Two charges + 4e and + e are at a distance | vedclass

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Question

(c) For equilibrium of $q$
|$F_1$| = |$F_2$|
Which gives ${x_2} = \frac{x}{{\sqrt {\frac{{{Q_1}}}{{{Q_2}}}} + 1}} = \frac{x}{{\sqrt {\frac{{4e}}{e}}

Vedclass · Accepted Answer

$x/3$

Vedclass · Answer

(c) For equilibrium of $q$
|$F_1$| = |$F_2$|
Which gives ${x_2} = \frac{x}{{\sqrt {\frac{{{Q_1}}}{{{Q_2}}}} + 1}} = \frac{x}{{\sqrt {\frac{{4e}}{e}} + 1}} = \frac{x}{3}$

Two charges $ + 4e$ and $ + e$ are at a distance $x$ apart. At what distance, a charge $q$ must be placed from charge $ + e$ so that it is in equilibrium

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