A) \[\left( \frac{r}{\sqrt[3]{2}} \right)\]
B) \[\left( \frac{2r}{\sqrt{3}} \right)\]
C) \[\left( \frac{2r}{3} \right)\]
D) \[{{\left( \frac{r}{\sqrt{2}} \right)}^{2}}\]
Correct Answer: A
Solution :
[a] From figure, \[\tan \theta =\frac{{{F}_{e}}}{mg}\Rightarrow \frac{r/2}{y}=\frac{\frac{k{{q}^{2}}}{{{r}^{2}}}}{mg}\] [\[\because F=\frac{k{{q}^{2}}}{{{r}^{2}}}\] from coulomb?s law] \[\Rightarrow {{r}^{3}}\propto y\Rightarrow r{{'}^{3}}\propto \frac{y}{2}\Rightarrow \frac{r'}{r}=\frac{1}{{{2}^{1/3}}}\Rightarrow r'=\frac{r}{\sqrt[3]{2}}\]You need to login to perform this action.
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