Two charges q and -3q are placed fixed on x-a | vedclass

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Question

At $P:$ on $2 \mathrm{q}$, Force due to $\mathrm{q}$ is to the left and that due to $-3 \mathrm{q}$ is to the right.

$	herefore \frac{2 q^{2}}{4 \

Vedclass · Accepted Answer

$\frac{d}{2}\left( {1 + \sqrt 3 } \right)$ from $q$

Vedclass · Answer

At $P:$ on $2 \mathrm{q}$, Force due to $\mathrm{q}$ is to the left and that due to $-3 \mathrm{q}$ is to the right.

$	herefore \frac{2 q^{2}}{4 \pi \varepsilon_{0} x^{2}}=\frac{6 q^{2}}{4 \pi \varepsilon_{0}(d+x)^{2}}$

$	herefore(d+x)^{2}=3 x^{2}$

$	herefore 2 x^{2}-2 d x-d^{2}=0$

$x=\frac{d}{2} \pm \frac{\sqrt{3} d}{2}$

($-ve$ sign would be between $\mathrm{q}$ and $-3 \mathrm{q}$ and hence is unacceptable.)

$x=\frac{d}{2}+\frac{\sqrt{3} d}{2}=\frac{d}{2}(1+\sqrt{3})$ to the left of $q$

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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