Answer
(a) $q_{1}=1.08\times10^{-9}C$, $q_{2}=3.24\times10^{-9}C$
(b) $F\approx4.2\times10^{-6}N=4.2\mu N$
Work Step by Step
Let us assume two negative charges $q_{1}$ and $q_{2}$ are separated by a distance $r$
given that $r =10.0cm=10.0\times10^(-2)m=1.0\times10^{-1}m$
given $q_{2}=3\times q_{1}$
Repulsive force force between the charges is $F=3.15\mu N=3.15\times10^{-6}N$
(a) calculation of charges
$F=k\frac{q_{1}\times q_{2}}{r_{}^2}$
putting $F=3.15\times10^{-6}N$, $r =1.0\times10^{-1}m$,
,$q_{2}=3\times q_{1}$ , $k=9.0\times10^{9}N.m^2/C^2$
$3.15\times10^{-6}N=9.0\times10^{9}N.m^2/C^2\times \frac{q_{1}\times3\times q_{1} }{(1.0\times10^{-1}m)^2}$
$3 q_{1}^2= 0.35\times10^{-17}C^2$
$q_{1}^2=0.11667\times10^{-17}C^2=1.1667\times10^{-18}C^2$
as the charges are negative
$q_{1}=-1.08\times10^{-9}C$,
other charge will be $q_{2}=3\times q_{2}=3\times-1.08\times10^{-9}C=-3.24\times10^{-9}C$
(b)
total charge $q=q_{1}+q_{2}=-1.08\times10^{-9}C+(-3.24\times10^{-9})C=-4.32\times10^{-9}C$
if charges are equally distributed $q_{1}=q_{2}=\frac{q}{2}$=$\frac{-4.32\times10^{-9}C}{2}=-2.16\times10^{-9}C$
$F=k\frac{q_{1}\times q_{2}}{r_{}^2}$
putting $F=3.15\times10^{-6}N$, $r =1.0\times10^{-1}m$,
,$q_{2}= q_{1}=-2.16\times10^{-9}C$ , $k=9.0\times10^{9}N.m^2/C^2$
$F=9.0\times10^{9}N.m^2/C^2\times\frac{2.16\times10^{-9}C\times2.16\times10^{-9}C}{(1.0\times10^{-1}m)^2}$
$F=4.199\times10^{-6}N\approx4.2\times10^{-6}N$
