Question

2 equally charged pith balls of 3cm apart repel each other with a force of 4*10^-5N the charge on each ball is

Answer 1

Say the charge on each ball is $q$
$r=3*10^{-2}m$
$F=4*10^{-5}N$

From coulomb's law we have $F=9*10^9\frac{q_1q_2}{r^2}$

Plugin the given values and get
$4*10^{-5}=9*10^9\frac{q^2}{(3*10^{-2})^2}$
$q^2=4\times\,10^{-18}$
$q=\pm 2\times\,10^{-9}C$

Answer 2

The Coulomb's law of force between two electrically charged particles :
         q₁ , q₂ = electric charges on the two particles.
           r = distance between two particles.
      the particles are assumed to be point sized.

$Force\ F=\frac{1}{4\pi \epsilon_0}\ \frac{q_1\ q_2}{r^2}\\\\4*10^{-5}N=9*10^9*\frac{q^2}{0.03^2\ m^2}\\\\q^2=\frac{4*10^{-5}\ *\ 0.03^2}{9*10^9}\\\\q=+ 2*10^{-9}C\ \ or\ \ -2*10^{-9}C\\\\q=+2\ nano\ Coulombs\ \ or,\ \ -2\ nC$

The charges can either both be positive or negative.