Surface charge on a conductor due to a charged rod 'q'

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SUMMARY

The total surface charge on a perfect conductor due to a charged rod with charge 'q' at a distance 'd' is exactly -q. This occurs because the electric field generated by the charged rod induces a negative charge on the conductor's surface until the electric field at a point charge at infinity approaches zero. The distance 'd' does not affect the total induced charge, as the induced charge will always equal -q regardless of the distance, provided the conductor is perfect. For non-perfect conductors, while the process may take longer due to finite conductivity, the final induced charge remains -q.

PREREQUISITES
  • Understanding of electrostatics and electric fields
  • Familiarity with the concept of induced charge on conductors
  • Knowledge of perfect vs. non-perfect conductors
  • Basic grasp of charge density and its implications
NEXT STEPS
  • Study the principles of electrostatics in detail
  • Learn about the behavior of electric fields around conductors
  • Explore the differences between perfect and non-perfect conductors
  • Investigate the concept of charge density and its effects on induced charges
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Students of physics, particularly those studying electrostatics, educators teaching electric field concepts, and anyone interested in the behavior of charges on conductors.

iVenky
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Homework Statement


A charged rod of charge 'q' is at a distance 'd' from a perfect conductor as shown below.
What's the total surface charge on the conductor?
upload_2018-12-24_23-54-41.png

2. Homework Equations

I tried to solve this without equations.

The Attempt at a Solution


[/B]
Basically, as long as there is E field due to +q, it can draw a negative charge from the conductor to the surface. This would stop once the total E field on a point charge at infinity goes to 0. This should happen when the total charge on the surface is exactly -q. Is this explanation correct?

However, what's bothering me is the non-dependence on distance 'd'. If d-> ∞, then I wouldn't expect 'q' draw any charge on the conductor since E field would be zero at d-> ∞. Then what's wrong here?
Also, does this explanation work even if this conductor is not perfect (has a σ≠∞)
 

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iVenky said:
If d-> ∞, then I wouldn't expect 'q' draw any charge on the conductor since E field would be zero at d-> ∞. Then what's wrong here?
The charge density becomes lower as the induced charge will be spread over a larger portion of the conductor.

iVenky said:
Also, does this explanation work even if this conductor is not perfect (has a σ≠∞)
The problem only deals with the static situation, i.e., the situation after a long time has passed. With finite conductance that will take longer to achieve for a non-perfect conductor, but the end result would be the same.
 
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