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Question Number 153959 by DELETED last updated on 12/Sep/21

Answered by DELETED last updated on 12/Sep/21

$${\mathrm{F}}_{\mathrm{AC}}=\mathrm{k}.\frac{{\mathrm{q}}_{\mathrm{a}}.{\mathrm{q}}_{\mathrm{c}}}{{\mathrm{r}}_{\mathrm{AC}}^2}$$$$=9.{10}^9.\frac{2.{10}^{-6}.2.{10}^{-4}}{{(30.{10}^{-2})}^2}$$$$=\frac{36}{9}\times {10}^{9-6-4+4-2}$$$$=4\times {10}^1=40\mathrm{N}$$$${\mathrm{F}}_{\mathrm{BC}}=\mathrm{k}.\frac{{\mathrm{q}}_{\mathrm{b}}.{\mathrm{q}}_{\mathrm{c}}}{{\mathrm{r}}_{\mathrm{BC}}^2}$$$$=9.{10}^9.\frac{2.{10}^{-6}.2.{10}^{-4}}{{(30.{10}^{-2})}^2}$$$$=\frac{36}{9}\times {10}^{9-6-4+4-2}$$$$=4\times {10}^1=40\mathrm{N}$$$${\mathrm{F}}_{\mathrm{R}}=\sqrt{{\mathrm{F}}_{\mathrm{ac}}^2+{\mathrm{F}}_{\mathrm{bc}}^2+{2\mathrm{F}}_{\mathrm{ac}}.{\mathrm{F}}_{\mathrm{bc}}\cos 60°}$$$$=\sqrt{{40}^2+{40}^2+2.40.40.\frac{1}{2}}$$$$=\sqrt{1600+1600+1600}$$$$=\sqrt{4800}\mathrm{N}=40\sqrt{3}\mathrm{N}//$$

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