# Field of a dipole (was: Hello, people)

1. Aug 28, 2009

### staetualex

Hy, i am back with my annoying questions about this and that. Could everyone tell me why the frigging dipole have an electric field that drops with 1/r^3? i mean, if i am far away , i should feel no electric field, you know, q2 = -q1 , so they cancel, but i feel (q1 q2 k)/r^3.What the HECK?

2. Aug 28, 2009

### HallsofIvy

Re: Hello, people

Why do you say "you know, $q_2 = q_1$ , so they cancel"? The field due to one pole depends on 1/r2 and the two distances are not the same so while $q_2- q_1= 0$, $q_2/r_2^2- q_1/r_1^2$] is not 0. It true that because of the equal charges in the denominator, that will drop off faster than just $1/r^2$- which is exactly what $1/r^3$ does.

3. Aug 28, 2009

### staetualex

Re: Hello, people

well, dipole have 2 charges equal but opposite. if i am far away, and i am a proton, i feel the force of a charge pushing me with a force equal x, but i also feel a force attracting me also of x magnitude. but the dipole's formula states that i should feel q1 x q2 x k divided by distance cubed. I am confused

4. Aug 28, 2009

### Born2bwire

HallsofIvy already explained it pretty well. The close proximity of the opposite charges changes the normal field equation. In the near field, we take the summation of the fields from the two charges, but in the far-field, the two charges can be approximated as a single dipole moment. This simplifies the math and introduces little error, the cancellation effect is shown in the fact that the dipole moment drops off as 1/r^3 here instead of 1/r^2.

5. Aug 28, 2009

### maverick_starstrider

Consider the potential (just take the gradient (times -1) to get the force):

$V(r)=kq(\frac{1}{r_{+}}-\frac{1}{r_{-}}$ if we apply cosine law and take d as the distance between the two charges and r as the distance from the point exactly between the two charges to somewhere in the distance we get $r_{\pm}^2=r^2+(\frac{d}{2})^2+2 r d cos \theta / 2 = r^2(1+\frac{d}{r}cos \theta + \frac{d^2}{4r^2})$ now if r is much greater than d then we neglect the d^2/r^2 term and get (with some reworking) $\frac{1}{r_{\pm}}=\frac{1}{r}(1 \pm \frac{d}{2r}cos \theta)$ so we have $\frac{1}{r_{+}}-\frac{1}{r_{-}}\approx \frac{d}{r^2}cos \theta$ so, finally, we have in the limit that r >> d V(r) is approximately $qk\frac{cos \theta}{r^2}$. Take the grad of that and you have a 1/r^3 relation.