Answer
The magnitude of the force exerted on the charge at the origin is $1.61~N$ and this force is directed in the positive x-direction.
Work Step by Step
The electric force exerted on the charge at the origin due to the $-5.00~\mu C$ is directed in the positive x-direction. We can find the magnitude of this force:
$F_1 = \frac{k~\vert q_1\vert \vert q_2 \vert}{r^2}$
$F_1 = \frac{(9.0\times 10^9~N~m^2/C^2)~(3.00~\times 10^{-6} C)(5.00 \times 10^{-6}~C)}{(0.20~m)^2}$
$F_1 = 3.375~N$
The electric force exerted on the charge at the origin due to the $8.00~\mu C$ is directed in the negative x-direction. We can find the magnitude of this force:
$F_2 = \frac{k~\vert q_1\vert \vert q_2 \vert}{r^2}$
$F_2 = \frac{(9.0\times 10^9~N~m^2/C^2)~(3.00~\times 10^{-6} C)(8.00 \times 10^{-6}~C)}{(0.35~m)^2}$
$F_2 = 1.763~N$
We can find the net force exerted on the charge at the origin:
$F = F_1-F_2$
$F = (3.375~N)-(1.763~N)$
$F = 1.61~N$
The magnitude of the force exerted on the charge at the origin is $1.61~N$ and this force is directed in the positive x-direction.
