A) \[2.8\times {{10}^{-7}}N\]
B) \[3.7\times {{10}^{-7}}N\]
C) \[6.2\times {{10}^{-7}}N\]
D) \[9.1\times {{10}^{-7}}N\]
Correct Answer: B
Solution :
\[F=kq\frac{{{q}_{1}}{{r}^{2}}}{{{r}_{2}}}\]\[F=\frac{9\times {{10}^{9}}\times 1.6\times {{10}^{-19}}\times 1.6\times {{10}^{-19}}}{{{(2.5\times {{10}^{-11}})}^{2}}}\]\[=3.7\times {{10}^{-7}}N\]
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