# two point charges of equal magnitude Q are placed at the separation of 5 cm such that the force acting between them is 50 Newton then find the magnitude of the charge

Dear Student
$F=\frac{QQ}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{r}}^{2}}=\frac{{Q}^{2}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{r}}^{2}}\phantom{\rule{0ex}{0ex}}50=\frac{{Q}^{2}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\left(0.05\right)}^{2}}\phantom{\rule{0ex}{0ex}}50=9×{10}^{9}×{Q}^{2}×400\phantom{\rule{0ex}{0ex}}{Q}^{2}=\frac{50}{400×9×{10}^{9}}=\frac{0.0013}{{10}^{8}}\phantom{\rule{0ex}{0ex}}Q=0.037×{10}^{-4}=3.7\mu C\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
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• 0
Magnitude = 10
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