A charge Q is placed at a distace of 4R above the centre of a disc of radius R. The magnitude of flux through the disc is ϕ. Now a hemispherical shell of radius R is placed over the disc such that it forms a closed surface. The flux through the curved surface (taking direction of area vector along outward normal as positive), is -

A charge Q is placed at a distace of 4R above the centre of a disc of

1.	A charge Q is placed at a distace of 4R above the centre of a disc of radius R. The magnitude of flux through the disc is ϕ. Now a hemispherical shell of radius R is placed over the disc such that it forms a closed surface. The flux through the curved surface (taking direction of area vector along outward normal as positive), is -
Answer» A charge Q is placed at a distace of 4R above the centre of a disc of radius R. The magnitude of flux through the disc is $ϕ$ . Now a hemispherical shell of radius R is placed over the disc such that it forms a closed surface. The flux through the curved surface (taking direction of area vector along outward normal as positive), is -

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