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A charge Q is placed at a distance of 4R above the centre of disc of radius R. The magnitude of flux through the disc is fi. 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 is?

A charge Q is placed at a distance of 4R above the centre of disc of radius R. The magnitude of flux through the disc is fi. 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 is?

Grade:11

1 Answers

Rituraj Tiwari
askIITians Faculty 1792 Points
9 months ago
Here, A charge q is placed at a distance of 4r above the centre of a disc of radius r.

=> If the magnitude flux through the disc is Ф. A hemisphere shell of radius R is placed over the disc such that it forms a closed surface. then Find the flux through the curved surface.

So, Фshell + Фdisc = 0

and According to question, No charge is found inside the closed surface.

Thus, Фshell = -Фdisc = -Ф

Therefore, the flux through the curved surface is -Ф.

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