A) \[90{}^\circ \]
B) \[0{}^\circ \]
C) \[180{}^\circ \]
D) \[45{}^\circ \]
Correct Answer: A
Solution :
[a] \[\vec{\tau }=\vec{P}\times \vec{E}=PE\sin \theta \] \[\text{For }q=90{}^\circ ,{{\vec{\tau }}_{\max }}=PE\]You need to login to perform this action.
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