A) 0.13 V
B) 1.3 V
C) 13 V
D) 130 V
Correct Answer: A
Solution :
\[V=9\times {{10}^{9}}.\frac{p}{{{r}^{2}}}\] \[=9\times {{10}^{9}}\times \frac{(1.6\times {{10}^{-19}})\times 1.28\times {{10}^{-10}}}{{{(12\times {{10}^{-10}})}^{2}}}\]= 0.13VYou need to login to perform this action.
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