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, \[{{\phi }_{E}}=0\]You need to login to perform this action.
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