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Q.

A charge Q is placed at a distance 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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a

zero

b

2ϕ

c

−ϕ

d

ϕ

answer is C.

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Detailed Solution

For the closed surface made by disc and hemisphere
qin = 0

∴ϕnet = 0

ϕdisc +ϕH.S = 0

∴ϕH.S =−ϕdisc =−ϕ 

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