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Electric Flux & Gauss's Law

A charged particle $q$ is placed at the centre $O$ of cube of length $L(ABCDEFGH).$ Another same charge $q$ is placed at a distance $L$ from $O$ . Then the electric flux through $ABCD$ is

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Answer: (b)

Intended reading: plane $ABCD$ contains the line joining $O$ and the outside charge $q$ .
Field of each point charge is radial from it, so at every point of $ABCD$ the total field lies in that plane.
$\vec{E} \perp d\vec{S}$ everywhere, so $\phi_{ABCD} = \oint \vec{E}\cdot d\vec{S} = 0$ .
So the answer is (b) zero.

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Electric Flux & Gauss's Law question from JEE Main 2002 (AIEEE)

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