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If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change ? places all are follow me |
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Answer» radius of the Gaussion SURFACE enclosing a CHARGE q is halved, how does the electric flux through the Gaussion surface CHANGE? SOLUTION : As ϕ=q∈0, so flux does not depend upon the radius of Gaussion surface, it will remain unchanged |
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