Question
Two charges q and -3q are placed fixed on x-axis separated by distance 'd'. Where should a third charge 2q be placed such that it will not experience any force?
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Answer
The force on any charge will be zero only if net electric field at the position of charge is zero. Let electric field is zero at a distance x from charge q.
At point $\text{P},\vec{\text{E}_\text{A}}+\vec{\text{E}_\text{B}}=0\Rightarrow|\vec{\text{E}_\text{A}}|=|\vec{\text{E}_\text{B}}|$
$\Rightarrow\ \frac{\text{q}}{4\pi\epsilon_0\text{x}^2}=\frac{3\text{q}}{4\pi\epsilon_0(\text{x}+\text{d})^2}$
$\Rightarrow\ (\text{x}+\text{d})^2=3\text{x}^2$
$\Rightarrow\ \text{x}^2+\text{d}^2=2\text{dx}=3\text{x}^2$
$\therefore\ 2\text{x}^2-2\text{dx}-\text{d}^2=0$
or $\text{x}=\frac{\text{d}}{2}\pm\frac{\sqrt{3\text{d}}}{2}$
(Negative sign lies between q and -3q and hence is unaceptable.)
Hence $\text{x}=-\frac{\text{d}}{2}+\frac{\sqrt{3\text{d}}}{2}=\frac{\text{d}}{2}(1+\sqrt{3})$
Hence if charge 2q is placed at a distance $\frac{\text{d}}{2}(1+\sqrt{3})$ to the left of q.
At point $\text{P},\vec{\text{E}_\text{A}}+\vec{\text{E}_\text{B}}=0\Rightarrow|\vec{\text{E}_\text{A}}|=|\vec{\text{E}_\text{B}}|$
$\Rightarrow\ \frac{\text{q}}{4\pi\epsilon_0\text{x}^2}=\frac{3\text{q}}{4\pi\epsilon_0(\text{x}+\text{d})^2}$
$\Rightarrow\ (\text{x}+\text{d})^2=3\text{x}^2$
$\Rightarrow\ \text{x}^2+\text{d}^2=2\text{dx}=3\text{x}^2$
$\therefore\ 2\text{x}^2-2\text{dx}-\text{d}^2=0$
or $\text{x}=\frac{\text{d}}{2}\pm\frac{\sqrt{3\text{d}}}{2}$
(Negative sign lies between q and -3q and hence is unaceptable.)
Hence $\text{x}=-\frac{\text{d}}{2}+\frac{\sqrt{3\text{d}}}{2}=\frac{\text{d}}{2}(1+\sqrt{3})$
Hence if charge 2q is placed at a distance $\frac{\text{d}}{2}(1+\sqrt{3})$ to the left of q.
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