Q1 = 2.6*10-6 Coulombs (on the Top)
Q2 = -3.75*10-6 C (Right)
Q3 = 2.6*10-6 C (Left)
d = .250 m
So these 3 charges lie in the pattern of an equilateral triangle (with sides d). If we're only considering the y-part of the forces acting on Q2 , then Q3 doesn't matter (assuming the x-axis does lie between Q2 and Q3). All we need to consider is Q1's interaction with Q2. Because we've set our x-axis between Q2 and Q3, we'll have to modify our usual force equation. The magnitude for the force equation is
|F21 |= k*Q2 * Q1/d2
where k is the electrostatic constant. But remember, this gives us the total magnitude of the force of Q2 on Q1, where we only want the y-component. We need to represent F as a vector
F21 = |F21x|x^ + |F21y| y^)
where x^ is a unit vector in the x direction, y^ a unit vector in the y direction, and |F21x/y| dictate how much of the force F21 goes in either the x or y direction. Because we only care about the y direction, we can rewrite the above as
F21 ycomponent = |F21y| y^
We can the find magnitude of this ( |F21y|) by considering half of our equilateral triangle
Q1
| -
| - |F21|
||F21y| -
| ---------------- Q2
where the hypotenuse is the magnitude of the force between the two charges and |F21y| is the y-component of that force (same as the equation above, represented by the broken vertical line). Here, the angle directly below Q1 (between F21 and the vertical) is what will tell us how much of our force is in the y-direction. Because its a right triangle and we know the angle closest to Q2 is 60degrees (equilateral triangle), we know the angle near Q1 is 30 degrees (triangle angles = 180deg=90deg+30deg+60deg)
Looking at that angle, we know that
Cos[30deg] = |F21y|/ |F21| (adjacent/hypotenuse)
--> |F21y| = Cos[30deg]*|F21|
--> F21 ycomponent = Cos[30deg]*|F21| y^
=Cos[30deg]*kQ2Q1/d2 y^
Tamia L.
Thank you07/11/21