Answer
$m = 2.2 \times 10^{-6} kg$
Work Step by Step
$Q_1 = 30 \times 10^{–9} C$ at $x = 0 m$
$Q_2 = -40 \times 10^{–9} C $ at $ x =0.72$
$Q_3 = 42 \times 10^{–6} C$ at $x = 0.28 m$
Now we want to find out what is the $F_{3 \space net}$
$F_{3 \space net} = F_{31} +F_{32} $
Where
$F_{31} = \frac{kq_3q_1}{(r_{31})^2}$ and $F_{32} = \frac{kq_3q_2}{(r_{32})^2}$
$ r_{31} = ( 0.28 m- 0m) = 0.28 m $
$ r_{32} = (0.72 m - 0.28 m = 0.44 m$
SO,
$F_{3 \space net} = F_{31} +F_{32} $
$F_{3 \space net} = \frac{kq_3q_1}{(0.28)^2} +\frac{kq_3|q_2|}{(0.44)^2} $
$F_{3 \space net} = k q_3 [\frac{q_1}{(0.28)^2} +\frac{|q_2|}{(0.44)^2}] $
$F_{3 \space net} = (8.99 \times 10^9N . m^2 / C^2) (42 \times 10^{–6} C) [\frac{30 \times 10^{–9} C}{(0.28)^2} + \frac{40 \times 10^{–9} C}{(0.44)^2}] $
$F_{3 \space net} = 0.22N$
From Newton's Second Law,
$m = \frac{F_{net}}{a}$
$m = \frac{0.22N}{100000 m/s^2}$
$m = 2.2 \times 10^{-6} kg$
