- #1

Ammora

- 17

- 0

## Homework Statement

Three point charges are fixed in place in the right triangle shown below, in which q1 = 0.71 µC and q2 = -0.67 µC. What is the electric force on the +1.0-µC (let's call this q3) charge due to the other two charges?

image: http://www.webassign.net/grr/p16-15alt.gif

magnitude ____ N

direction ______°, measured counterclockwise from the +x-axis

## Homework Equations

kq/r^2

## The Attempt at a Solution

The only thing I don't know how to do is calculate the x and y components of the charge on q3 due to q1.

Here's what I did:

First I found the missing side, 6 cm. Then I converted all the cm to m by multiplying it by .1 m and then I converted all the uC to C by multiplying it by X10^-6.

q1: Fx= k(q1)(q3)/r^2

= 8.99e9 (0.71e-6) (1.0e-6)/(0.10)^2 * 6/10

Fy=k(q1)(q3)/r^2

= 8.99e9 (0.71e-6) (1.0e-6)/(0.10)^2 * 8/10

q2: Fx=k(q2)(q3)/r^2

=8.99e9 (-0.67e-6) (1.0e-6)/(0.10)^2

Total force= Fx added together. And then sqrt. (Fx^2 + Fy^2)

As for the direction, tan theta= (Fy/Fx) and then add 180 to it?

Please help me, it's due tonight, thank you so much in advance.