What Forces Act on a Charge at the Corner of a Rectangle?

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Homework Statement


Four charges are placed on the corners of a rectangle. What is the resultant force on the positive charge (a = 1.4 m, b = 0.9 m, q = 2.7 × 10-9C)?

the rectangle has a charge
-q____-q
-q____+q on the 4 corners of the rectangle ant a line r from the +q to the -q diagonally
the bottom is labeled a and the left side is labeled b

Homework Equations


kq/ r^2
Fx and Fy components


The Attempt at a Solution



i started by using k and q 8.99*10^9 * 2.7*10^-9

then i divided it by the values of a and b and r which i got by using tan inverse of 1.4/ .9 i got my three numbers

-8.091*10^-8 x component
-3.34*10^-8 y component
-2.3499*10^-8 both

and since there is only 2 to add together for each i found F of x and y by multiplying by sin theta which is 32.74 degrees for the y components and cos theta for x components

next i found those numbers and plugged it into the formula F = square root of Fx^2 + Fy^2 and i got 7.143*10^-8 which is wrong i don't know where i went wrong

thanks
 
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The Coulomb froce law incorporates a product of two charges; you only have one.

The diagonal distance should be obtained from the Pythagorean theorem; inverse tangent will give you an angle.

I have no idea what you did after that.

There are three Coulomb forces acting on the positive charge. Determine them, and then add them up vectorially. Try to work it out symbolically, first.