A) \[9\times {{10}^{3}}\ Newton\]
B) \[9\times {{10}^{-3}}\ Newton\]
C) \[1.1\times {{10}^{-4}}\ Newton\]
D) \[{{10}^{4}}\ Newton\]
Correct Answer: A
Solution :
\[F=\frac{k{{Q}^{2}}}{{{r}^{2}}}=9\times {{10}^{9}}\times {{1}^{2}}\times \frac{1}{{{(1000)}^{2}}}=9\times {{10}^{3}}N\]
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