Question
Suppose the second charge in the previous problem is $-1.0 \times 10^{-6}C$. Locate the position where a third charge will not experience a net force.
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Answer
By coulomb's Law, force,
$\text{F}=\frac{1}{4\pi\in_0}\frac{\text{Q}_1\text{Q}_2}{\text{r}_2}$
So, force on charge q due to $q_1,$
$\text{F}=\frac{9\times10^{9}\times2.0\times10^{-6}\times\text{q}}{\text{x}^2}$
Force on charge q due to $q_2$
$\text{F}'=\frac{9\times10^9\times10^{-6}\times\text{q}}{(10-\text{x})^2}$
According to the question,
$\Rightarrow\text{F}-\text{F}'=0$
$\Rightarrow\frac{9\times10^{9}\times2.0\times10^{-6}\times\text{q}}{\text{x}^2}=\frac{9\times10^9\times10^{-6}\times\text{q}}{(\text{x}-10)^2}$
$\Rightarrow\text{x}^2=2(\text{x}-10)^2$
$\Rightarrow\text{x}^2-40\text{x}+200=0$
$\Rightarrow\text{x}=20\pm10\sqrt{2}\text{m}$
$\Rightarrow\text{x}=34.14\text{cm}$ $\Big(\because\ \text{x}\neq20-10\sqrt{2}\Big)$
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