Lightning Rod Tip Radii: Why So Small?

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    Lightning Rod
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Discussion Overview

The discussion revolves around the small radii of the tips of lightning rods, exploring the implications of tip curvature on electric fields and charge distribution. Participants engage in theoretical reasoning and mathematical considerations related to electrostatics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that the electric field is stronger at points with smaller radii of curvature.
  • One participant elaborates that at a point of increasing radius of curvature, charges on opposite sides are at equal distances along the surface, but the charge on the side with smaller curvature is effectively closer, affecting equilibrium.
  • Another participant proposes that the amount of charge in an arc between fixed distances is less for larger radii, leading to a higher charge density at smaller radii.
  • A participant introduces a case of a conducting sphere and questions how the electric field at the surface depends on the radius, prompting further inquiry into the relationship between radius and electric field strength.

Areas of Agreement / Disagreement

Participants express various viewpoints regarding the relationship between tip curvature and electric field strength, with no consensus reached on the exact dependency of electric field on curvature or the implications of these relationships.

Contextual Notes

Some discussions involve assumptions about charge distribution and equilibrium that may not be fully articulated. The mathematical dependencies on curvature remain unresolved.

amaresh92
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greetings,
why the radii of the tip of lightning rod is small(few mm)?
thanks.
 
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hiamaresh92! :smile:

put a lightning rod near a flat metal plate …

what will happen if you place a charge near them? :wink:
 
The electric field is greater where the radius of curvature of the end is smaller.
 
Meir Achuz said:
The electric field is greater where the radius of curvature of the end is smaller.

because at a point of increasing radius of curvature, if two charges on opposite sides are at equal distances measured along the surface, the one on the side of smaller radius of curvature will be actually nearer …

so in equilibrium, …

ooh, that's the wrong result, isn't it? :cry:

ah … let's start again o:)

at radius of curvature r, the amount of charge in an arc between fixed distances s and s + ds (measured in 3D space, not along the surface) will be less for greater r (by a factor cos(s/r), the radius of the arc), and its component along the surface will also be less (by a factor cos(s/2r)) …

so for smaller r, s/r is larger and so cos(s/r) and cos(s/2r) are smaller, and so the amount of charge and its "component" (loosely speaking) at a fixed distance is smaller, so the repulsive force will be smaller on the side of decreasing r

so in equilibrium, there must be a higher charge density at decreasing r! :smile:

(does anyone know the exact dependency on curvature?)
 
A simple case is a conducting sphere of radius R with a charge Q.
How does the field at the surface depend on R?
 
Meir Achuz said:
A simple case is a conducting sphere of radius R with a charge Q.
How does the field at the surface depend on R?

?? :confused: R is constant.
 

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