Answer
(a)Magnitude of force on each sphere is $F=0.8285N$
(b) Attractive
Work Step by Step
Distance between the spheres is $r=0.50m$
$1$ electron $= 1.6\times10^{-19}C $ charge
so$ 3.0\times10^{13}$ electrons is equal to $ 3.0\times10^{13}\times1.6\times10^{-19}C=4.8\times10^{-6}C $ charge
so sphere from where electron is removed will get positively charged due to electron deficiency
let us assume it $q_{1}=+4.8\times10^{-6}C $
similarly second sphere where these electron are placed will get negatively charged due to excessive electrons
let us assume it $q_{2}=-4.8\times10^{-6}C $
(a)
from coulombs law we know that magnitude of force between two charged objects is given by
$F=k\frac{|q_{1}|\times|q_{2}| }{r^2}$
in our problem it is given that magnitude of force
$|q_{1}|=4.8\times10^{-6} C$
$ |q_{2}| =4.8\times10^{-6} C$
$k=8.99\times10^{9}N.m^2/C^2$
$r=0.5m$
so magnitude of force will be
$F=8.99\times10^{9}N.m^2/C^2\times\frac{4.8\times10^{-6} C|\times4.8\times10^{-6} C }{(0.5m)^2}$
$F=828.518\times10^{-3}N$=$0.8285N$
magnitude of force on each sphere is $F=0.8285N$
(b) due to opposite polarity on spheres force will be attractive in nature.