Question

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 $

Answer 1

Answer: (B)

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 $