A) \[2\times {{10}^{-2}}C~\]
B) \[2\times {{10}^{-3}}C\]
C) \[2\times {{10}^{-4}}C\]
D) \[2\times {{10}^{-5}}C\]
Correct Answer: B
Solution :
\[{{E}_{\max }}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{\max }}}{{{R}^{2}}}\] \[3\times {{10}^{6}}=\frac{9\times {{10}^{9}}\times {{Q}_{\max }}}{2.5\times 2.5}\] or \[{{Q}_{\max }}=\frac{3\times {{10}^{6}}\times 2.5\times 2.5}{9\times {{10}^{9}}}\] \[=2\times {{10}^{-3}}C\]
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