Finding Electric Field Outside a Charged Conductor

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SUMMARY

The discussion focuses on calculating the electric field outside a spherically shaped uncharged conductor with an irregularly shaped hole containing a charge. Utilizing Gauss's Law, the correct approach involves recognizing that the electric field (E) is not uniform but varies as 1/r². The electric field outside the conductor is determined by the surface charge density (σ), which is calculated as σ=q/(4πr²), leading to the expression E=σ/ε₀. It is crucial to remember that the electric field inside the conductor is zero, and the charges on the outer surface distribute evenly.

PREREQUISITES
  • Understanding of Gauss's Law and its application in electrostatics
  • Knowledge of electric field concepts and vector properties
  • Familiarity with surface charge density calculations
  • Basic principles of electrostatics in conductors
NEXT STEPS
  • Study the implications of Gauss's Law in different geometries
  • Learn about electric field calculations in non-uniform charge distributions
  • Explore the concept of electric field lines and their behavior in conductors
  • Investigate the role of ε₀ (permittivity of free space) in electrostatic equations
USEFUL FOR

Students and professionals in physics, particularly those studying electrostatics, electrical engineers, and anyone involved in understanding electric fields around conductors.

Suziii
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Homework Statement




In case of having a conductor (spherically shaped and uncharged) and an irregularly shaped hole inside it with a charge at an arbitrary point, I am asked to find the electric field right outside the conductor sphere.


Homework Equations



Gauss's law ∫EdA= Q(enclosed) / (a constant)

The Attempt at a Solution


I assumed that the electric field outside of the conductor is uniform so I had:
E*(4πr^2)= q/ the constant - we can get E from this, but it turns out to be a wrong answer, maybe just taking E to be constant is not correct...
 
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Your answer is almost correct. However, you must remember that E is a vector.
Also, you should note that the electric field is not constant but falls off as [itex]\frac{1}{r^2}[/itex].
You can see this from your answer.
 
The magnitude of E is the same all over the outer surface of the sphere, and E*(4πr^2)= q/ε0 according to Gauss' Law. Haven't you forgotten ε0?
As the electric field is zero inside the metal, the electric field lines starting from the inner charge do not go through the metal wall. Therefore the charges accumulated on the outer surface "do not know" anything about the field inside, so they arrange themselves evenly, as far from each other as possible. The surface charge density is homogeneous and equal to σ=q/(4πr2). The electric field can be expressed by the surface charge density as E=σ/ε0.

ehild
 
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