A) 1.34 V/m
B) 2.68 V/m
C) 5.36 V/m
D) 9.37 V/m
Correct Answer: A
Solution :
\[{{S}_{av}}=\frac{1}{2}{{\varepsilon }_{0}}cE_{0}^{2}=\frac{P}{4\pi {{R}^{2}}}\] Þ \[{{E}_{0}}=\sqrt{\frac{P}{2\pi {{R}^{2}}{{\varepsilon }_{0}}C}}\] \[=\sqrt{\frac{3}{2\times 3.14\times 100\times 8.85\times {{10}^{-12}}\times 3\times {{10}^{8}}}}\] = 1.34 V/m
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