# Homework Help: Dipoles and electric field expressions

1. Apr 20, 2012

### melissaisusi

Okay. So my assignment is due in an hour and i'm stuck on this last question. If I can get some help, I would greatly appreciate it.

So the problem reads: The electric field of a dipole can be calculated by assuming the positive charge q is at z = a/2 and the negative charge -q is at z = -a/2 (x=y=0). The electric field along the z axis found by vector addition of the electric field from the individual charges. FInd an expression for the electric field. (Hint: 1 / (1+x)^2 ) is approximately equal to 1 - 2x for small x.) By what power of z does the electric field fall off?

Okay. So from the problem, I know that I can solve using the equation:
|E1| = (1/4∏ε0)(q1/r^2)

but how do i associate z and r?

2. Apr 20, 2012

### collinsmark

Hello melissaisusi,

Welcome to physics forums!

Here are a few hints and guidance.

1. For this problem, don't bother figuring out the electric field for all space. Restrict yourself calculating the electric field to the z-axis. As a matter of fact, for the second part of this problem, the extreme positive z-axis is all that is necessary. (By that I mean you could break the problem into three parts and calculate the electric field on the positive and negative z-axis, and the portion of the z-axis in between the charges [and maybe you should do that for the first part, where you find an expression for E]. But sticking with the positive z-axis should be sufficient for the second part of this particular problem.)
2. Since you are only concerned about the z-axis, just set r = z.
3. And what is r (and thus z)? The distance r is the distance to the origin. But it's not the distance to the charge in this case. Every previous time you have used the
|E1| = (1/[4πε0])(q1/r2)
formula, the assumption was that the charge was centered at the origin. All that changes now. You need to find |E1| as a function of the distance from the charge q1. So you'll need to subtract off an a/2 somewhere from the distance to the origin.

Good luck!