A) 8 mC
B) 2 mC
C) 5 mC
D) \[2\mu C\]
Correct Answer: B
Solution :
\[\tau =PE\,\sin \,\theta \] \[\Rightarrow \tau =qIE\,\sin \theta \] \[\Rightarrow q=\frac{\tau }{IE\,\sin \theta }\] \[=\frac{4}{2\times {{10}^{-2}}\times 0.5\times 2\times {{10}^{5}}}=2mc\]You need to login to perform this action.
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