# Point Charges, Maginitudes and Electric Fields.

1. Sep 14, 2012

### Northbysouth

1. The problem statement, all variables and given/known data

Point charge 2.5 microC is located at x = 0, y = 0.30 m, point charge -2.5 microC is located at x = 0 y = -0.30 m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.5 microC at x = 0.40 m, y = 0?

2. Relevant equations
pythagorean theorem

E = kq/r^2
F = qE =

3. The attempt at a solution

I drew the charges on an x-y graph. Then I tried to find the electric field of a single charge:

E1 = (9x10^9 Nm^2/C)(2.5x10^-6 C)(4.5x10^-6 C)/(0.5^2 m)(0.4/0.5))

E1 = 0.50625 m

I then doubled this value to get 1.0125 m. Which is says is wrong. Any suggestions?

2. Sep 14, 2012

### voko

Why do you think you can just double the value?

3. Sep 14, 2012

### Northbysouth

Because there's two point charges.

4. Sep 14, 2012

### Muphrid

Vectors, man. Doing what you did is like saying you can add 1 mile north and 1 mile east and get 2. It's nonsense.

5. Sep 14, 2012

### Northbysouth

I've managed to get the right answer but I don't quite understand all of it. Here's what I did:

I labelled the point at (0, 0.30) as charge 1. I then found the force of this point using this fomrula:

F1 = k*q1*Q

where Q is the charge of the third point (4.5 micrC)

F1 = (9x10^9 Nm^2/C^2)(2.5x10^-6 C)(4.5x10^-6 C)/(0.5^2)
F1 = 0.405 N *(3/5)
F1 = 0.243 N

I then doubled this value to account fro the charge at point 2 as well, thereby giving me an F = 0.486 N

It says this is correct, but I'm a little unsure why I multiple 0.405 by (3/5). Could someone explain this to me?

6. Sep 14, 2012

### Muphrid

Why don't you find the x and y components of the force separately and see?

7. Sep 14, 2012

### Northbysouth

Is it equivalent to sin(θ) where θ is equal to the angle between point 3 (4.5 microC) and point 1 (2.5 microC)?

My understanding is that the equation:

F = K11q2/r^2 gives me the force of one charge acting on another. Thus this would mean that the force of charge 1, which I found to be 0.405 N would be acting from charge 1 to charge 3 (point Q with a charge of 4.5 microC)

Thus:

F1x = (0.405N)(0.3/0.5) = 0.243

F1y = (0.405N)(0.4/0.5) = 0.324

But I'm not sure I understand this either. I can take the magnitude of these two values and it'll give me 0.405N, which doesn't surprise me. Can you give me some more hints?

8. Sep 14, 2012

### Muphrid

When you add the forces from two points, you should add the components separately.

For one of the charges, you should get negative force component for one of the directions.