Question

Calculate the electric field strength 30 cm from a 5 nC charge.

Answer 1

According to the Coulomb's law, the electric field strength of a point charge is given by:

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

where $q = 5\times 10^9C$ is the charge, $r = 30cm = 0.3m$ is the distance to point at which we consider the field and $k = 9\times 10^9N\cdot m^2/C^2$ is the constant.

Substituting the numerical values, obtain:

$$E = 9\times 10^9\cdot \dfrac{5\times 10^{-9}}{0.3^2}=500V/m$$

Answer. 500 V/m.