An average human weighs about 650 N . If each of two average humans could carry 1.0 C of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 650 N weight?

According to the Coulomb’s law, the force between two charges spaced at a distance is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.

Mathematically,

$\mathbf{F}\mathbf{=}\mathbf{k}\frac{\left|q_1{q}_2\right|}{{\mathbf{r}}^{\mathbf{2}}}$ ….. (1)

Here, F is the force, ${\mathrm{q}}_1$and localid="1668317559492" ${\mathrm{q}}_2$are the charges, k is the Coulomb’s charge, and r is the distance between two charges.

Consider the given data as below.

The repulsive force, F = 650 N

The Coulomb’s constant, k =$9\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2$

The charges, q = 1.0 C

Rewrite equation (1) as below.

localid="1668317682291" $\begin{array}{r}F=\frac{k{q}^2}{r^2}\\ \left|r\right|=\sqrt{\frac{k{q}^2}{F}}\end{array}$

Substitute known values in the above equation, and you have

localid="1668317685578" $\begin{array}{r}r=\sqrt{\frac{9\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}\times (1\mathrm{C}{)}^2}{650\mathrm{N}}}\\ =3.7\times {10}^3\mathrm{m}\\ =3.7\mathrm{km}\end{array}$

Hence, the distance between the two humans is 3.7 km.