Question

The electric field has a magnitude of 3.0 N/C at a distance of 30 cm from a point charge. What is the charge? 

Answer 1

The electric field due point charge

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

Hence, the charge

$$q=Er^2/k\\ =3.0*0.3^2/9*10^9=3*10^{-11}\:\rm C$$