Gauss's law

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VARUN RAJ (1635)

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i am assuming that the area of both the sides are equal assuming

charge on outer side of the x to be q1 and on the inner side of q-q1

charge on the inner side of y is -(q-q1) because if we draw a gaussian surface passing thru the conductore the electric field inside should be -(coz of conduction properties)

charge on outer side=q-q1

the electric field at any pnt x or y =0 as it is inside the conductor

so q-q1=q solving we get q1=q2

so now we can get the all the answers easily

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