- #1

whitejac

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## Homework Statement

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.7 cm . Two of the particles have a negative charge: q1 = -6.0 nC and q2 = -12.0 nC. The remaining particle has a positive charge, q3= 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

## Homework Equations

Fcos(θ) = Fx

Fsin(θ) = Fy

θ = Tanh(Fy/Fx)

F = (Fx^2 + Fy^2)^1/2

F(2on3) = K(q1q2)/r^2)

## The Attempt at a Solution

Using the program's help function, I have found the components

(F1on3)x = 2.96x10^-4

(F1on3)y = 5.13x10^-4

So the only thing left was to find F(2on3) in the same way I found F(1on3).

Arranging the problem so that q3 was at the origin, q1 was at the top of the triangle and q2 was along the x-axis, I found this:

Fy = 0 (sin0=0)

Fx = (9x10^9)(12x10^-9)(8x10^-9) / (2.7x10^-2)^2

= 1.19x10^-3

This makes sense given that it should attract.

Here's where the problem falls off of the rails...

I need to find Theta

So Tanh (5.13x10^-4)/((2.96x10^-4)+(1.19x10^-3)) = 0.006

I really need this to be in the vicinity of 19degrees. I do not understand what mistake I am making... Does anybody else?