A) \[8.5\times {{10}^{-5}}V{{m}^{-1}}\]
B) \[8.5\times {{10}^{-4}}V{{m}^{-1}}\]
C) \[8.5\times {{10}^{-3}}V{{m}^{-1}}\]
D) \[8.5\times {{10}^{-2}}V{{m}^{-1}}\]
Correct Answer: C
Solution :
From symmetry considerations and also from theory \[\frac{{{V}_{a}}}{{{V}_{d}}}=\frac{{{V}_{b}}}{{{V}_{c}}}\] \[E=\frac{V}{l}=\frac{IR}{l}=\frac{I\rho l}{Al}=\frac{I\rho }{A}\] \[=\frac{1\times 1.7\times {{10}^{-8}}}{2\times {{10}^{-6}}}\] \[=8.5\times {{10}^{-3}}V{{m}^{-1}}\]
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