Question

Calculate the strength and direction of the electric field  E  due to a point charge point charge of 5.0  nC   at a distance of 8.00 mm from the charge. 

Answer 1

The electric field due to a point charge $q = 5n C = 5\times 10^{-9}C$ at distance $r = 8mm = 0.008m$ is given as follows:

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

where $k = 9\times 10^{9}N\cdot m^2/C^2$ is the Coulomb's constant. Thus, obtain:

$$E = 9\times 10^{9}\dfrac{N\cdot m^2}{C^2}\cdot \dfrac{5\times 10^{-9}C}{(0.008m)^2} \approx 7\times 10^5N/C$$

Answer. $7\times 10^5N/C$.