Shielding an off-center charge with a conducting shell

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SUMMARY

A conducting, charged spherical shell effectively shields the electric field of an off-center point charge located inside it. By applying Gauss's law, it is established that the electric field outside the shell is zero due to the symmetry of the charge distribution. The outer surface of the shell remains an equipotential surface, and the induced charge distribution on the inner surface cancels any asymmetry caused by the internal charge. This conclusion is supported by the uniqueness of the electric field geometry in this symmetric configuration.

PREREQUISITES
  • Understanding of Gauss's law
  • Familiarity with electric fields and equipotential surfaces
  • Knowledge of charge distribution concepts
  • Basic principles of electrostatics
NEXT STEPS
  • Study the implications of Gauss's law in electrostatics
  • Explore the concept of electric field symmetry in spherical conductors
  • Investigate charge distribution on conductors in electrostatic equilibrium
  • Learn about dipole moments and their effects on electric fields
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Students of physics, electrical engineers, and anyone interested in understanding electrostatic shielding and charge distribution in conductors.

greypilgrim
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Hi.

I'd like to show that a conducting, charged spherical shell can shield the field of an inside opposite point charge even if this charge is not at the center. I was thinking about a Gaussian surface just outside the sphere, such that if there were electric field vectors they would be perpendicular to the surface.

I'd now like to show that all those field vectors either point inwards or outwards. Then I could use Gauss's law and show that the field vectors are in fact zero (since the total charge inside is zero). But I can't see why this is true. Couldn't the inside charge maybe induce a dipolar charge distribution on the outside of the sphere if it were really close to sphere?
 
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greypilgrim said:
Couldn't the inside charge maybe induce a dipolar charge distribution on the outside of the sphere if it were really close to sphere?
It does! This is exactly what cancels the asymmetry from the charge inside.

The outer surface of the shell is an equipotential surface. You can use symmetry to find an electric field that fits.
 
Hi.

I assume you're talking about placing a mirror charge outside to make the potential constant on the sphere? I know this method, but I was looking for a simpler way and hoped Gauss might help. I only want to show that the outside field vanishes, not how the inside field looks like.

mfb said:
You can use symmetry to find an electric field that fits.
But the electric field produced by the outside mirror charge is only meaningful on the inside of the sphere, not on the outside, isn't it? At least this is what Wikipedia says, for the potential though.

mfb said:
It does! This is exactly what cancels the asymmetry from the charge inside.
I might be asking for too much, but would you mind drawing a rough sketch about how the charges distribute on the outer and inner surface of the sphere?
 
greypilgrim said:
I assume you're talking about placing a mirror charge outside to make the potential constant on the sphere?
No.

You don't need any mirror charges here.

Outside the spherical shell, the situation is symmetric and you know the total charge. There is exactly one possible electric field geometry that fits to this situation. If you don't care about the inside you don't have to find the charge distribution there.
 
Ah okay, you mean the fact that the potential is constant on the sphere makes the situation symmetric and defines the field by uniqueness.
mfb said:
It does! This is exactly what cancels the asymmetry from the charge inside.
You agreed that there will be a non-constant (even dipolar) charge distribution on the outside of the sphere. How can I find this distribution?
 
greypilgrim said:
You agreed that there will be a non-constant (even dipolar) charge distribution on the outside of the sphere.
Sorry, misread the quoted part. The non-constant charge distribution is on the inside of the sphere.
 

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