If coulomb's law innvolves 1/(r cube) dependence, instead of
1/(r square) would Gauss's Theorem be still the same?
If coulomb's law innvolves 1/(r cube) dependence, instead of
1/(r square) would Gauss's Theorem be still the same?

Last Activity: 3 Years ago

Last Activity: 3 Years ago


Last Activity: 3 Years ago

Last Activity: 3 Years ago