Finding the net force on a charge in the center of 4 other charges.

Cmertin

1. 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]

2. Homework Equations
F=$$\frac{kQq}{r^{2}}$$
F=qE

3. 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 $$\frac{\sqrt{d^{2}+d^{2}}}{2}$$ 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=$$\frac{kQq}{r^{2}}$$
F=$$\frac{8.988x10^9\frac{N*m^{2}}{C^{2}}*5*5x10^{-7}C*-5*5x10^{-7}C}{(.5869m)^{2}}$$*cos(45°)+0=-1.153x10$$^{-1}$$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!

Last edited by a moderator:
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gneill

Mentor
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 $$\frac{\sqrt{d^{2}+d^{2}}}{2}$$ 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=$$\frac{kQq}{r^{2}}$$
F=$$\frac{8.988x10^9\frac{N*m^{2}}{C^{2}}*5*5x10^{-7}C*-5*5x10^{-7}C}{.5869m^{2}}$$*cos(45°)+0=-1.153x10$$^{-1}$$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!
There are two components to the net force. You've calculated for one of them. Watch your signs.

Last edited by a moderator:

Cmertin

There are two components to the net force. You've calculated for one of them. Watch your signs.
Yes, there is going to be a y-component and an x-component however this question is only asking for
first post said:
What is the net force, in the x-direction, on this charge?
Thus, I'm confused as to what mistake I am making in finding the x-component of the vector.

gneill

Mentor
Looking at you diagram, what do you think the sign should be of the x-component of the force?

Cmertin

Looking at you diagram, what do you think the sign should be of the x-component of the force?
It should be positive then, right?

Mentor
Yup.

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