$(a)\;8 \mu C\qquad(b)\;0.8 \mu C\qquad(c)\;0.08 \mu C\qquad(d)\;80 \mu C$

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Answer : (c) $\;0.08 \;\mu C$

Explanation : The electric field at the surface of the sphere is Aa and being radia it is along the outward normal . Therefore

$\phi=\oint\;E\;ds\;cos\;\theta=Aa\;(4\;\pi\;a^2)$

$Q=\in_{0}\phi=4\;\pi\;\in_{0}\;Aa^3$

$=(\large\frac{1}{9\times10^{9}})\;90\times8$

$Q=8\times10^{-8}\;C$

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