ElectrostaticsMedium
Question
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 -
Options
A.zero
B.φ
C.− φ
D.2φ
Solution
For the closed surface made by disc and hemisphere
qin = 0 ∴ fnet = 0
fdisc + fH.S = 0
∴ fHS = − fdisc = − φ
qin = 0 ∴ fnet = 0
fdisc + fH.S = 0
∴ fHS = − fdisc = − φ
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