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Dipole interaction 
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#1
Feb2513, 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)/(rd/2+D/2) + (Q)(q)/(r+d/2+D/2) + (Q)(q)/(rd/2D/2) + (Q)(q)/(r+d/2D/2)] , but this doesn't work, could someone point me in the right direction. 


#2
Feb2513, 05:05 PM

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P: 12,081

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
Feb2513, 06:23 PM

P: 164




#4
Feb2613, 07:10 AM

Mentor
P: 12,081

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


#5
Feb2613, 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)/(rd/2+D/2)] take r = 1/(rd/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.



#6
Feb2613, 12:32 PM

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P: 12,081

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
Feb2613, 12:55 PM

P: 164




#8
Feb2613, 01:04 PM

Mentor
P: 12,081

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
Feb2613, 01:38 PM

P: 164




#10
Feb2613, 02:46 PM

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Thanks
PF Gold
P: 11,868

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



#11
Feb2613, 02:47 PM

P: 164




#12
Feb2613, 03:15 PM

P: 164




#13
Feb2613, 03:19 PM

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PF Gold
P: 11,868

Nope, that's fine. If you look at the result you're trying to arrive at, you should notice it has r^{3} on the bottom, so you want the 1/r and 1/r^{2} terms to cancel out.



#14
Feb2613, 03:22 PM

P: 164




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