A) \[E{{L}^{2}}\]
B) \[E{{L}^{2}}\,\cos \theta \]
C) \[E{{L}^{2}}\,sin\theta \]
D) zero
Correct Answer: D
Solution :
Electric flux, \[\phi =EA\cos \theta ,\] where \[\theta =\] angle between E and normal to the surface. Here \[\theta =\frac{\pi }{2}\] \[\Rightarrow \,\,\,\phi =0\]You need to login to perform this action.
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