Voltage at Halfway Point Between Point Charges in an Equilateral Triangle

[rissa_rue13]

Homework Statement

There is an equilateral triangle with one point charge at each vertex. The point charges have charges of +Q, +2Q, and -4Q. The length of one side of the triangle is L. Determine an expression in simplest form for the voltage at a point halfway between the +Q and +2Q point charges.

Homework Equations

V=kQ/r
V_net=V₁+V₂+V₃

The Attempt at a Solution

Going off of V_net=V₁+V₂+V₃:
V_net= (kQ/r) + (k2Q/r) - (k4Q/r)
I think once substituting the values for L in place of r, you get:
V_net= (kQ/(L/2)) + (k2Q/(L/2)) - (k4Q/Lcos30)
Am I heading in the right direction?

 

[Redbelly98]

Yes, that looks good.

 

[rissa_rue13]

I think I made a mistake, I think it should be:
V_net= (kQ/(L/2)) + (k2Q/(L/2)) - (k4Q/(L/cos 30))

This can then be simplified:
V_net=kQ[(2/L)+(4/L)-(4cos30/L)]

Am I on the right track?

 

[LowlyPion]

I think I made a mistake, I think it should be:
V_net= (kQ/(L/2)) + (k2Q/(L/2)) - (k4Q/(L/cos 30))

This can then be simplified:
V_net=kQ[(2/L)+(4/L)-(4cos30/L)]

Am I on the right track?

The distance from the far vertex is L(cos 30). I think you were right the first time.