Question

if the electric field is 100N/C at a distance of 50 cm from a point charge is the value of q? 

Answer 1

By definition, the electric field of a point charge is given as follows:

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

where $q$ is the charge, $r = 50cm = 0.5m$ is the distance from the charge, $k = 9\times 10^9N\cdot m^2/C^2$ is the Coulomb's constant. Substituting $E = 100N/C$ and expressing $q$, obtain:

$$q = \dfrac{Er^2}{k} = \dfrac{100\cdot 0.5^2}{9\times 10^9} \approx 2.8\times 10^{-9}C$$

Answer. $2.8\times 10^{-9}C$.