Charge q is enclosed by a gaussian sperical surface of radius r if the radius is doubled then the outward electric flux will
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Answer:
q/ε where 'ε' is the permittivity of electric field in free space
Explanation:
flux of a charge q through a closed gaussian surface is always constant irrespective of its size,shape i.e q/ε
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Explanation:
When dipole moment vector is parallel to electric field vector.
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