JEE MAIN - Physics (2002 - No. 49)
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
$$q/4\,\pi \,{ \in _0}L$$
zero
$$q/2\,\pi \,{ \in _0}L$$
$$q/3\,\pi \,{ \in _0}L$$
Explanation
Both the charges are identical and placed symmetrically about $$ABCD$$. The flux crossing $$ABCD$$ due to each charge is $${1 \over 6}\left[ {{q \over {{ \in _0}}}} \right]$$ but in opposite directions. Therefore the resultant is zero.
Comments (0)
