The electric flux through a closed surface enclosing a charge of $q=8.85\times 10^{-12}\text{C}$ is (take ${ϵ}_0=8.85\times 10^{-12}\text{F/m}$):

Solution

1. Gauss's Law states $\Phi =\frac{q}{{ϵ}_0}$.
2. Given $q=8.85\times 10^{-12}\text{C}$ and ${ϵ}_0=8.85\times 10^{-12}\text{F/m}$.
3. Calculate $\Phi =\frac{8.85\times 10^{-12}}{8.85\times 10^{-12}}$.
4. The result is $\Phi =1\text{V}\cdot \text{m}$.
5. Hence the answer is (A).