Inverse square & cube laws for various geometries

In summary, the force magnitude for near field geometries for monopole and dipole material is PP/r4 for point charges, and λλ/r0 for line charges.
  • #1
fleem
440
0
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 [itex]1/r^2[/itex] for monopole-based objects, and [itex]1/r^4[/itex] for dipole-based objects (yes, field for dipoles drops with [itex]1/r^3[/itex], 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.
 
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  • #2
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.
 
  • #3
Bill_K said:
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.
 

Related to Inverse square & cube laws for various geometries

1. What is the inverse square law?

The inverse square law is a physical law that states that the intensity of a physical quantity, such as light or gravity, is inversely proportional to the square of the distance from the source of that quantity.

2. How does the inverse square law apply to light?

In the context of light, the inverse square law states that the brightness of a light source decreases as the distance from the source increases. This is because the same amount of light is spread out over a larger area as it travels away from the source.

3. What is the inverse cube law?

The inverse cube law is a physical law that states that the intensity of a physical quantity is inversely proportional to the cube of the distance from the source. This law applies to phenomena such as sound, radiation, and gravity.

4. How does the inverse cube law apply to sound?

The inverse cube law applies to sound by stating that the perceived loudness of a sound decreases as the distance from the source increases. This is because the sound waves spread out in all directions, decreasing in intensity as they travel away from the source.

5. How do the inverse square and cube laws apply to different geometries?

The inverse square and cube laws apply to different geometries by describing how the intensity of a physical quantity changes as the distance from the source increases. These laws are based on the concept that the intensity is spread out over a larger area or volume as it travels away from the source, resulting in a decrease in intensity. This applies to various shapes and configurations, such as spherical, cylindrical, and conical geometries.

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