Answer
a. $-1.1\times10^5 N \cdot m^2/C$
b. 0
Work Step by Step
a. Use Gauss’s law to find the flux .
$$\Phi_E=\frac{Q_{enc}}{\epsilon_o}$$
$$\Phi_E=\frac{-1.0\times10^{-6}C}{8.85\times10^{-12}C^2/(N \cdot m^2)}$$
$$=-1.1\times10^5 N \cdot m^2/C$$
b. There is no enclosed charge so the flux is zero.
