Lightning Rod Tip Radii: Why So Small?

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The small radii of lightning rod tips enhance the electric field strength due to the relationship between curvature and charge distribution. A smaller radius of curvature results in a higher charge density at the tip, which increases the electric field intensity. When a charge is placed near a lightning rod, the electric field is stronger at points with smaller radii, leading to more effective charge attraction. This principle explains why lightning rods are designed with pointed tips. Understanding these dynamics is crucial for optimizing lightning rod efficiency.
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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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