Explanation:
We can represent the situation with the following figure
Therefore, we need to calculate the magnitude of F1 and F2.
The magnitude of electric force between two charges is equal to
[tex]F=k\frac{q_1q_2}{r^2}[/tex]Where k = 8.98 x 10^9, q1 and q2 are the charges and r is the distance between the charges. The distance between the object of charge -4 x 10^(-3) C and the object of charge +1C can be calculated using the Pythagorean theorem as
[tex]\begin{gathered} \text{ Pythagorean Theorem: c=}\sqrt{a^2+b^2} \\ \\ \text{ Replacing a = 50 and b = 50} \\ c=\sqrt{50^2+50^2} \\ c=\sqrt{2500+2500} \\ c=\sqrt{5000} \\ c=70.71\text{ m} \end{gathered}[/tex]Therefore, the distance between the charges is 70.71 m.
Then, replacing q1 = -4 x 10^(-3) C, q2 = +1C and r = 70.71 m, on the initial equation, we get:
[tex]\begin{gathered} F1=(8.98\times10^9)\frac{(-4\times10^{-3})(+1)}{70.71^2} \\ \\ F1=7184\text{ N} \end{gathered}[/tex]To calculate F2, we need to replace q1 = +2 x 10^(-3) C, q2 = +1C, and r = 50 m, so
[tex]\begin{gathered} F2=(8.98\times10^9)\frac{(2\times10^{-3})(1)}{50^2} \\ \\ F2=7184\text{ N} \end{gathered}[/tex]Now, we need to calculate the resultant force, so we need to identify the x and y coordinates of each force and add them
[tex]\begin{gathered} F1x=F1\cos(45)=7184\cos45=5079.86 \\ F1y=F1\sin(45)=7184\sin45=-5079.86 \\ F2x=0 \\ F2y=F2=7184 \\ \\ \text{ Resultant force} \\ Fx=F1x+F2x \\ Fx=5079.86+0=5079.86\text{ N} \\ \\ Fy=F1y+F2y \\ Fy=-5079.86+7184=2104.14\text{ N} \end{gathered}[/tex]Finally, we can calculate the magnitude and direction of the force as follows
[tex]\begin{gathered} \text{ magnitude } \\ F=\sqrt{(Fx)^2+(Fy)^2} \\ F=\sqrt{(5079.86)^2+(2104.14)^2} \\ F=5498.40\text{ N} \\ \\ \text{ Direction} \\ \theta=\tan^{-1}(\frac{Fy}{Fx}) \\ \\ \theta=\tan^{-1}(\frac{2104.14}{5079.86}) \\ \\ \theta=\tan^{-1}(0.41) \\ \theta=22.5° \end{gathered}[/tex]Therefore, the magnitude of the electric force is 5498.40 N and the direction is 22.5 degrees.
