# Question about multipole expansion of electrostatic potential?

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?

• 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? – G. Smith Mar 23 at 21:35
• @G.Smith i am sorry I was voice typing that came out wrong so I edited it. – Nobody recognizeable Mar 23 at 21:37
• @G.Smith is it ok now? – Nobody recognizeable Mar 23 at 21:41
• 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. – Feynmans Out for Grumpy Cat Mar 23 at 22:01
• @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. – Nobody recognizeable Mar 23 at 22:06

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.