# Homework Help: Dipole interaction

1. Feb 25, 2013

### Lengalicious

1. The problem statement, all variables and given/known data

See attachment

2. Relevant equations

3. The attempt at a solution

Thus far I believe I am supposed to use calculate the interaction with kQq/Δr? I have tried summing over interactions so that 1/4∏ε*[(-Q)(-q)/(r-d/2+D/2) + (-Q)(q)/(r+d/2+D/2) + (Q)(-q)/(r-d/2-D/2) + (Q)(q)/(r+d/2-D/2)] , but this doesn't work, could someone point me in the right direction.

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Last edited: Feb 25, 2013
2. Feb 25, 2013

### Staff: Mentor

With taylor series for those factors of 1/(r+d/2) and similar (and assuming d/2 << r), you should get the correct result.

3. Feb 25, 2013

### Lengalicious

Ok, thanks, is kQq/Δr the correct expression for the interaction energy between 2 charges, and more importantly is that what i'm supposed to be using? Because if so I don't see how I get any 1/(r-d/2) terms, I just get stuff like 1/(r-d/2+D/2) unless I neglect the D/2 because r >> D?

4. Feb 26, 2013

### Staff: Mentor

This is correct.

And right, you do not get "exactly" 1/(r+d/2), but you get similar expressions.

5. Feb 26, 2013

### Lengalicious

Ok thanks for the help, I can't seem to figure out what point i'm supposed to be taylor expanding about? For instance for the interaction between -Q and -q, 1/4∏ε*[(-Q)(-q)/(r-d/2+D/2)] take r = 1/(r-d/2+D/2) and that r >> D & d, d/2 vanish & D/2 vanish? Then I am just left with 1/r, bit confused, even if i try to expand that without making the approximation mentioned I still am not quite sure about what point I am expanding about.

Last edited: Feb 26, 2013
6. Feb 26, 2013

### Staff: Mentor

The difference between 1/r and your terms is the point of the expansion.

$$\frac{1}{r+x} = \frac{1}{r} \cdot \frac{1}{1+\frac{x}{r}} \approx \frac{1}{r} (1 \pm \dots)$$

7. Feb 26, 2013

### Lengalicious

Yeh I got that bit but I mean if f(x) = 1/r * 1/(1 + x/r) and this expands to 1/r [f(a) + f'(a)(x-a) . . .] what is "a" supposed to be?

Last edited: Feb 26, 2013
8. Feb 26, 2013

### Staff: Mentor

That is the point where you develop your taylor expansion. It depends on the definition of the f you choose, but I would expect 1 there.

9. Feb 26, 2013

### Lengalicious

Sorry I think I'm confused because I'm not entirely sure what the point is i'm developing the taylor expansion from, because if r = (r-d/2+D/2) that's over a length not a point :S? so kQq/(r-d/2+D/2) where i take f(x) = 1/r*(1/(1-x/r)) where x = d/2-D/2, still can't see what the point i'm expanding about is? Is it a=d/2 or a=D/2 or even a=r? so f(x) ≈ 1/r*[1/(1-a/r) + 1/r*(1/(1-a/r)2)*(x-a)] but so confused about the a :(.

Last edited: Feb 26, 2013
10. Feb 26, 2013

### vela

Staff Emeritus
You're expanding about x/r = 0. To be a bit more explicit, the function you're expanding is $f(z) = \frac{1}{1-z}$ about $z=0$ and then plugging in $z=x/r$.

11. Feb 26, 2013

### Lengalicious

Yey, thanks very much to both of you for the help. :!!)

12. Feb 26, 2013

### Lengalicious

Actually I just tried that and my expansion ≈ 1/r + (d/2 + D/2)/r^2, then i tried for the remainder terms and the final result came to (2rQq - 2rQq)/r^2 = 0, i imagine my expansion came out wrong, is it wrong?

Last edited: Feb 26, 2013
13. Feb 26, 2013

### vela

Staff Emeritus
Nope, that's fine. If you look at the result you're trying to arrive at, you should notice it has r3 on the bottom, so you want the 1/r and 1/r2 terms to cancel out.

14. Feb 26, 2013

### Lengalicious

Ok brilliant, thanks a bunch for the help, I'll take it from here just got to get the manipulation right :D.