# solve q -13 please solve with detail explaination

$m={10}^{-6}kg,l=50cm,letthechargebeq\phantom{\rule{0ex}{0ex}}letthetensionebTinstring\phantom{\rule{0ex}{0ex}}T\mathrm{sin}\theta =mg\phantom{\rule{0ex}{0ex}}\mathrm{sin}\theta =\sqrt{0.{5}^{2}-0.{1}^{2}}/0.5=\frac{\sqrt{0.24}}{0.5},\mathrm{cos}\theta =\frac{0.1}{0.5}\phantom{\rule{0ex}{0ex}}T\times \frac{\sqrt{0.24}}{0.5}={10}^{-6}*10\phantom{\rule{0ex}{0ex}}T={10}^{-5}*0.5/\sqrt{0.24}N\phantom{\rule{0ex}{0ex}}T\mathrm{cos}\theta =\frac{9*{10}^{9}*{q}^{2}}{0.{2}^{2}}\phantom{\rule{0ex}{0ex}}\frac{{10}^{-5}*0.5}{\sqrt{0.24}}\times \frac{1}{5}=\frac{9*{10}^{9}*{q}^{2}}{0.{2}^{2}}\phantom{\rule{0ex}{0ex}}q=2.98*{10}^{-9}C\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}regards$

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