# How can I split a resultant force into its $x$ and $y$ components?

Point charge 3.5μC is located at x = 0, y = 0.30 m, point charge -3.5μC 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 = 5.0μC at x = 0.40 m, y = 0?

I first drew a diagram, showing that it can be broken down into two .3, .4, .5 triangles, leaving me with a theta of 36.87°. I then calculated that because the charges are equal, each individual force enacted on charge Q is .63N. I know I need to split the force into it's x and y components to find the resultant vector using R=√Rx^2 + Ry^2, but I do not know how to split these forces up.

• I see that the resultant will be going straight down now, but how do I calculate the x component of each force to find the magnitude of the force on Q?
– Cre
Sep 2 '14 at 7:03
• Be careful, the charges are equal in magnitude but opposite in sign. This means that the force will change directions as well (as is indicated by the direction of the arrows in your diagram). Oct 3 '14 at 16:42

• In particular note that $\sin\theta = 0.3/0.5$ Sep 2 '14 at 7:26