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## Homework Statement

Four charges Aq,Bq,Cq, and Dq (q = 5.00 × 10-7C) sit in a plane at the corners of a square whose sides have length d = 83.0 cm, as shown in the diagram below. A charge, Eq, is placed at the origin at the center of the square.

DATA: A = 2, B = 4, C = 7, D = 4, E = -5. Consider the charge at the center of the square. What is the net force, in the x-direction, on this charge?

[PLAIN]http://img151.imageshack.us/img151/8150/cleancopycopy.png [Broken]

## Homework Equations

F=[tex]\frac{kQq}{r^{2}}[/tex]

F=qE

## The Attempt at a Solution

What I did, is I found the net force for each side. Since Force=qE, it would be simple subtraction of the numbers. This gave the vector direction and magnitude of each force. The resulting force of Bq and Dq was 0 while the resulting force of Aq and Cq was 5q SouthEast (represented in the picture below).

[PLAIN]http://img14.imageshack.us/img14/9906/cleancopy.png [Broken]

The distance between Dq and Eq would be equal to [tex]\frac{\sqrt{d^{2}+d^{2}}}{2}[/tex] since it would only be half the distance of the triangle created by using Pythagoras's theorem (which I got as a value of .5869m).

Now I used the formula F=[tex]\frac{kQq}{r^{2}}[/tex]

F=[tex]\frac{8.988x10^9\frac{N*m^{2}}{C^{2}}*5*5x10^{-7}C*-5*5x10^{-7}C}{(.5869m)^{2}}[/tex]*cos(45°)+0=-1.153x10[tex]^{-1}[/tex]N

*Note: the "0" is in there because the force vector of Bq to Cq resulted as 0.*

I tried this but it says that it's the wrong answer. If you could help me, that would be great.

Thanks!

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