Solving Dipole Interaction Problems

[Lengalicious]

Homework Statement

See attachment

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.

 

[mfb]

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]

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]

This is correct.

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

 

[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.

 

[mfb]

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]

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]

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]

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]

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]

You're expanding about x/r = 0.

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

 

[Lengalicious]

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]

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]

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.