Answer
(a) $F=1.4384\times10^{4}N$ attractive
(b)$F=3.2364\times10^{3}N$ repulsive
Work Step by Step
charge on sphere $1$ is $q_{1}= -20.0\mu C=-20.0\times10^{-6} C$
charge on sphere $2$ is $q_{2}= +50.0\mu C=+50.0\times10^{-6} C$
Distance between the spherical objects is $r=2.50cm=2.5\times10^{-2}m$
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
$|q_{1}|=20.0\times10^{-6} C$
$ |q_{2}| =50.0\times10^{-6} C$
$k=8.99\times10^{9}N.m^2/C^2$
$r=2.5\times10^{-2}m$
so magnitude of force will be
$F=8.99\times10^{9}N.m^2/C^2\times\frac{20.0\times10^{-6} C|\times50.0\times10^{-6} C }{(2.5\times10^{-2}m)^2}$
$F=1438.4\times10^{1}N$=$1.4384\times10^{4}N$
Since both charges are of opposite polarity they will experience attractive force.
(b)
when sphere are brought in contact charge distribution will occur and net charge will be
$q=q_{1}+q_{2}=-20\mu C+50\mu C=+30\mu C$
and then if they are separated they will have equal charges of
$q_{1}=q_{2}=\frac{q}{2}=\frac{+30\mu C}{2}=+15\mu C=+15\times10^{-6}C$
force between these sphere will be
$F=k\frac{|q_{1}|\times|q_{2}| }{r^2}$
in our problem it is given that magnitude of force
$|q_{1}|=15.0\times10^{-6} C$
$ |q_{2}| =15.0\times10^{-6} C$
$k=8.99\times10^{9}N.m^2/C^2$
$r=2.5\times10^{-2}m$
so magnitude of force will be
$F=8.99\times10^{9}N.m^2/C^2\times\frac{15.0\times10^{-6} C|\times15.0\times10^{-6} C }{(2.5\times10^{-2}m)^2}$
$F=323.64\times10^{1}N$=$3.2364\times10^{3}N$
since the charges on both the objects are of same polarity (positive) forces will be repulsive.
