Question

What  is the electric field due to a point charge of  45  μC   at a distance of 1 meter away from it?   

Answer 1

The electric field due to a point charge $q = 45\mu C = 45\times 10^{-6}C$ at distance $r = 1m$ 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:

$$E = 9\times 10^{9}\dfrac{N\cdot m^2}{C^2}\cdot \dfrac{45\times 10^{-6}C}{(1m)^2} = 4.05\times 10^5N/C$$

Answer. $4.05\times 10^5N/C$.