Is this easy charge problem correct?

[fallen186]

Homework Statement

-----------------------------------------
A charge of -1.0 µC is located at the origin, a second charge of 2.6 µC is located at x = 0, y = 0.1 m, and a third charge of 13 µC is located at x = 0.2 m, y = 0. Find the forces that act on each of the three charges.
I need to find the y forces on the charge 2.6 µC. I have one last try.

Homework Equations

-----------------------------------------
F=k*((q₁)(q₂)/r²)
K = 9E9 Nm²/c²
q₁= charge 1 in columbs
q₂ = charge 2 in columbs
r = distance between the two charges (in meters)

a² + b² = c²
sin(xº)=O/H

#₁E#₂ = #₁*10^#₂

The Attempt at a Solution

-----------------------------------------
F₁ = 9E9 * ((2.6µC)*( -1.0µC))/.1²
F₁ = -2.34 N
*It's negative because its being pulled down due to attraction

F₂ = 9E9 * ((2.6µC)*( 13.0µC))/.224²
F₂ = 5.6 N
**.1m² +.2m² =.05 , .05^.5 = .224m **
**.1m and hyptonuse angle: sin^-1(.2m/.224) = 63.23º **
** Other angle is 26.76º (90º-63.23º)**
sin(xº)=O/H --> H*sin(xº)=O
5.6 N sin (26.76º) = +2.54 N

+2.54 N - 2.34 N = +.2 N
Is this the correct answer? +.2N
------------------------------------
I tried using 5.6cos(26.56º) = 5.009N (Hcos(xº) = A) for the x forces and it told me I was wrong. I plugged in 5 N & -5 N :|

 

[Delphi51]

The physics looks good. I didn't check the calculator work, except for the angle - and I got 26.56 degrees instead of your 26.76 degrees. Better check that again. Use tan instead of sin so you don't get any rounding error due to the hypotenuse calc.