A) 1.062 amp
B) \[1.062\times {{10}^{-2}}\]amp
C) \[1.062\times {{10}^{-3}}\]amp
D) \[1.062\times {{10}^{-4}}\]amp
Correct Answer: B
Solution :
\[{{I}_{D}}={{\varepsilon }_{0}}\frac{d{{\varphi }_{E}}}{dt}={{\varepsilon }_{0}}\frac{EA}{t}={{\varepsilon }_{0}}\left( \frac{V}{d} \right).\frac{A}{t}.\] \[=\frac{8.85\times {{10}^{-12}}\times 400\times 60\times {{10}^{-4}}}{{{10}^{-3}}\times {{10}^{-6}}}=1.602\times {{10}^{-2}}amp\]
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