## College Physics (4th Edition)

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.
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.