Work done moving a test charge into conducting shell

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SUMMARY

The discussion focuses on calculating the work done in moving a test charge q from infinity to the center of a spherical shell with thickness T, radius R, and surface charge density σ. The participant correctly applies Gauss's Law, concluding that the electric field E(r) is zero inside the shell, leading to a potential difference of zero. Consequently, the total work done in this scenario is determined to be zero. The participant seeks clarification on the implications of the electric field and potential relationship, emphasizing the distinction between zero electric field and zero potential.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric fields and potentials
  • Knowledge of spherical charge distributions
  • Basic calculus for evaluating potential differences
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  • Study the implications of electric fields in conductors, particularly in spherical geometries
  • Explore the relationship between electric field and potential in electrostatics
  • Investigate the concept of induced electric fields and their effects on charge distributions
  • Review advanced topics in electrostatics, such as potential energy in electric fields
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Graduate students in physics, electrical engineers, and anyone studying electrostatics and electric field theory.

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Homework Statement



Ok, so I've read many of the threads on here and they all say the same thing. I think I understand the Gauss Law and the theory behind the spherical shell.

The question is this:

Find the WORK done bringing a test charge q from infinity to the center of a spherical shell of thickness T, radius R, and surface charge \sigma. Assume the charge passes through an infinitesimal hole in the shell.

Homework Equations



W = q[\phi(inf)-\phi(r)]

The Attempt at a Solution



W_{\infty, R+T} = q[\phi(\infty)-\phi(R+T)] = -\phi(R+T)
and
W_{R+T,0} = q[\phi(R+T)-\phi(0)] = \phi(R+T)

My second equation takes the 'fact' that E(r) = 0 inside the shell, resulting in a zero potential.
The final result is that the total WORK = 0
This is for a graduate class, and this result seems somewhat trivial. My other assumption is that the test charge induces an electric field inside the shell, but I do not think work can be done by moving the test charge through its own electric field? I could be totally wrong, and that's why I'm asking this question.

Any help clarifying my result would be greatly appreciated.
 
Physics news on Phys.org
The electric field being zero inside the shell does not mean the potential is zero. Recall that electric field is the negative gradient of potential.
 
Last edited:

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