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(D) According to Gauss's Law,the total electric flux through a closed surface is $\Phi = \frac{q_{enclosed}}{\varepsilon_{0}}$.For a charge $q$ placed

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$0$

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(D) According to Gauss's Law,the total electric flux through a closed surface is $\Phi = \frac{q_{enclosed}}{\varepsilon_{0}}$.For a charge $q$ placed exactly at the centre of the flat circular base of a hemisphere,the electric field lines originating from the charge $q$ are parallel to the flat surface at every point on it.Since the electric field vector $\vec{E}$ is perpendicular to the area vector $\vec{A}$ of the flat surface (i.e.,$\vec{E} \cdot \vec{A} = EA \cos(90^{\circ}) = 0$),the electric flux passing through the flat surface is $0$.However,the flux passing through the curved hemispherical part of the surface is $\frac{q}{2 \varepsilon_{0}}$.

If a charge $q$ is placed at the centre of the flat surface of a closed hemispherical non-conducting surface,what is the total electric flux passing through the flat surface?

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