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Dipole interaction

by Lengalicious
Tags: dipole, interaction
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Lengalicious
#1
Feb25-13, 01:02 PM
P: 164
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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mfb
#2
Feb25-13, 05:05 PM
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P: 11,617
With taylor series for those factors of 1/(r+d/2) and similar (and assuming d/2 << r), you should get the correct result.
Lengalicious
#3
Feb25-13, 06:23 PM
P: 164
Quote Quote by mfb View Post
With taylor series for those factors of 1/(r+d/2) and similar (and assuming d/2 << r), you should get the correct result.
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?

mfb
#4
Feb26-13, 07:10 AM
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P: 11,617
Dipole interaction

This is correct.

And right, you do not get "exactly" 1/(r+d/2), but you get similar expressions.
Lengalicious
#5
Feb26-13, 11:51 AM
P: 164
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.
mfb
#6
Feb26-13, 12:32 PM
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P: 11,617
The difference between 1/r and your terms is the point of the expansion.

$$\frac{1}{r+x} = \frac{1}{r} \cdot \frac{1}{1+\frac{x}{r}} \approx \frac{1}{r} (1 \pm \dots)$$
Lengalicious
#7
Feb26-13, 12:55 PM
P: 164
Quote Quote by mfb View Post
The difference between 1/r and your terms is the point of the expansion.

$$\frac{1}{r+x} = \frac{1}{r} \cdot \frac{1}{1+\frac{x}{r}} \approx \frac{1}{r} (1 \pm \dots)$$
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?
mfb
#8
Feb26-13, 01:04 PM
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P: 11,617
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.
Lengalicious
#9
Feb26-13, 01:38 PM
P: 164
Quote Quote by mfb View Post
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.
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 :(.
vela
#10
Feb26-13, 02:46 PM
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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##.
Lengalicious
#11
Feb26-13, 02:47 PM
P: 164
Quote Quote by vela View Post
You're expanding about x/r = 0.
Yey, thanks very much to both of you for the help.
Lengalicious
#12
Feb26-13, 03:15 PM
P: 164
Quote Quote by vela View Post
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##.
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?
vela
#13
Feb26-13, 03:19 PM
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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.
Lengalicious
#14
Feb26-13, 03:22 PM
P: 164
Quote Quote by vela View Post
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.
Ok brilliant, thanks a bunch for the help, I'll take it from here just got to get the manipulation right :D.


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