# Homework Help: Electrodynamic question

1. Jul 24, 2005

### asdf1

Can someone explain how to solve the following problem?
A physical electric dipole consists of two equal and opposite charges (+/-)q
separated by a distance d. Find the approximate potential at points far
from the dipole.

2. Jul 24, 2005

### marlon

What have you done up till now ?
Besides, you should have posted this in the HOMEWORK section.
This site gives the formula's you need, which you should also find in your textbook:

Dipole

regards
marlon

3. Jul 24, 2005

yeah, these problems were a bit of a pain, i thought.

find the potential at a far away point from the first charge. then find the potential at the same point for the second charge. add'em.

then you use a ton of approximations to get a neat solution. iirc correctly, be prepared to use taylor expansions... and then you can throw away all but a few terms.

4. Jul 24, 2005

### mukundpa

If the point is far from the dipole and not on the axis of the dipole then you can resolve the dipole along the line joining the mid-point of the dipole to the point and at rightangle
to it. potential due to perpendicular component is zero and the point is on the axis of the other component. You can use the formula for axil position.

5. Jul 25, 2005

### asdf1

opps~
new user...
thank you!
but i'm not too clear on how to make the approximations?

6. Jul 25, 2005

### mukundpa

-q.....M.......q....................................P
*.......d........* .
.........< ---------------r-------------->
If the point is at distance r from mid point of the dipole along the axis of the depole then the potential is
V = kq/(r - 0.5d) - kq/(r + 0.5d) = kqd/(r^2 - 0.25d^2)

then what if r >> d

Last edited: Jul 25, 2005
7. Jul 26, 2005

### asdf1

i see~
that clears a lot of things up!
thank you~
so if r>>d, then it V would be approximately 0?

8. Jul 26, 2005

### mukundpa

here if r>>0.25d but r is not very large we can neglect only 0.25d^2 as compared to r^2
say r =0.20 m and d= 0.001 m then r^2 = 0.04 m^2 and 0.25d^2 = 0.00000025 m^2, so as compared to r^2 we can neglect0.25 d^2, but can't say that the potential at 0.20 m is zeroo. Right?

9. Jul 26, 2005

### Jelfish

I think the confusion might be if you assume
$$r-0.5d\approx r$$ and $$r+0.5d\approx r$$
Then, it seems that the equation becomes
$$V= \frac{kq}{r-0.5d}-\frac{kq}{r+0.5d}\approx \frac{kq}{r}-\frac{kq}{r}=0$$

10. Jul 26, 2005

### mukundpa

If we are approximating, V is nearly equal to zero not exactly zero and we know that the potential, due to a point charge, is inversly proportional to squre of the distance, will be small enough and due to a dipole will be still smaller.

11. Jul 26, 2005

### Allday

This is a little video explaining the approximation techniques that you could use here. Taylor series and binomial expansion.

http://physicsstream.ucsd.edu/review/taylor.html [Broken]

Last edited by a moderator: May 2, 2017