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

AI Thread Summary
A charged rod 'q' induces a surface charge on a perfect conductor, which becomes -q as the electric field from the rod draws negative charge to the surface until the field at infinity reaches zero. The discussion raises concerns about the role of distance 'd', noting that as 'd' approaches infinity, the electric field should also approach zero, suggesting no charge should be induced. However, it is clarified that the induced charge density will decrease as it spreads over a larger area of the conductor. The analysis remains valid for non-perfect conductors, though achieving the final charge distribution will take longer due to finite conductivity. Ultimately, the static situation after sufficient time will yield the same total induced charge of -q.
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


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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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