Solving Gauss' Law Problem: Zero E Field?

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    Gauss Gauss' law Law
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SUMMARY

The discussion centers on the application of Gauss' Law to a system of two concentric hollow spheres: an inner sphere with charge +q and an outer sphere with charge -q. According to Gauss' Law, the total enclosed charge is zero, leading to an electric field of zero outside the spheres. Participants debate the implications of this conclusion, particularly regarding the influence of the closer positive charge on a test charge placed outside the spheres, ultimately affirming that the electric field remains zero due to the symmetry of the charge distribution.

PREREQUISITES
  • Understanding of Gauss' Law and its mathematical formulation
  • Familiarity with electric fields and charge distributions
  • Knowledge of Coulomb's Law and its implications
  • Basic grasp of spherical symmetry in physics
NEXT STEPS
  • Study the derivation and applications of Gauss' Law in electrostatics
  • Explore the Shell Theorem and its implications for spherical charge distributions
  • Investigate the use of simulations, such as those from PhET, to visualize electric fields
  • Examine advanced topics in electrostatics, including potential theory and Green's functions
USEFUL FOR

Students of physics, educators teaching electrostatics, and anyone interested in understanding the principles of electric fields and charge interactions through Gauss' Law.

  • #31
BvU said:
And what is capital G according to you ?

I meant gravitational field not the gravitational constant in the formula, bad choice of letters, il edit that.
 
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  • #32
David112234 said:
So referring to gravity, for the inner sphere to increase the gravity by 2X that would mean that the smaller sphere would have to have the same magnitude of gravity at a certain distance as the outer sphere, correct?
The gravitational force ##\ {GMm\over r^2}\ ## is the same, yes. Same integral as before -- or same Gauss theorem.

David112234 said:
Gravity is dependent on the radius?
Nonsense. The ##r## in ##\ {GMm\over r^2}\ ## is the distance, not the radius. Same masses, same distance, same gravitational field, same gravitational force.

I think you've got it by now.
 
  • #33
BvU said:
The gravitational force ##\ {GMm\over r^2}\ ## is the same, yes. Same integral as before -- or same Gauss theorem.

Nonsense. The ##r## in ##\ {GMm\over r^2}\ ## is the distance, not the radius. Same masses, same distance, same gravitational field, same gravitational force.

I think you've got it by now.

For the most part.
The distance from the test charge to the center of the sphere, so since both spheres have the same center the distance is the same, is that correct?
 
  • #34
I still don't understand, where's the problem to follow one of the calculations, I offered above. All these debates must confuse you more than they help. Math helps, you just have to learn it anyway! Vector calculus is usually not easy to grasp, and thus you have to get used to it by doing a lot of problems and in this way get an intuition on this math. All these qualitative arguments won't help you to understand physics!
 
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  • #35
It was not about a specific problem but a Gauss law situation where I could not understand how it was true. Now i get it and learned to fully trust Gauss law. Thank you for all the help.
 
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