I Solving Gauss' Law Problem: Zero E Field?

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In the discussion, participants analyze the electric field produced by two concentric hollow spheres, one with charge +q and the other with charge -q. They apply Gauss's Law, concluding that the total enclosed charge is zero, leading to an electric field of zero outside the spheres. Despite this, some participants struggle with the intuition that a test charge placed outside should feel a force from the outer sphere. The conversation highlights the importance of symmetry in applying Gauss's Law and the need for careful consideration of charge distribution and distances when calculating electric fields. Ultimately, the discussion emphasizes that while Gauss's Law may seem counterintuitive, it holds true under the right conditions.
  • #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 Guass 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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