Inverse square & cube laws for various geometries

[fleem]

I would like to compile a short list of inverse-exponent force magnitude fall-off laws for several simple geometries for material made of monopoles and for material made of parallel dipoles (of negligible length).

Far field force for any two objects is, of course, proportional to $1/r^2$ for monopole-based objects, and $1/r^4$ for dipole-based objects (yes, field for dipoles drops with $1/r^3$, but I'm interested in force between two objects that are both made of dipoles. The definition for field strength is related to the force on a monopole).

So what is the force magnitude fall-off (both monopole and dipole material) for the following near-field geometries?:

a point near a line
a line near a parallel line
a line near a perpendicular line

(Side note: the field above a plane made of dipoles is uniform, so there will be zero force (except torque) on dipoles above it)

Thanks.

 

[Bill_K]

fleem, You can figure out things like this using just dimensionality arguments. For two point charges the force is QQ/r². Now a dipole moment is P ~ Qd, so the force must be?
Ans: The P's give you two more lengths in the numerator, so the force must be PP/r⁴
A line charge works the other way: λ = charge per unit length, and the force between two line charges is λλ/r^??
Ans: You now have two lengths less in the numerator, so the force must be λλ/r⁰, i.e. independent of r.

 

[fleem]

fleem, You can figure out things like this using just dimensionality arguments. For two point charges the force is QQ/r². Now a dipole moment is P ~ Qd, so the force must be?
Ans: The P's give you two more lengths in the numerator, so the force must be PP/r⁴
A line charge works the other way: λ = charge per unit length, and the force between two line charges is λλ/r^??
Ans: You now have two lengths less in the numerator, so the force must be λλ/r⁰, i.e. independent of r.

Good grief I didn't even think of doing that. Thank you! This answers all my questions.