Explanation

(c) Let two identical positive charges (+q) are placed on both sides at a distance $a$ from the origin $O$ and negative charge (-q’) is placed at a large distance $x$from$\ O$.
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Here, $F_{1} = F_{2} = \frac{Kqq}{(a^{2} + x^{2})}\ \ \ \ \ \ldots.(i)$
The situation is shown in figure. Here the vertical components $\ F_{1}\sin\theta$ and $F_{2}\sin\theta$
are equal and opposite. So they cancel each other. The resultant force on the
charge $q''''''''$ is
[$AC = BC = \sqrt{a^{2} + x^{2}}\ and\cos\theta = \frac{x}{\sqrt{a^{2} + x^{2}}}\rbrack$
$F = F_{1}\cos{\theta + F_{2}\cos{\theta = 2Fcos\theta}}$
$(\because F_{1} = F_{2})$
$\Rightarrow$ $F = \frac{2Kqq}{{(a}^{2} + x^{2})}.\frac{x}{\sqrt{a^{2} + x^{2}}}$
$\ \ \ \ \ = \frac{2Kqqx}{{{(a}^{2} + x^{2})}^{3/2}\ }\ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
But $F = ma(m\ $mass of the charge q’, and a=acceleration)
$\ \ \ \ \ \ \ \ \ ma = \frac{2Kqqx}{\left( a^{2} + x^{2} \right)^{3}/^{2}\ }$
for $a = \frac{2Kqq}{m}.\frac{x}{\left( a^{2} + x^{2} \right)^{3}/^{2}}\ \ \ \ \ldots(iii)$
When the particle starts to move, its acceleration is zero and after then the acceleration of the particle is given by Eq. (iii).
So it is clear, when $x = 0,$ then $a = 0$ and when $x = \pm \frac{a}{\sqrt{2}}$ then the acceleration is maximum. Hence the graph between acceleration and its$\ x$-coordinate is given as
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