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