# At what distance of separation must two 1.00-microCoulomb charges be positioned in order for the repulsive force between them to be equivalent to the weight (on Earth) of a 1.00-kg mass?

Nov 3, 2015

$30 , 3 m m$

#### Explanation:

From Coulomb's Law of Electrostatic forces between charged objects we get that

$F = k \frac{{Q}_{1} {Q}_{2}}{r} ^ 2$

$\therefore r = \sqrt{\frac{k {Q}_{1} {Q}_{2}}{F}}$

$= \frac{\sqrt{\left(9 \times {10}^{9}\right) {\left(1 \times {10}^{- 6}\right)}^{2}}}{9 , 8}$

$= 30 , 3 m m$

(Note that the weight force is given by $W = m g = 1 \times 9 , 8 = 9 , 8 N$)