
The net charge of the shell and rod is $-3.40\times 10^{-12}~C$
We can use Equation (23-12) to find the electric field:
$E = \frac{\lambda}{2\pi~\epsilon_0~r}$
$E = \frac{-3.40\times 10^{-12}~C/11.00~m}{(2\pi)~(8.854\times 10^{-12}~F/m)~(0.0260~m)}$
$E = -0.214~N/C$
The magnitude of $E$ is $~~0.214~N/C$
