Point charge 2.5μC is located at x = 0, y = 0.30 m, point charge -2.5μC is located at x = 0 y = -0.30 m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.0μC at x = 0.40 m, y = 0?

Question

anonymous · Answer

I solved half of the problem , and left the rest for you to complete by just repeating the steps I used and finding the Force magnitude for each and then add them together :
$$\left| F \right|= \sqrt{(x-component)^{2}+(y-component)^{2}}$$  
do this for each force and then add the results together and you will get the force magnitude ! 
as for the direction , you just need to add the Forces vectors together ( each component is added to the corresponding component from the other force ) and then you shall have a vector like this : 
|dw:1392637683926:dw|