Charge and Electric Field Intensity

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SUMMARY

The discussion centers on the differences in electric field intensity between a sphere and a thin rod, both possessing identical charges. It concludes that the curvature of the surface significantly influences electric field intensity, with stronger fields present at points of greater curvature. The repulsive forces among charges on a curved surface act at angles rather than parallel to the surface, which alters the resultant electric field. This understanding clarifies why the sphere exhibits a greater electric field intensity compared to the thin rod.

PREREQUISITES
  • Understanding of electric fields and charge distribution
  • Knowledge of electrostatics and conductors at equilibrium
  • Familiarity with the concepts of force and work in physics
  • Basic principles of surface curvature and its effects on electric fields
NEXT STEPS
  • Research the mathematical formulation of electric field intensity for different geometries
  • Explore the principles of electrostatic equilibrium in conductors
  • Study the effects of surface curvature on charge distribution
  • Learn about the relationship between electric field intensity and distance from charged bodies
USEFUL FOR

Students and professionals in physics, electrical engineering, and anyone interested in understanding the principles of electric fields and charge interactions.

Soaring Crane
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If there are two bodies, such as a sphere and a thin rod (each with exactly the same charge), why does the sphere have a greater electric field intensity? Are there any other variables that influence the electric field intensity if the charge is not an ultimate dependent factor?

If the charges are the same, would the intensity difference have anything to do with the force, work, or distance involved?
 
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Soaring Crane said:
If there are two bodies, such as a sphere and a thin rod (each with exactly the same charge), why does the sphere have a greater electric field intensity? Are there any other variables that influence the electric field intensity if the charge is not an ultimate dependent factor?

If the charges are the same, would the intensity difference have anything to do with the force, work, or distance involved?

Well the answer to your question has to do with the surface. Particularly the curvature of the surface. Conductors at electrostatic equilibrium have stronger electric field where the surface is most curved.

Consider a flat surface, there are n charges (electrons). Being that they have a negative charge, they will repel each other. The repulsive force is directly parallel to the surface.

Now if we have n charges along a curved surface, the repulsive force is no longer parallel to the surface, but at an angle.

Do you get the idea what's happening here?
 
Yes, I think I have a clearer idea now.
 

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