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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?

Grade:12

1 Answers

Prajwal kr
49 Points
9 years ago

No. Any deviation from Couloumb''s law indicates the departure from Gauss''s law and vice versa.

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