Confused about the electric field at the surface of a conductor

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SUMMARY

The electric field at the surface of a conductor has a normal component defined by the formula E = ρ/ε, where ρ represents charge density and ε is the permittivity of the medium. While the electric field due to a point charge is given by E = Kq/r, it does not become infinite at the surface of the conductor. Instead, the charges within the conductor respond to external fields, ensuring that the total electric field remains finite at the surface. The key takeaway is that the gradient of the electric field at the surface is what maintains the finite value.

PREREQUISITES
  • Understanding of electrostatics and electric fields
  • Familiarity with the concepts of charge density (ρ) and permittivity (ε)
  • Knowledge of Coulomb's law and the formula for electric fields (E = Kq/r)
  • Basic principles of conductors in electrostatic equilibrium
NEXT STEPS
  • Study the behavior of electric fields in conductors and insulators
  • Explore the concept of electric field gradients and their implications
  • Investigate the role of charge distribution in capacitors
  • Learn about the mathematical derivation of electric fields from charge distributions
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Students and professionals in physics, electrical engineering, and anyone interested in understanding electrostatics and the behavior of electric fields around conductors.

parsa7parsa
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Hi
We know that the electric field at the surface of a conductor only have a normal component equal to ρ /ε (finite number).
But let’s consider the point P (at the surface of a conductor ) . Assume that there is a charge at an infinitesimal distance from the point p . we can obtain the field at the P by the fourmula (E=Kq/r) .obviously, E ~1/r. so the normal component of the field is infinite. Now if we add the field due to other charges, it will remain infinite. So where could I be possibly wrong?
 
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We know that the electric field at the surface of a conductor only have a normal component equal to ρ /ε (finite number).
It's the gradient of the electric field that has that value.

The charges in the conductor will respond to the electric field of the small charge close to the conductor - affecting the way the total field comes out. How do they respond?

Note - at a very small distance from a point charge, the field is not infinite.
If the charge is actually at point P, then it is part of the conductor. Inside a conductor, the charges are infinitesimally small (in this model).
 
parsa7parsa said:
we can obtain the field at the P by the fourmula (E=Kq/r)

You should ask yourself what exactly is "q" in that equation going to be for your capacitor with a given charge density (*cough*) ρ
 
please note that the charge is <within> the conductor
 

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