A) \[2\times {{10}^{-2}}C\]
B) \[2\times {{10}^{-3}}C\]
C) \[2\times {{10}^{-4}}C\]
D) 0
Correct Answer: B
Solution :
\[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{r}^{2}}}\] \[_{q\max }=E.4\pi {{\varepsilon }_{0}}{{r}^{2}}\] \[\begin{align} & =3\times {{10}^{6}}\times \frac{1}{3\times {{10}^{9}}}\times {{\left( \frac{5}{2} \right)}^{2}} \\ & =2\times {{10}^{-3}}C \\ \end{align}\]
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