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Can Gauss's law in electrostatics tell u...

Can Gauss's law in electrostatics tell us exactly where the charge is located within the Gaussian surface ?

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No, charge cannot lie on a Gaussian surface.
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Gauss's law of electrostatics would be invalid if

Which of the following statements is//are correct? a. Electric field calculated by Gauss law is the field due to only those charges which are enclosed inside the Gaussian surface. b. Gauss law is applicable only when there is a symmetrical distribution of charge. c. Electric flux through a closed surface is equal to total flux due to all the charges enclosed within that surface only.

Gauss's law and Coulomb's law , although expressed in different forms , are equivalent ways of describing the relation between charge and electric field in static conditions . Gauss's law is epsilon_(0) phi = q_(encl) ,when q(encl) is the net charge inside an imaginary closed surface called Gaussian surface. The two equations hold only when the net charge is in vaccum or air . A Gaussian surface encloses two of the four positively charged particles. The particles that contribute to the electric field at a point P on the surface are

When no charge is confined with in the Gauss's surface, it implies that -

Consider Gauss's law oint vecE * vecd s = q/epsilon_0 Then, for the situation shown in figure at the Gaussian surface

A+q_(1) charge is at centre of an imaginary spherical Gaussion surface 'S', and -q_(1) charge is placed nearby this +q_(1) charge inside 'S'. A charge +q_(2) is located outside this Gaussian surface. Then electric field on Gaussian surface will be :

A charge Q is enclosed by a spherical Gaussian surface of radius R. If the radius of the sphere is double, how will the outward electric flux charge ? If a charge -Q is brought outside the Gaussian surface, will the electric flux reduce to zero ?

MODERN PUBLICATION-ELECTRIC CHARGES AND FIELDS -Conceptual Questions
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