What must be the distance between point charge ${\mathit{q}}_1\mathbf{=}\mathbf{26}\mathbf{.}\mathbf{0}\mathit{\mu }\mathit{c}$ and point chargerole="math" localid="1661869629566" ${\mathit{q}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{47}\mathbf{.}\mathbf{0}\mathit{\mu }\mathit{c}$ for the electrostatic force between them to have a magnitude of 5.70 N ?

1. Charges on the point charges,$q_1=26.0\mu \text{C},q_2=-47\mu \text{C}$
2. Magnitude of the force, F = 5.70 N

The electrostatic force is defined as the attractive or the repulsive force acting between the particles due to their charge. According to Coulomb's law, the magnitude of the force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the separation between them.

The magnitude of the electrostatic force on a body by another charged body according to Coulomb’s law,

$\mathbf{F}\mathbf{=}\frac{\mathbf{k}\left|{\mathbf{q}}_{\mathbf{1}}\right|\left|{\mathbf{q}}_{\mathbf{2}}\right|}{{\mathbf{r}}^{\mathbf{2}}}································\left(1\right)$

Using equation (1), the distance between the two point-charges can be given as follows:

$$\begin{gathered} r=\sqrt{\frac{k\left|q_1\right|\left|q_2\right|}{F}} \\ =\sqrt{\frac{\left(9\times {10}^9{\text{N.m}}^2{\text{/C}}^2\right)\left(26\times {10}^{-6}\text{C}\right)\left(47\times {10}^{-6}\text{C}\right)}{\left(5.70\text{N}\right)}} \\ =1.39\text{m} \end{gathered}$$

Hence, the value of the separation is 1.39 m.