A) \[{{10}^{55}}\]
B) \[{{10}^{42}}\]
C) \[{{10}^{28}}\]
D) \[{{10}^{12}}\]
Correct Answer: B
Solution :
\[\frac{{{F}_{e}}}{{{F}_{g}}}=\frac{k{{q}^{2}}}{G{{m}^{2}}}\] \[=\frac{9\times {{10}^{9}}\times {{(1.6\times {{10}^{-19}})}^{2}}}{6.67\times {{10}^{-11}}\times {{(9.1\times {{10}^{-31}})}^{2}}}\] \[={{10}^{42}}\]
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