A) 1.23 mV
B) 1.23 mV
C) 1.23 V
D) 12.3 mV
Correct Answer: A
Solution :
\[I=\frac{1}{2}{{\varepsilon }_{0}}CE_{0}^{2}\] Þ \[{{E}_{0}}=\sqrt{\frac{2I}{{{\varepsilon }_{0}}C}}=\sqrt{\frac{2\times 5\times {{10}^{-16}}}{8.85}}=0.61\times {{10}^{-6}}\frac{V}{m}\] Also \[{{E}_{0}}=\frac{{{V}_{0}}}{d}\] Þ \[{{V}_{0}}={{E}_{0}}d=0.61\times {{10}^{-6}}\times 2=1.23\mu V\]
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