question_answer

    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 Then the electrons flux through ABCD is:

A)  \[q/4\pi {{\varepsilon }_{0}}L\]                

B)  zero

C)  \[q/2\pi {{\varepsilon }_{0}}L\]                

D)  \[q/3\pi {{\varepsilon }_{0}}L\]

Correct Answer: B

Solution :

                Electric flux\[{{\phi }_{E}}=\oint{\overrightarrow{E}}.d\overrightarrow{s}\] \[d\overrightarrow{s}\]is the areal vector and\[\overrightarrow{E}\]is the electric field vector. Therefore,\[\overrightarrow{E}.d\overrightarrow{s}=Eds\cos \theta \] \[\theta \]is angle between\[\overrightarrow{E}\]and\[d\overrightarrow{s}\] But           \[\theta =90{}^\circ \] Thus,        \[{{\phi }_{E}}=0\]