Answer
The magnitude of the electric field is $~~4.5\times 10^5~N/C$
Work Step by Step
The charge on the inner shell is $5.0\times 10^{-6}~C/m$
The charge on the outer shell is $-7.0\times 10^{-6}~C/m$
At a radius of $8.0~cm$, both shells are enclosed within a cylindrical Gaussian surface.
We can use Equation (23-12) to find the electric field:
$E = \frac{\lambda_{net}}{2\pi~\epsilon_0~r}$
$E = \frac{-2.0\times 10^{-6}~C/m}{(2\pi)~(8.854\times 10^{-12}~F/m)~(0.080~m)}$
$E = -4.5\times 10^5~N/C$
The magnitude of the electric field is $~~4.5\times 10^5~N/C$.
