Forces between 3 charged particles

[Badger]

Homework Statement

What is the force F_vec on the 1 nC charge?
Give answer as magnitude in Newtons.

Homework Equations

F = K qq/r^2
K = 1/4(pi)epsilon_knot; = 8991804694 N m^2/C

The Attempt at a Solution

A. Convert it all. cm to m. nC to C.
B. Vectors for Forces. The two x-component forces from the two 2 nC charges are equal and opposite (both repelling the 1 nC charge) so they cancel.
So F(x_net) = 0 Newtons
C. Similarly the y-component forces are the equal and in the same direction. So I merely found one and doubled it.
To obtain y, I used the ole: y = sqroot[ (.01m)^2 - (.01m/2)^2) = .00866m

F = Fnet_y = 2F_y = [2 * (8.99 Nm^2/C^) * (2E-9 C)(1E-9 C)] / (.00866m)^2
F = 4.80 * 10^-4 N

where did i go wrong? thanks

 

[rootX]

C. Similarly the y-component forces are the equal and in the same direction. So I merely found one and doubled it.
To obtain y, I used the ole: y = sqroot[ (.01m)^2 - (.01m/2)^2) = .00866m

F = Fnet_y = 2F_y = [2 * (8.99 Nm^2/C^) * (2E-9 C)(1E-9 C)] / (.00866m)^2
F = 4.80 * 10^-4 N

where did i go wrong? thanks

F = Fnet_y = 2F_y = [2 * (8.99 Nm^2/C^) * (2E-9 C)(1E-9 C)] / (.00866m)^2

means that two charges are separated by 0.00866 m, according to definition of this formula. But are they really separated by that distance?

You are saying here that move both 2 nC charges to the midpoint of the bottom line!

 

[Badger]

cool i got it now. thanks