two point charges of magnitude Q are placed at the separation of 3 cm such that force between them is equal to weight of 50 kg person find q Share with your friends Share 0 Decoder_2 answered this $Dearstudent\phantom{\rule{0ex}{0ex}}F=50\times 9.8=490N\phantom{\rule{0ex}{0ex}}F=\frac{k{q}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}{q}^{2}=\frac{F{r}^{2}}{k}=\frac{490\times 3\times 3\times {10}^{-4}}{9\times {10}^{9}}\phantom{\rule{0ex}{0ex}}q=7\mu C\phantom{\rule{0ex}{0ex}}Regards\phantom{\rule{0ex}{0ex}}$ 0 View Full Answer