Question

You measure an electric field of 2.52×106 N/C at a distance of 1.196 m from a point charge. What is the magnitude of the charge?

Answer 1

Solution. The electric field of a point charge can be found behind the formula

$$E=\frac{kQ}{r^2}$$

where r=1.196 m is a distance; Q is charge; k=9×10^9 N m²C^-2 is Coulomb's constant.

Therefore

$$Q=\frac{Er^2}{k}=\frac{2.52\times 10^6\times 1.196^2}{9\times10^9}\approx 4\times10^{-4}C$$

Answer.

$$Q=4\times10^{-4}C$$