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Question Number 131684 by aurpeyz last updated on 07/Feb/21

Answered by ajfour last updated on 07/Feb/21

$$⊝(-2\mu C)\oplus \left(10\mu C\right)$$$$\frac{1}{4\pi {ϵ}_0}\frac{\mid q_3{q}_2\mid }{r^2}=\frac{1}{4\pi {ϵ}_0}\frac{\mid q_3{q}_1\mid }{{(r+1m)}^2}$$$$\Rightarrow \frac{r+1m}{r}=\sqrt{\mid \frac{q_1}{q_2}\mid }$$$$1+\frac{1m}{r}=\sqrt{5}$$$$r=\frac{1m}{\sqrt{5}-1}=\frac{1m(\sqrt{5}+1)}{4}$$$$r=\frac{1m(3.236)}{4}=0.809m$$$$x=-0.809m\left(A\right)$$

Commented by aurpeyz last updated on 08/Feb/21

$$thankyou$$

Answered by Dwaipayan Shikari last updated on 07/Feb/21

$$Considerthepointis(x,0).charge=q$$$$\mid \frac{{qq}_1}{4\pi {ϵ}_0{x}^2}\mid =\mid \frac{{qq}_2}{4\pi {ϵ}_0{(1+x)}^2}\mid \Rightarrow \frac{1+x}{x}=\sqrt{5}\Rightarrow x=\frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{4}=0.809$$

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