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A charge Q is placed at a distance $D( > R)$ above the centre of disk of radius R The magnitude of flux through disk is $\phi$ . Now a hemispherical shell of radius R is put above the disk to form a closed surface. The flux through the closed surface, if direction of area vector along outward normal is +ve, is :

$(A)\;-\phi \\ (B)\;\frac{-\phi D}{R} \\ (C)\; \phi \\ (D)\;\frac{\phi R}{D} $

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$\phi _{shell} +\phi _{disc}=0$
($\therefore$ there is no change inside the closed surface )
=> $ \phi _{shell} =-\phi _{disc}=- \phi$
Hence A is the correct answer.
answered Jan 10, 2014 by meena.p

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