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Sahay Sir > Question Answers > Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) > A positive point charge Q is placed (on the axis of disc) at a distance of 4 R . Above the centre of a disc of radius R as shown in situation I. the magnitude of electric flux through the disc is (ϕ). Now a hemis pherical shell of radius R is placed over the disc. such that it forms a closed surface as shown in situation II. The flux through the curved suface in situation II taking direction of area vector along outward normal as positive is

A positive point charge Q is placed (on the axis of disc) at a distance of 4 R . Above the centre of a disc of radius R as shown in situation I. the magnitude of electric flux through the disc is (ϕ). Now a hemis pherical shell of radius R is placed over the disc. such that it forms a closed surface as shown in situation II. The flux through the curved suface in situation II taking direction of area vector along outward normal as positive is

0 ) and ( 1m 0) is 2 m ) respectively. The resultant electric field due to the two dipoles at the point (1 m At each point on the surface of the cube shown in the figure Two short dipoles pk and p/2 k are located at ( 0