Question

At what distance should two charges, each equal to 1 C, be placed so that the force between them equals your weight?

Answer 1

Force of attraction between the two charges

$=\frac{1}{4\pi {\in }_0}\frac{q_1{q}_2}{r^2}$

$=\frac{9\times {10}^9\times 1\times 1}{r^2}$

But force of attraction is equal to the weight.

$\therefore mg=\frac{9\times {10}^9}{r^2}$

$\Rightarrow r^2=\frac{9\times {10}^9}{m\times 10}=\frac{9\times {10}^8}{m}$

[Taking g = 10 $m/s^2$]

$\Rightarrow r^2=\frac{9\times {10}^8}{m}$

$\Rightarrow r=\frac{3\times {10}^4}{m}m$

e.g. Assuming, m = 64 kg,

$r=\frac{3\times {10}^4}{\sqrt{64}}$

$=\frac{3}{8}\times {10}^{4}m$

= 3750 m