Solving Dipole Interaction Problems

In summary, this person is trying to solve for the interaction energy between two charges. They are using the calculate the interaction with kQq/Δr equation, but they are not getting the 1/(r-d/2+D/2) terms. They are also not sure what point they are expanding about.
  • #1
Lengalicious
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Homework Statement



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Homework Equations





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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  • #2
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
mfb said:
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?
 
  • #4
This is correct.

And right, you do not get "exactly" 1/(r+d/2), but you get similar expressions.
 
  • #5
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.
 
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  • #6
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
mfb said:
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?
 
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  • #8
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
mfb said:
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 :(.
 
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  • #10
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
vela said:
You're expanding about x/r = 0.

Yey, thanks very much to both of you for the help. :!)
 
  • #12
vela said:
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?
 
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  • #13
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
vela said:
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.
 

1. What is a dipole-dipole interaction?

A dipole-dipole interaction is a type of intermolecular force between two polar molecules. It occurs when the positive end of one molecule is attracted to the negative end of another molecule, resulting in a slight alignment of the molecules.

2. How do you calculate the strength of a dipole-dipole interaction?

The strength of a dipole-dipole interaction can be calculated using Coulomb's law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

3. What factors affect the strength of a dipole-dipole interaction?

The strength of a dipole-dipole interaction is affected by the magnitude of the dipole moment, the distance between the molecules, and the orientation of the dipoles relative to each other. The greater the dipole moment and the closer the molecules are, the stronger the interaction will be.

4. How does temperature affect dipole-dipole interactions?

At higher temperatures, molecules have more kinetic energy and are able to overcome the attractive forces of dipole-dipole interactions. As a result, dipole-dipole interactions become weaker and less significant at higher temperatures.

5. How can dipole-dipole interactions affect the physical properties of a substance?

Dipole-dipole interactions can contribute to the overall intermolecular forces present in a substance, affecting its boiling point, melting point, and viscosity. Substances with stronger dipole-dipole interactions will have higher boiling points, melting points, and viscosities compared to substances with weaker dipole-dipole interactions.

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