A) \[8\times {{10}^{11}}C\]
B) \[2\times {{10}^{-12}}C\]
C) \[3\times {{10}^{-11}}C\]
D) \[6\times {{10}^{-10}}C\]
Correct Answer: A
Solution :
Given \[E=2N/C,\] \[r=60\times {{10}^{-2}}m\] \[\because \] \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{r}^{2}}}\] or \[q=4\pi {{\varepsilon }_{0}}E{{r}^{2}}=\frac{2\times 60\times {{10}^{-2}}}{9\times {{10}^{9}}}\] \[=8\times {{10}^{-11}}C\]
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