Answer
$q_1=14.5\mu C; q_2=47.5\mu C$
Work Step by Step
$q_1q_2=\frac{Fr^2}{K}$
We plug in the known values to obtain:
$q_1q_2=\frac{(85.0N)(0.270m)^2}{9.0\times 10^9m^2/C^2}$
$q_1q_2=6.885\times 10^{-10}C^2$
We also know that
$(q_2-q_1)^2=(q_1+q_2)^2-4q_1q_2$
$(q_2-q_1)^2=(62.0\times 10^{-6}C)^2-4(6.885\times 10^{-10}C)$
This simplifies to:
$q_2-q_1=3.3\times 10^{-5}C$....eq(1)
The total charge of the system is $q_1+q_2=62.0\times 10^{-6}C$....eq(2)
Adding eq(1) and eq(2), we obtain:
$(q_1+q_2)+(q_2-q_1)=62.0\times 10^{-6}C+3.3\times 10^{-5}C$
This simplifies to:
$q_2=47.5\mu C$
Using eq(2), we obtain:
$q_1+4.75\times 10^{-5}C=62.0\times 10^{-6}C$
This simplifies to:
$q_1=14.5\mu C$
