Q. Two point charges are $3\, m$ apart and their combined charge is $8 \mu C$. The force of repulsion between them is $0.012\, N$. Charges are
Solution:
$q_1 + q_2 = 8\, \mu C \, \, \, \, \, \, \, \, \, \, \, \, \, $ ....(i)
$F = 0.012 \: N$
or, $k \frac{q_1 q_2}{r^2} = 0.012$
or, $9 \times 10^9 \times \frac{q_1 q_2}{3^2} = 0.012$
or, $q_1q_2 = 12 \times 10^{-12} C \, \, \, \, \, \, \, \, \, \, \, \, \, $ ...(ii)_
Now, $(q_1 - q_2)^2 = (q_1 + q_2)^2 - 4 q_1 q_2$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, = 8^2 - 4 \times 12 \, \, \, \, \, \, $ [Using eqn (i) arid (iii)]
$ \, \, \, \, \, \, \, \, \, \, \, \, \, = 16$
$q_1 - q_2 = \pm 4 \mu C \, \, \, \, \, \, \, \, \, \, \, \, \, $ ...(iii)
Solving eqn (i) and (iii), we get, charges are $6 \, \mu C$ and $2 \, \mu C$
