Electrostatics - Negative charge in sphere

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SUMMARY

The discussion focuses on the electrostatics of a grounded metal sphere with a negative charge located inside it. The participant utilized the method of image charges to determine the electric field and force acting on the charge. They applied the uniqueness theorems to establish that the potential inside the sphere is uniquely determined, leading to the conclusion that the total charge on the sphere remains zero. The participant also explored the implications of changing the surface potential from zero to V, revealing that the total surface charge on the inner surface must equal the negative of the internal charge.

PREREQUISITES
  • Understanding of electrostatics principles, including Gauss's law
  • Familiarity with the method of image charges
  • Knowledge of electric field and potential concepts
  • Ability to apply uniqueness theorems in electrostatics
NEXT STEPS
  • Study the application of Gauss's law in various geometries
  • Learn about the method of image charges in more complex configurations
  • Investigate the relationship between electric field and surface charge density
  • Explore the implications of varying surface potentials on charge distribution
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Students and professionals in physics, particularly those specializing in electrostatics, electrical engineering, and anyone involved in solving problems related to electric fields and potentials in conductive materials.

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Homework Statement



a metal grounded sphere of radius R in vacuum has a negative charge inside at a distance a from the centre. a) Draw the e-field in the sphere, b) find the force on the charge, c) and determine the maximal and minimal surface charge density, d) as well as the magnitude of the charge on the in,- and outside of the sphere, when we change the surface potential from zero to V.

The Attempt at a Solution



I did a-b-c, and I don't know d yet.

a-b:

I was thinking Image charges here.

if I put a charge [tex]q' = q \frac{b}{R}[/tex] at distance [tex]b=\frac{R^{2}}{a}[/tex] from the centre in a straight line from the centre and charge q, the potential is zero on the surface, just like in the problem. the first uniqueness theorem now tells me that the potential inside the sphere is uniquely determined.

The second uniquess theorem tells me the E-field is uniquely determined, because the total charge on the sphere is given (=0).

The force is now just a matter of putting symbols in formulas.

c: the surface charge is minimal/maximal on the places of the sphere closest and furthest from the image charge. So [tex]\nabla E = \frac{\rho}{e_0}[/tex] and you compute [tex]\rho[/tex]

But what is the total charge when the potential is changed?
 
Last edited:
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"because the total charge on the sphere is given (=0)."

When the sphere is grounded, the total surface charge on the inner surface of the spherical shell must be the negative of the charge q. This is because V=0, means E=0 just outside. By Gauss's law the net charge must be zero.
If the sphere's surface is at V, E just outside equals V/R.
Gauss's law will then give the charge on the outer surface of the sphere.
 

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