Electric Potential with Non-conducting sphere and conducting shell

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SUMMARY

The discussion centers on calculating the electric potential at radius b just outside a conducting shell surrounding a non-conducting sphere with charge +Q. Initially, the potential at b before the conducting shell is introduced is given by V = Q/(4πε0b). After the conducting shell is introduced but not grounded, the potential at b equals V = Q/(4πε0c). Once the conducting shell is grounded, it acquires a charge of -Q, resulting in a net electric potential of 0 at radius b after the grounding wire is removed.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with Gauss's Law
  • Knowledge of conductors and insulators in electrostatics
  • Proficiency in using the formula for electric potential V = Q/(4πε0r)
NEXT STEPS
  • Study the implications of grounding conductors in electrostatic scenarios
  • Learn about the behavior of electric fields in conductive materials
  • Explore Gauss's Law applications in spherical symmetry
  • Investigate the concept of electric potential energy in electrostatic systems
USEFUL FOR

This discussion is beneficial for physics students, electrical engineers, and anyone studying electrostatics, particularly in understanding the behavior of electric potential in systems involving conductors and non-conductors.

electronvolt
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Homework Statement


A spherical nonconductor of radius a carries charge +Q uniformly spread through its volume. 2 hemispherical conducting shells of inner radius b and outer radius c are placed concentrically with the nonconducting sphere to form a single conducting sphere.

The conducting shell is momentarily connected by a wire to 0 potential (grounded) and then the wire is removed.
Determine the electric potential at r=b just outside the conducting shell
a) Before the conducting shell is introduced
b) After the conductor is introduced but before it is grounded
c) After the conductor is grounded.[/B]

Homework Equations


Potential difference =
V= ##{\frac{Q}{4πε_0r}} ##. [/B]

The Attempt at a Solution


For a) The nonconducting sphere can be modeled as a point charge, so the potential at b is simply ##{\frac{Q}{4πε_0b}} ##.

For b) The electric field is the same at all points except that when b<r<c, where the electric field is 0. So the potential at b should be the same as the potential at c, so V = ##{\frac{Q}{4πε_0c}} ##.

For c) While the conducting shell is grounded it will gain negative charge -Q due to the positive charge on the nonconducting sphere. Once the wire is removed the net charge of both the nonconducting sphere and shell will be 0, and so the electric field for r>c will be 0. The electric field inside of the conducting will also be 0. Since the field is zero from infinitely far away to b, the electric potential is 0. I'm not at all confident about this answer, though.

Any assistance would be greatly appreciated.
[/B]
 
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(b) check your reasoning ... the conducting sphere is electrically neutral, so what happens to the charges when it is brought close to the charged sphere?
(c) think of it as charge conducted away.
 

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