Question

The electric field has a magnitude of 3.0 N/C at a distance of 60 cm from a point charge..What is the charge?

Answer 1

$$E=k\frac{q}{r^2}$$

$$q=\frac{Er^2}{k}=\frac{3.0*0.6^2}{9*10^9}=1.2*10^{-10}\:\rm C$$