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Grade 11Electrostatics

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?

Profile image of Madiha
9 Years agoGrade 11
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1 Answer

Profile image of Rituraj Tiwari
5 Years 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.