A) 3.8
B) 4.5
C) 6.2
D) 8.5
Correct Answer: D
Solution :
\[A={{10}^{-4}}{{m}^{2}}\] \[{{E}_{\max }}={{10}^{6}}V/m\] \[C=15\mu F\] \[C=\frac{k{{\varepsilon }_{0}}A}{d}\] \[\frac{Cd}{{{\varepsilon }_{0}}A}=k\] \[k=\frac{15\times {{10}^{-12}}\times 500\times {{10}^{-6}}}{8.86\times {{10}^{-12}}\times {{10}^{4}}}\] \[=\frac{15\times 5}{8.86}=8.465\] \[k\approx 8.5\]You need to login to perform this action.
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