# Homework Help: Simple coloumb's law problem

1. Feb 2, 2008

### awvvu

1. The problem statement, all variables and given/known data
4 charges are arranged in the corners of a square of lengths L as follows:

(-Q) --------- (+q)
|
|
|
(+4Q) ------- (-Q)

What is the magnitude of the force on the (+q)?

3. The attempt at a solution
My answer doesn't match up with my book's.

x-component of the force is, where K is $\frac{1}{4 \pi \epsilon_0}$:

$$F_x = K q(\frac{- Q}{L^2} + \frac{4Q}{2 L^2}) = K q \frac{Q}{L^2}$$

The x and y components of the force are equal by symmetry:

$$\vec{F} = K q \frac{Q}{L^2} (\hat{x} + \hat{y})$$

Therefore the magnitude is:

$$F = \sqrt{2} K q \frac{Q}{L^2}$$

However, my textbook says the answer is:

$$F = (2 - \sqrt{2}) K q \frac{Q}{L^2}$$

I have no idea where they got the "2" term from. I probably just made a careless mistake somewhere, but I can't see it.

Last edited: Feb 2, 2008
2. Feb 2, 2008

### awvvu

Bump -- anyone?

3. Feb 2, 2008

### Littlepig

so, in F_x,when computing F(4Q)

you must 1 compute L, and then compute projection in x, i guess that wasn't made...
i made $$Kq \frac{4Q}{\frac{L}{cos(45º)}^{2}}*cos(45º)$$

gretts
littlepig

P.s, Sorry, but i give up, 10m to try making the equation and i still couldn't do it...

Last edited: Feb 2, 2008
4. Feb 2, 2008

### awvvu

Ah, that's exactly it! I always forget to do that. Thanks.