Question

a. What is the net electric flux through the surface is it encloses a net charge of 3.60μC?

b. The electric flux through a closed surface is found to be 7.80Nm2

/C. What quantity

of charge is closed by the surface?

Answer 1

(a)

$$\phi=\frac{q}{\epsilon_0}=\frac{3.60*10^{-6}}{8.85*10^{-12}}=4.07*10^5\:\rm N\cdot m^2/C$$

(b)

$$q=\phi\epsilon_0=7.80*8.85*10^{-12}=6.90*10^{-11}\:\rm C$$