A) \[{{10}^{9}}V/m\]
B) \[{{10}^{11}}V/m\]
C) \[{{10}^{10}}V/m\]
D) \[{{10}^{8}}V/m\]
Correct Answer: D
Solution :
Potential gradient is \[k=\frac{V}{l}=\frac{iR}{l}\] \[(\therefore \,V=iR)\] \[=\frac{i\times \rho \frac{l}{A}}{l}\] \[\left( \because R=\rho \frac{l}{A} \right)\] \[=\frac{i\rho }{A}\] \[\therefore \] \[k=\frac{0.01\times {{10}^{-3}}\times {{10}^{9}}\times {{10}^{-2}}}{{{10}^{-2}}\times {{10}^{-4}}}={{10}^{8}}V/m\]
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