Question

If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change ?
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Answer 1

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If the 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

Answer 2

Answer:

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Explanation:

When dipole moment vector is parallel to electric field vector.