# Electric Charge

1. Jul 4, 2006

### sophzilla

Hello -

I worked out this problem but I got a wrong answer.

First, I used the Pythagorean theorem to find the radial distance between A and each charges. So 1.2m divided by 2 (= 0.6), then the square root of .6 squared + .6 squared = .849, which is the radius.

Then I used Coulomb's Law to calculate the net force:

kqAq1/R2 + kqAq2/R2 + ... and so forth.

I took out the kqA/R2, which is the same for all, and came up with:

kqA/R2 (q1 + q2 + q3 + q4).

But it so happens that the numbers inside the parenthesis turns out to be 0.

What did I do wrong?

2. Jul 4, 2006

### neutrino

I think you're forgetting the vector nature of force. Simply adding up the numbers won't do any good. You have to add them vectorially.

3. Jul 4, 2006

### Kurdt

Staff Emeritus
First your pythagorean theorem is a bit off. it should be:

$$a^2=b^2+c^2$$

EDIT: You got it right I misinterpreted what you had done originally.

Then you have neglected to take any directions when working out the force so try setting up a reference freame and adding the directions into your equation.

4. Jul 4, 2006

### nazzard

Hello sophzilla,

I think you don't take into account that the forces are vectors.

You can start applying Coulomb's law for one diagonal at a time. For instance $$Q_4, q, Q_2$$. Do you agree that the 2 forces will add up to a net force pointing from $$q$$ towards $$Q_2$$?

Regards,

nazzard

5. Jul 4, 2006

### sophzilla

I would appreciate any help. Thanks.

6. Jul 4, 2006

### Kurdt

Staff Emeritus
Those are not vectors. That is why it is not working. Consider the unit vectors i and j and how they would add up to pointin the directions you require to the charges from the centre. The magnitudes are then as you have calculated.