Question:
At what distance should two charges, each equal to $1 \mathrm{C}$, be placed so that the force between them equals your weight?
Solution:
$F=W$
$\frac{k q_{1} q_{2}}{r^{2}}=m g$
$\frac{9 \times 10^{9} \times 1 \times 1}{r^{2}}=m \times 10$
$r=\frac{3 \times 10^{4}}{\sqrt{m}}$ meters
If $m=49 \mathrm{~kg}$
$r=4285.7 \mathrm{~m}$