# Electric Force: Equilateral Triangle

1. Jul 17, 2014

### Sirmeris

1. The problem statement, all variables and given/known data
Three charges, +Q, +Q, and -Q are placed at the vertices of an equilateral triangle of length "s" on a side. Find the magnitude and direction of the force on one of the +Q charges.

2. Relevant equations
ForceElectric=(K*q1*q2/d2)

3. The attempt at a solution
I've set the triangle up so that the bottom side is parallel to the x-axis. My +Q charges are in both the bottom left (charge 1) and bottom right (charge 3) corners, while the -Q charge (charge 2) is at the top. Charge 3 is the charge that I have chosen to do the calculation for.

Force2 on 3 = (2QK/s2)

Force2 on 3 = Force1 on 3, so Force1 on 3 = (2QK/s2)

x-component: Forcex:2 on 3 = (2QK/s2)*cos(60°) = (QK/s2)

y-component: Forcey:2 on 3 = (2QK/s2)*sin(60°)

∑Fx = Force1 on 3 - Forcex:2 on 3

∑Fx = (2QK/s2) -(QK/s2)

∑Fy = Forcey:2 on 3

∑Fy = (2QK/s2)*sin(60°)

Fnet = ∑Fy + ∑Fx (vector equation)

Fnet = √((∑Fy)2 + (∑Fx)2)

Simplified this (hopefully) goes to:

Fnet = (2QK/s2)

This was my final answer for the magnitude, but I'm not even sure if it is correct, or if my procedure is right. I'm also not sure how to find the direction. Any help would be greatly appreciated! Thank-you!

2. Jul 17, 2014

### Staff: Mentor

Hi Sirmeris. Welcome to Physics Forums.

Q x Q = Q2

3. Jul 17, 2014

### Staff: Mentor

Once you have ∑Fx and ∑Fy you can determine the magnitude using Pythagoras, as you show, and the angle, θ, of that vector is given as

θ = tan-1 (∑Fy / ∑Fx)

Always wise to sketch these, so you can be sure to give your final answer in the correct quadrant.

4. Jul 18, 2014

### Satvik Pandey

Hi Sirmeris
Use this diagram to find net force on +Q.
Net.F=√(F^2+F^2+2F*Fcosθ)
θ=120.In this problem force on +Q due to -Q and other +Q charge is equal.

5. Jul 18, 2014

### Satvik Pandey

this is the figure.
What should be the direction of the resultant.