- #1

wrobel

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##V(\boldsymbol r)\sim-\frac{1}{|\boldsymbol r|^2}, \quad |\boldsymbol r|\to 0##

in physics ?

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- Thread starter wrobel
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- #1

wrobel

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##V(\boldsymbol r)\sim-\frac{1}{|\boldsymbol r|^2}, \quad |\boldsymbol r|\to 0##

in physics ?

- #2

jbriggs444

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Just spit-balling here... An inverse square potential would imply an inverse cube force law. What do we have for inverse cube forces?

##V(\boldsymbol r)\sim-\frac{1}{|\boldsymbol r|^2}, \quad |\boldsymbol r|\to 0##

in physics ?

How about the electrostatic force on (or from) a dipole. That should scale as the differential of an inverse square, i.e. as an inverse cube.

- #3

wrobel

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##V\sim-\frac{1}{|\boldsymbol r|^n},\quad n\ge 2##

- #4

Dale

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Excellent response. You could do arbitrary order multipoles to get any n>0 desiredHow about the electrostatic force on (or from) a dipole. That should scale as the differential of an inverse square, i.e. as an inverse cube.

- #5

jbriggs444

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I am not sure how the suggestion of a dipole fails to satisfy that requirement.

##V\sim-\frac{1}{|\boldsymbol r|^n},\quad n\ge 2##

It is a net attractive force and consequently has a negative potential everywhere referenced to zero at infinity. So the sign is right. It approximates an inverse square potential. So the approximation is right (when r >> size of dipole). And n=2 which satisfies n>=2.

- #6

DrClaude

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You get a ##-1/R^2## potential for ion-dipole interaction, ##-1/R^3## for dipole-dipole, ##-1/R^4## for ion-induced dipole, and so on.

- #7

wrobel

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It seems I still have some stupid questions. I have been thinking that dipole is the potential of the type ##V=\frac{\cos\varphi}{r^2}## (in polar coordinates) but this potential changes sign.

Could you please be more detailed?

- #8

jbriggs444

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In simple terms, a "dipole" would be a pair of equal and opposite charges with some fixed separation. For example, a positive charge and an equal negative charge on opposite ends of an insulating stick.It seems I still have some stupid questions. I have been thinking that dipole is the potential of the type ##V=\frac{\cos\varphi}{r^2}## (in polar coordinates) but this potential changes sign.

Could you please be more detailed?

The net charge of this dipole is zero. And we can consider its location to be the midpoint between the two charges.

Now add a fixed positive point charge at the origin of your coordinate system and have this dipole floating in space somewhere. One could use a negative point charge instead. It changes nothing. What is the force of the point charge on the dipole?

Well,

Edit: Apologies for the length and the simple mindedness of the response. I was talking my way through it until I finally got to the point of understanding how you'd arrived at your formulation.

Last edited:

- #9

hutchphd

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Are you looking asymptotically for a spherically symmetric potential from a localized source ?

It seems I still have some stupid questions. I have been thinking that dipole is the potential of the type ##V=\frac{\cos\varphi}{r^2}## (in polar coordinates) but this potential changes sign.

Could you please be more detailed?

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