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If I take a certain dipole and by dipole I mean that two charges of opposite sign differentiated by very small distance. And if I take the formula of multipole expansion of potential then I see no other term except the dipole moment exists because everything except the dipole becomes zero. Now if I have a sphere whose charge density is not quite uniform (ie $\rho(r,\theta)= krcos\theta$, where $r$ and $\theta$ are spherical polar coordinates) but I know that it's two sides have same but opposite amount of charges then should I treat it as a dipole and by the expansion only the survives or do I have to make precise calculations as the result is unpredictable?

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  • $\begingroup$ What do you mean by “I see movement”? And can you please break your second very long sentence down into multiple sentences that are easier to understand? $\endgroup$ – G. Smith Mar 23 at 21:35
  • $\begingroup$ @G.Smith i am sorry I was voice typing that came out wrong so I edited it. $\endgroup$ – Nobody recognizeable Mar 23 at 21:37
  • $\begingroup$ @G.Smith is it ok now? $\endgroup$ – Nobody recognizeable Mar 23 at 21:41
  • $\begingroup$ No charge distribution can truly have only a dipole moment term in its multipole expansion--the higher moment terms might get as small as you want but one cannot have a charge distribution with only a dipole moment term. $\endgroup$ – Feynmans Out for Grumpy Cat Mar 23 at 22:01
  • $\begingroup$ @DvijMankad are you sure an exact dipole has higher order terms. Can you give reference to me as i know to compute till quadrupole moment. $\endgroup$ – Nobody recognizeable Mar 23 at 22:06
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This distribution has no net charge, but it has a dipole moment, a quadrupole moment, an octupole moment, etc. If you are approximating the potential, you can keep as few or as many of these as you need for the accuracy you desire. If you want an exact result, don’t use the multipole expansion.

The result is not “unpredictable”. The dipole term will decrease as $1/r^2$, the quadrupole as $1/r^3$, etc. Far away, the terms get smaller and smaller. But close in, they are all important.

If you take oppositely charged point particles and bring them together while letting the product of the charge and the separation stay constant, you get a pure dipole with no other moments. But your spherical distribution is not a pure dipole. It is dipole + higher moments.

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  • $\begingroup$ If we have a dipole then quadrupole and monopole moments have zero values ; dipole moment exists but does that say octupole moment and so on all are zero? $\endgroup$ – Nobody recognizeable Mar 23 at 21:49
  • $\begingroup$ If you take oppositely charged point particles and bring them together while letting the product of the charge and the separation stay constant, you get a pure dipole with no other moments. But your spherical distribution is not a pure dipole. It is dipole + higher moments. $\endgroup$ – G. Smith Mar 23 at 21:53
  • $\begingroup$ great ! You should add that in the answer $\endgroup$ – Nobody recognizeable Mar 23 at 21:54
  • $\begingroup$ You asked, you got. Perhaps you would consider marking the answer as accepted? $\endgroup$ – G. Smith Mar 23 at 21:55
  • $\begingroup$ thanks for all the help. $\endgroup$ – Nobody recognizeable Mar 23 at 21:58

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