# Electric dipole charges/Electric Field

## Homework Statement

Two point charges likes those in the figure below are called an electric dipole. Show that the electric field at a distant point along the x-axis is given by $$E_{x}=\frac{4k_{e}qa}{x^3}$$
Figure: http://img300.imageshack.us/my.php?image=58ag9.png

## Homework Equations

Electric field equation: $$E=\frac{k_{e}q}{r^2}$$
Anyothers?

## The Attempt at a Solution

I'm unsure of how to apply the electric field equation to this problem (if it is even going to be used). I'm unfamiliar with electric dipoles and certaintly haven't been dealing with the electric fields of them. Could someone give me a hint as to where I should start on this problem? I appreciate it, thanks!

P.s. I think that this may be in relation with this problem: http://www.sciforums.com/showthread.php?t=62789 but I'm unsure of what they mean and why the distance on the bottom of the fraction is cubed.

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hage567
Homework Helper
That's the right equation to use. The "r" in the equation means the distance between the charge and the point you are looking at. On your picture, put a point out somewhere on the x-axis at a distance x. What is the distance between each charge and that point?
The electric field is a vector quantity, so you have to sum the relevant components at x due to each charge.

(You'll see how the x^3 comes in later when you make use of the fact that it is at a "distant" point.)

Well, at point x the positive charge on the right creates a field kq/(x-a)^2, this field is directed in the right direction. The negative charge creates a field of magnitude kq/(x+a)^2, and this field is pointed to the left. To get the net field, subtract the field created by negative charge from the field created by the positive charge: (kq/(x-a)^2)-(kq/(x+a)^2). After some algebraic manipulations you should get: 4axkq/(x^2-a^2)^2. Since x is large compared to 'a', the 'a' in the denominator can be ignored, so you get: 4axkq/(x^2)^2, after some minor manipulations you get: 4akq/x^3.

Gotcha! Thanks!