Electric Force: Equilateral Triangle

[Sirmeris]

Homework Statement

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.

Homework Equations

Force_Electric=(K*q₁*q₂/d²)

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.

Force_{2 on 3} = (2QK/s²)

Force_{2 on 3} = Force_{1 on 3}, so Force_{1 on 3} = (2QK/s²)

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

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



∑F_x = Force_{1 on 3} - Force_{x:2 on 3}

∑F_x = (2QK/s²) -(QK/s²)


∑F_y = Force_{y:2 on 3}

∑F_y = (2QK/s²)*sin(60°)




F_net = ∑F_y + ∑F_x (vector equation)

F_net = √((∑F_y)² + (∑F_x)²)

Simplified this (hopefully) goes to:



F_net = (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!

 

[NascentOxygen]

Hi Sirmeris. Welcome to Physics Forums.

Force2 on 3 = (2QK/s2)

Check your algebra ...

Q x Q = Q²

 

[NascentOxygen]

Fnet = ∑Fy + ∑Fx (vector equation)

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.

 

[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.

 

[Satvik Pandey]

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

 

[TaTyl]

View attachment 71444 this is the figure.
What should be the direction of the resultant.

East