Answer
$q_{2}=-7.3046\mu C$
Work Step by Step
There are two charges $q_{1}=+3.5 \mu C $, and and $q_{2}$
distance between the charges is $r=0.26 m$
magnitude of charge $|q_{1}|=3.5 \mu C= 3.5\times10^{-6}C$
suppose magnitude of unknown charge is $q_{2}$
magnitude of force on charge $q_{2}$ due to charge $q_{1} $ is $F= 3.4N$
From coulombs law magnitude of force is given by $F=k\frac{|q_{1}|\times |q_{2}|}{r^2}$
putting the values in above equation
$k=8.99\times10^{9}N.m^2/C^2$,
$F= 3.4N$,
$r=0.26 m$
$|q_{1}|=3.5 \mu C $
$3.4N=8.99\times10^{9}N.m^2/C^2\times\frac{ 3.5\times10^{-6}C\times |q_{2}|}{(0.26m)^2}$
$|q_{2}|=\frac{3.4N\times(0.26m)^2}{8.99\times10^{9}N.m^2/C^2\times3.5\times10^{-6}C}=7.3046\times10^{-6}C$
$|q_{2}|=7.3046\mu C$
since particle 1 experiences attractive force $q_{2}$ must be a negative charged particle.
$q_{2}=-7.3046\mu C$
