# Homework Help: Point charges magnitude and direction

1. Sep 2, 2010

### J89

1. The problem statement, all variables and given/known data
Three point charges are located at the corners of an equilateral triangle as in the figure below. Find the magnitude and direction of the net electric force on the 0.40 µC charge. (A = 0.20 µC, B = 6.60 µC, and C = -3.80 µC.). Diagram below...

2. Relevant equations

F= Ke |q1||q2|/r^2, cos and sin symbols

3. The attempt at a solution

a) .20*10^-6*(6.60*10^-6)*8.99*10^9/(.500)^2 = .0474672
b) .20*10^-6*(3.80*10^-6)*8.99*10^9/(.500)^2 =.0273296

.0474672*cos(240) = .015463926386
.0474672*sin(240) = .044877634258

= .015463926386 + .0273296 = .042793526386

.042793526386^2 + .044877634258^2 = .003845287957 = sqrt(.003845287957 ) = 0.062 N. Direction is 275 degrees. However, according the real solution, the answer is not 0.062! Can someone find what I did wrong here?

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2. Sep 2, 2010

### Subdot

Where is the 0.40 µC charge located in the picture?

Edit: At any rate, I found the problem. You had your calculator in radians mode. In degrees mode, 0.0474672sin(240) = -0.0411.

Last edited: Sep 2, 2010
3. Sep 2, 2010

### J89

it is supposed to be .20, sorry, but thanks :)