Answer
$a) \space q_1=q_2=7.416\times 10^{-7}C$
$b) \space q_1 =3.71\times 10^{-7}C,$ $q_2 = 14.84\times 10^{-7}C$
Work Step by Step
Force between two charges $q_1$ and $q_2$ separated at distance $r$ is given by $F = \frac{k q_1 q_2}{r^2} $
$q_1q_2 = {Fr^2\over k}$ with $k = 9\times 10^9Nm^2C^{-2}, F=0.220N, r = 15cm = 0.15m$
$a) \space$ If the two charges are equal i.e $q_1= q_2$
$q^2 = {Fr^2\over k}={0.220 \times (0.15)^2 \over 9\times 10^9} = 5.5\times 10^{-13}$
$q = 7.416\times 10^{-7}C$
$q_1=q_2=q=7.416\times 10^{-7}C$
$b) \space$ If one sphere has four times the charge of other i.e $q_1= q$ then $q_2 = 4q$
$4q^2 = {Fr^2\over k}={0.220 \times (0.15)^2 \over 9\times 10^9} = 5.5\times 10^{-13}$
$q = 3.71\times 10^{-7}C$
$q_1=q=3.71\times 10^{-7}C$
$q_2 = 4q = 14.84\times 10^{-7}C$
