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Question Number 13832 by tawa tawa last updated on 24/May/17

Answered by ajfour last updated on 24/May/17

Commented by ajfour last updated on 24/May/17

$$F_1=F_2=\frac{Q_1{Q}_3}{4\pi {ϵ}_0{r}^2}=F\left(say\right)$$$$F_{net}=2F\cos \alpha inthedirectionshown$$$$r=\sqrt{a^2+b^2}=\sqrt{{(0.6)}^2+{(0.8)}^2}=1m$$$$\cos \alpha =\frac{a}{r}=0.6$$$$F=\frac{\left(6\times {10}^{-6}C\right)\left(4\times {10}^{-6}C\right)\left(9\times {10}^9{Nm}^2/C^2\right)}{1m^2}$$$$=6\times 4\times 9\times {10}^{-3}N$$$$=0.216N$$$$F_{net}=2(0.216N)(0.6)$$$${\mathit{F}}_{net}=0.26\mathit{N}.$$

Commented by tawa tawa last updated on 24/May/17

$$\mathrm{Wow},\mathrm{Am}\mathrm{really}\mathrm{greatful}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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