# Electric Field Due to Point Charges?

1. Dec 4, 2005

### ylem

Hello!

Just hoping for a bit of help on a homework problem I'm doing at the moment...

Basically this is the question:

Three point charges are placed along the y axis: a charge q at y=a, a charge -2q at the origin and a charge q at q=-a. Such an arrangement is called an electric quadrupole.
a. Find the magnitude and direction of the electric quadrupole at points along the positive x-axis.
b. Use the binomial expansion to find an approximate expression for the electric field valid for x>>a. Contrast this behaviour to that of the electric field of a dipole.

I've tried to do this loads of different ways...

I've basically used E = 1/4pi...q/r^2 (didn't know how do get a symbol for the permittivity of free space!) and then for the vector have put (xi - aj) for q at a and so on... But then when I add them up to come up with a total E I just get something that looks really messy - although it is to the power of a half and could be expanded using the binomial, it just looks wrong...

Any input on how I could try to solve the question would be greatly appreciated!

Thanks, Sam(antha)

2. Dec 4, 2005

### siddharth

What you did seems to be right. Just find the vector sum of the Electric field due to the three charges. The y-component of the field due to the charge at y=a and y=-a should cancel. So along the x-axis, your net Field should also point along the x direction.
If you post and show exactly where you are having difficulties (ie, which equations) it will be easier to help.

Last edited: Dec 4, 2005
3. Dec 4, 2005

### ylem

Thanks!

This is where I'm getting confused though, because if the y components cancel then when I find the magnitude of the field I will just be squaring the x components, so I won't have anything to the power of a half to expand using the Binomial?

4. Dec 5, 2005

### siddharth

You will have a power of 3/2. When you take the x-component due to the 2 charges at y=a and y=-a, the field will be
$$\frac{q \cos\theta }{4 \pi \epsilon (a^2 + x^2)}$$
So find $$\cos \theta$$ and substiute to find the Field.
So you will be able to use your approximation when x>>a

Last edited: Dec 5, 2005