# Force between electric dipoles along the dipole axis

## Homework Statement

Compute the force between two identical dipoles. See Problem P63 on page 544 to set up the problem. Explain why this result makes sense by comparing it to the force between two point charges and the force between a point charge and a dipole, in terms of the relationship to the distance between the two objects.
P64 shows the two dipoles that are parallel and in the same orientation, separated by a distance r from their centers and have a distance of s between them. The charges that make up the dipole are -q and +q.

## Homework Equations

E(dipole)=2kqs/r^3

## The Attempt at a Solution

I thought I would calculate the force on each charge of the dipole on the right by using the equation above and F=Eq. Then I would add the two. For r, I'm confused about weather I should use (r+s/2) and (r-s/2).When I tried this, I couldn't find a way to simplify to the desired answer which I know is 6kq^2s^2/r^4.
Thanks so much for any help in advance.

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TSny
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## The Attempt at a Solution

I thought I would calculate the force on each charge of the dipole on the right by using the equation above and F=Eq. Then I would add the two.
Sounds good.
For r, I'm confused about weather I should use (r+s/2) and (r-s/2). When I tried this, I couldn't find a way to simplify to the desired answer which I know is 6kq^2s^2/r^4.
I think you have a good approach. The "binomial approximation" might be of use.
https://en.wikipedia.org/wiki/Binomial_approximation

• NihalRi
Sounds good. I think you have a good approach. The "binomial approximation" might be of use.
https://en.wikipedia.org/wiki/Binomial_approximation
My teacher just told me to do it differently by calculating four forces using coulomb's law. How do binomial approximations work, I know how to do regular binomial expansions.

TSny
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My teacher just told me to do it differently by calculating four forces using coulomb's law. How do binomial approximations work, I know how to do regular binomial expansions.
Doing it this way, I think you will need to go beyond the binomial expansion (i.e., beyond first order in the small quantity s/r).

Let's see your expressions for the forces on each charge of the dipole on the right.

• NihalRi
Doing it this way, I think you will need to go beyond the binomial expansion (i.e., beyond first order in the small quantity s/r).

Let's see your expressions for the forces on each charge of the dipole on the right.

Doing it this way, I think you will need to go beyond the binomial expansion (i.e., beyond first order in the small quantity s/r).

Let's see your expressions for the forces on each charge of the dipole on the right. TSny
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OK, this looks correct for the net force on the dipole. You need to approximate this expression using the condition that s << r; or, equivalently, s/r << 1. Can you manipulate each of the last two terms in F so that they are expressed in terms of the small quantity s/r?

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OK, this looks correct for the next force on the dipole. You need to approximate this expression using the condition that s << r; or, equivalently, s/r << 1. Can you manipulate each of the last two terms in F so that they are expressed in terms of the small quantity s/r?  TSny
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OK. Looks good. Do you know how to approximate ##\frac{1}{(1+s/r)^2}##?

OK. Looks good. Do you know how to approximate ##\frac{1}{(1+s/r)^2}##?
unfortunately not, does it have something to do with a McLaurin expansion? I couldn't find any clear explanation online.

how many terms do you think I have to use? I tried to use the first three and I got,  TSny
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You're almost there. You have a sign error in the first equation of the second picture. The expansion of ##\frac{1}{(1+s/r)^2}## does not equal the expansion of ##\frac{1}{(1-s/r)^2}##.

I don't quite follow your cancellations in the second equation of the second picture.

You're almost there. You have a sign error in the first equation of the second picture. The expansion of ##\frac{1}{(1+s/r)^2}## does not equal the expansion of ##\frac{1}{(1-s/r)^2}##.

I don't quite follow your cancellations in the second equation of the second picture.
I'll fix the signs :D. I thought that since s is much smaller than r, the terms with s In them will be negligible.

Okay so I fixed the signs and just expanded and found that it simplified to the right answer XD XD XD, thank you so soooo soooo much

TSny
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Good work!

• NihalRi