Question

Calculate the electric field that a test charge will experience at a distance of 2.04 x 10^-3 m from the source charge of 5.02 x 10^-13 C.

Answer 1

The electric field produced by a point charge is given as follows:

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

where $k = 9\times 10^{9}Nm^2/C^2$ is the Coulomb constant, $q = 5.02\times 10^{-13}C$ is the charge, and $r = 2.04\times 10^{-3}m$ is the distance from the source charge. Thus, obtain:

$$E = 9\times 10^{9}Nm^2/C^2\cdot \dfrac{5.02\times 10^{-13}C}{(2.04\times 10^{-3}m)^2} \approx 1.09\times 10^3N/C$$

Answer. $1.09\times 10^3N/C$.