A) \[\text{1}{{0}^{\text{16}}}\]
B) \[\text{1}{{0}^{\text{18}}}\]
C) \[\text{1}{{0}^{\text{2}0}}\]
D) \[\text{1}{{0}^{\text{22}}}\]
Correct Answer: D
Solution :
\[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[=9\times {{10}^{9}}\times \frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[=9\times {{10}^{9}}\frac{(1.6\times {{10}^{-19}})(1.6\times {{10}^{-19}})}{{{(0.5\times {{10}^{-10}})}^{2}}}\] \[=\mathbf{9}\mathbf{.2\times 1}{{\mathbf{0}}^{\mathbf{-8}}}\mathbf{N}\] Weight,\[W=mg\] \[=(9.1\times {{10}^{-31}})(9.8)\] \[=8.9\times {{10}^{-30}}N\] \[\therefore \] \[\frac{F}{W}=\frac{9.2\times {{10}^{-8}}}{8.9\times {{10}^{-30}}}\approx {{10}^{22}}\]
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