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Howdoes theelectric flux due toa pointchargeenclosedby asphericalGaussiansurface get aaffected whenitsradiusis increased ?

Answer»

Solution :As per Gauss’s theorem,
Electric`PHI=q/epsilon_0`where q is the CHARGE enclosed by a closed SURFACE through which flux is `phi` Electric flux does not depend on the radius, hence it REMAINS unaffected.


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