A) \[2.3\times {{10}^{-8}}\]N
B) \[4.6\times {{10}^{-8}}\]N
C) \[1.5\times {{10}^{-8}}\]N
D) None of these
Correct Answer: A
Solution :
\[F=9\times {{10}^{9}}\times \frac{{{Q}^{2}}}{{{r}^{2}}}\]\[=9\times {{19}^{9}}\times \frac{{{(1.6\times {{10}^{-19}})}^{2}}}{{{({{10}^{-10}})}^{2}}}=2.3\times {{10}^{-8}}N\]You need to login to perform this action.
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