A) B mC
B) 4 mC
C) 6 mC
D) 2 mC
Correct Answer: D
Solution :
(d) 2 mC Here, \[E=2\times {{10}^{5}}\,N{{C}^{-1}},\,l=2\,cm\,,\,\tau =4\,Nm\] Torque, \[\overrightarrow{\tau }\times \overrightarrow{p}\times \overrightarrow{E};\,\tau =pE\,\sin \theta\] \[\therefore \,\,\,\,\,\,4=p\times 2\times {{10}^{5}}\times \sin \,30{}^\circ\] Or \[p=4\times {{10}^{-5}}Cm\] \[\therefore\] Charge, \[q=\frac{p}{l}=\frac{4\times {{10}^{-5}}Cm}{0.02\,m}\] \[=2\times {{10}^{-3}}C=2\,mC\]You need to login to perform this action.
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