Work done via induced charges in a grounded conductor

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SUMMARY

The discussion centers on the Method of Images as presented in Griffiths' textbook, specifically section 3.2, which addresses the energy calculations for a point charge near a grounded conducting plane. It highlights the discrepancy in energy calculations between the image charge system and the real system, where the former involves work done on both charges, while the latter only involves work on the point charge due to induced charges moving along an equipotential. The confusion arises regarding how induced charges can exist in a grounded conductor at zero potential when no force acts on them, and the implications of finite resistance in conductors are also discussed.

PREREQUISITES
  • Understanding of the Method of Images in electrostatics
  • Familiarity with concepts of electric potential and induced charges
  • Knowledge of Griffiths' "Introduction to Electrodynamics" textbook
  • Basic principles of electrostatics and conductors
NEXT STEPS
  • Study Griffiths' section 3.2 on the Method of Images in detail
  • Research the concept of electric potential and equipotential surfaces
  • Explore the implications of finite resistance in conductors
  • Investigate related problems in electrostatics involving grounded conductors
USEFUL FOR

Students of electromagnetism, physicists exploring electrostatics, and educators teaching the Method of Images and its applications in grounded conductors.

GeniVasc
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I'm currently studying Method of Images in Griffiths book and in section 3.2 he introduces the method of images for a point charge at a distance ##d## from a grounded conducting plane at potential ##V = 0##.

In subsection 3.2.3, Griffiths compute the energy of the real system and the image charge system and obtain results differing by a factor of 2, The explanation of griffiths is that in the image problem, we do work on both charges bringing them from infinity to a distance ##2d## apart from each other.

However in the real problem, griffiths says that we do work only on the point charge ##q##, since the induced charge on the conductor moves along an equipotential. My problem is: if the potential inside the grounded conductor still ##V=0## even with the external electric field due to the point charge, how charges can be induced, since no force acts on them? And what about the positive charges that go to the Earth? there's no work over them too?

I've tried to re-read conductors section in griffiths chapter 2, but as long as I remember, he makes no mention to this problem.

_I've already read some questions related to this topic but none of them answered my question_\\

_Also sorry for possible typos, I'm still learning english_
 
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This has to be true. Suppose I start with two real charges. They have a negative potential energy (zero at infinity). Call that energy U. I can insert a conductor for 'free" into the V=0 plane. No work no change in energy. If I remove either charge I get back half the energy. (Charge moves onto the plate but no work is done). Only if I remove both do I get U. QED
 
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GeniVasc said:
My problem is: if the potential inside the grounded conductor still V=0 even with the external electric field due to the point charge, how charges can be induced, since no force acts on them?
If there's no field then there's no work done but there's never 'no field'. This sort of confusion can be resolved if you think in terms of the conductor having some finite resistance. When you go through the process of moving your test charge to the test position, work will be done on the internal charges as they move by an infinitesimal distance through an infinitesimally low field.
There's an equivalent problem when dealing with balls bouncing on a solid concrete slab; you treat the mass of the slab as infinite. The difference between the ideal and practical situations are so small as to allow the approximation. A sensitive enough transducer could measure the amount by which the practical situation deviates from ideal theory.

Near enough is good enough for any theory that's properly stated.
 

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