David112234
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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.
The discussion revolves around the application of Gauss' Law to a scenario involving two concentric hollow spheres, one positively charged and the other negatively charged. Participants explore the implications of this configuration on the electric field outside the spheres, questioning whether a test charge placed outside would experience any electric force.
Participants express differing views on the application of Gauss' Law in this context, with no consensus reached regarding the behavior of the electric field or the forces acting on the test charge. The discussion remains unresolved, with multiple competing interpretations of the problem.
Participants note limitations in their understanding related to the assumptions of symmetry and the specific configurations of charges. The discussion highlights the complexity of applying Gauss' Law in non-ideal situations.
BvU said:And what is capital G according to you ?
The gravitational force ##\ {GMm\over r^2}\ ## is the same, yes. Same integral as before -- or same Gauss theorem.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?
Nonsense. The ##r## in ##\ {GMm\over r^2}\ ## is the distance, not the radius. Same masses, same distance, same gravitational field, same gravitational force.David112234 said:Gravity is dependent on the radius?
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.