A) \[1.36\times {{10}^{2}}N{{m}^{2}}{{C}^{-1}}\]
B) \[1.36\times {{10}^{4}}\,N{{m}^{2}}{{C}^{-1}}\]
C) \[0.515\,\times {{10}^{2}}\,N{{m}^{2}}{{C}^{-1}}\]
D) \[0.515\times {{10}^{4}}\,\,N{{m}^{2}}{{C}^{-1}}\]
Correct Answer: B
Solution :
(b) \[1.36\times {{10}^{4}}\,N{{m}^{2}}{{C}^{-1}}\] Here, r =10 cm = 0.1m, \[E=5\times {{10}^{5}}\,N{{C}^{-1}}\] As the angle between the plane sheet and the electric field is\[60{}^\circ \], angle made by the normal to the plane sheet and the electric field is \[Q=90{}^\circ -60{}^\circ =30{}^\circ \] \[{{\phi }_{E}}=ES\,\cos \theta =E\times \pi {{r}^{2}}\cos \theta \] \[=5\times {{10}^{5}}\times 3.14\times {{\left( 0.1 \right)}^{2}}\cos 30{}^\circ \] \[=136.\times {{10}^{4}}\,N{{m}^{2}}{{C}^{-1}}\]
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