Answer
The net electric force on charge B is $4.05\times 10^{-4}~N$ directed downward.
Work Step by Step
We can find the magnitude of the electric force on charge B from charge A:
$F = \frac{k~q_A~q_B}{r^2}$
$F = \frac{(9.0\times 10^9~N~m^2/C^2)(1.0\times 10^{-9}~C)(2.0\times 10^{-9}~C)}{(0.020~m)^2}$
$F = 4.5\times 10^{-5}~N$
A force of $4.5\times 10^{-5}~N$ is directed down.
We can find the magnitude of the electric force on charge B from charge C:
$F = \frac{k~q_B~q_C}{r^2}$
$F = \frac{(9.0\times 10^9~N~m^2/C^2)(2.0\times 10^{-9}~C)(2.0\times 10^{-9}~C)}{(0.010~m)^2}$
$F = 3.6\times 10^{-4}~N$
A force of $3.6\times 10^{-4}~N$ is directed down.
Since both forces are directed down, we can add the forces to find the net force.
$F_{net} = 4.5\times 10^{-5}~N+3.6\times 10^{-4}~N$
$F_{net} = 4.05\times 10^{-4}~N$
The net electric force on charge B is $4.05\times 10^{-4}~N$ directed downward.
