What's the minimum number of charges (with same charge in magn.) in a dielectric sphere so that if you are far away from this sphere the E. field falls with the inverse 4th power of the distance?

Question

anonymous · Answer

With the dipole you have the inverse cubic, and  by Gauss' comes out that $$\sum_{?}^{?}q$$ must be zero (if not it will result a net inverse 2nd power force). I guess there should e al least 4, but i can't prove it... on the contrary it comes out the inverse 3rd power!! Anyone has any idea?

festinger · Answer

I believe what you are looking for is the "multipole expansion" for electrostatics. In particular, you are looking for the quadrapole term, whose potential term can be found in the following link. Since $ \textbf{E}=\nabla \cdot V $, the $\frac{1}{r^{3}}$ potential term for quadrapole becomes a $\frac{1}{r^{4}}$ term. http://physicspages.com/2012/02/20/multipole-expansion-in-electrostatics/

festinger · Answer

$\textbf{E}=-\nabla V$

anonymous · Answer

Thanks for your reply, but I didn't do gradients yet.. so according to you what would be the answer (the minimum number of charges)?

anonymous · Answer

The functional behavior of the field due to a charge distribution does not depend on the absolute magnitude of the charge.  It depends on the distribution of the charge so doubling or halving the charge does not affect the relative amount of a 1/R^4 component compared to 1/R^2 or any other dependence.

anonymous · Answer

So to check if I've understood: on the x axis the E field falls off 1/r^2 and in other direction such y axis in different ways?|dw:1408296100154:dw|

anonymous · Answer

1/r^3 sorry

anonymous · Answer

For a uniformly polarized , linear, isotropic, homogeneous dielectric sphere the field varies as 1/R^3 in all directions. There is no 1/r^4 dependence at any distance.  There is of course a dependence on the angular distance from the axis of the polarization.

anonymous · Answer

Ok thanks you

anonymous · Answer

Ok I solved by myself, here's the solution I found:

First, the number has to be equal. That's because a odd number of charge means $$\sum_{0}^{n}\neq0$$, and by Gauss theorem you'll have a net charge, and so a 1/R^2 Electric field.

So the mimum could be 2, but it's easy to demonstrate that the field would fall with 1/R^3, and that's not our goal.

That I tried with 4 charges and I found the solution:

Put the charges on the vertex of a square, than measure E net on a point P which is on a line perpendicular to the square and passing through a vertex. P is D far from the nearest vertex and the side of the square is d.

Assume that D>>d

Here's the calculations:

|dw:1408522105360:dw|

$$|Enet|= |E_1|+|E_3|-|E_2|-|E_4|$$
$$|E_2|=|E_4|$$
$$|Enet|= |E_1|+|E_3|-2|E_2|$$
$$|Enet|= \frac{ q }{ 4\pi \epsilon_o }\left[ \frac{ 1 }{ D^2 }+\frac{ 1 }{ (D^2+2d^2) }+\frac{ 1 }{ (D^2+d^2) } \right]$$
Do Least Common Denominator:

$$|Enet|=\frac{ q }{ 4\pi \epsilon_o }\frac{ D^4+2D^2d^2+d^2D^2+2d^4+D^4+D^2d^2-2D^4-2D^2d^2 }{ D^6+2d^2D^4+d^2D^4+2d^4D^2}$$

$$d^n$$ terms are very small and u can forget about them

so simplify and you'll have:
$$|Enet|=\frac{ q }{ 4\pi \epsilon_o }\frac{ d^2 }{ D^4+3d^2D^2 }$$
  And here's the 1/R^4 falling dependence.
So the anwer s 4, thanks to anyone who answered :)

festinger · Answer

I didn't think of the problem like that. it's quite interesting! I think my lecturer made us to a proof like this once before but I forgot about it in favor slightly more powerful(and more complicated looking) multipole expansion. What you have described here is a quadrapole: http://www.bem.fi/book/07/fi/0701.gif Don't worry about $a_{20}$. I believe for the negative charge it should actually be -2 instead.

anonymous · Answer

Thanks you! For sure your version is more powerful and complete but I haven't done yet a lot of things so I could understand a very little :) you mean that in a20 -1 is  wrong and should be instead -2?