Explanation

(a) Following figures show the situations of charges fixed on the axis. An electron is placed to the left of these charges. The cases are as follows
Case I Let distance between $+ q$ and $- 4q = d$
$\therefore$ distance between $–e$ and $+ q = x$
$\therefore$ distance between $–e$ and $- 4q = (x + d)$
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Now, force between $–e$ and $+ q$
$F_{1} = - \frac{1}{4\pi\varepsilon_{0}}\frac{qe}{x^{2}}$ (attractive)
Force between $–e$ and $- 4q$
$F_{2} = \frac{1}{4\pi\varepsilon_{0}}\frac{4qe}{(x + d)^{2}}$ (repulsive)
Solving, we get
$\ \ \ x = d$
$\therefore F_{1} = - F_{2}$
$F_{net} = 0$
Hence no net force acts on the electron and so it will be in equilibrium.
Case II In this case force acting between $e$ and $–q$
$F_{1} = \frac{1}{4\pi\varepsilon_{0}}\frac{qe}{x^{2}}$ (repulsive)
[image:1_image286.emf]
and force between $–e$ and $+ 4q$
$F_{2} = - \frac{1}{4\pi\varepsilon_{0}}\frac{4qe}{(x + d)^{2}}$ (attractive)
Solving we get
$x = d$
$\therefore$ $F_{1} = - F_{2}$ (numerically)
$\therefore$ Net force on $–e$ is zero
Case III Again force between $–e$ and $4q$
[image:1_image287.emf]
Similarly, $F_{1} = - \frac{1}{4\pi\varepsilon_{0}}\frac{4qe}{x^{2}}$ (attractive)
$F_{2} = - \frac{1}{4\pi\varepsilon_{0}\ }\frac{qe}{(x + d)^{2}}\ $ (repulsive)
Since, electron is closer to $+ 4q$ than $–q,$ so $F_{1} > F_{2}$
In this case electron will not remain at rest and starts moving towards the system.
Case IV In this case force between $–e\ and - 4q$
[image:1_image288.emf]
$F_{1} = + \frac{1}{4\pi\varepsilon_{0}}\frac{4qe}{x^{2}}$ (repulsive)
Force between $–e\ and + q$
$F_{2} = - \frac{1}{4\pi\varepsilon_{0}}\frac{qe}{(x + d)^{2}}$ (attractive)
Since, electron is closer to $- 4q\ $than +q, then $F_{1} > F_{2}.$
Thus, electron will move away from the system. It means equilibrium stage cannot be obtained.