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(D) According to Gauss's Law,the total electric flux through a closed surface is $\phi_{	ext{total}} = \frac{q}{\epsilon_0}$.The closed surface consi

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$3$

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(D) According to Gauss's Law,the total electric flux through a closed surface is $\phi_{	ext{total}} = \frac{q}{\epsilon_0}$.The closed surface consists of two parts: the hemisphere and the conical surface.Therefore,$\phi_{	ext{hemisphere}} + \phi_{	ext{cone}} = \frac{q}{\epsilon_0}$.Since the charge $q$ is placed exactly at the center of the circular base,which is the common boundary between the hemisphere and the cone,the electric field lines are distributed symmetrically.The hemisphere covers a solid angle of $2\pi$ steradians,and the cone also covers a solid angle of $2\pi$ steradians (since the total solid angle around a point is $4\pi$ steradians).Thus,the flux is divided equally between the two surfaces:$\phi_{	ext{hemisphere}} = \frac{1}{2} \left( \frac{q}{\epsilon_0} ight) = \frac{q}{2\epsilon_0}$$\phi_{	ext{cone}} = \frac{1}{2} \left( \frac{q}{\epsilon_0} ight) = \frac{q}{2\epsilon_0}$We are given that the flux through the conical surface is $\frac{nq}{6\epsilon_0}$.Equating the two expressions for the flux through the cone:$\frac{nq}{6\epsilon_0} = \frac{q}{2\epsilon_0}$$\frac{n}{6} = \frac{1}{2}$$n = 3$.

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