Four equal point charges $Q$ each are placed in the $XY$ plane at $(0,2),(4,2),(4,-2)$ and $(0,-2)$. The work required to put the fifth $Q$ at the origin of the coordinate system will be:
A) $\dfrac{{{Q}^{2}}}{2\sqrt{2\pi {{\varepsilon }_{0}}}}$
B) $\dfrac{{{Q}^{2}}}{4\pi {{\varepsilon }_{0}}}(1+\dfrac{1}{\sqrt{5}})$
C) $\dfrac{{{Q}^{2}}}{4\pi {{\varepsilon }_{0}}}(1+\dfrac{1}{\sqrt{3}})$
D) $\dfrac{{{Q}^{2}}}{4\pi {{\varepsilon }_{0}}}$

Hint: We know that there is some potential energy at origin due to these four charges. So, when the fifth charge is put at origin, it requires work done equal to the potential energy at that place. If we assume that the initial position of the fifth charge is at infinity, there is not any contribution of fifth charge in potential energy at origin.

Complete step by step solution:
If the initial position of a charge is not given then it is assumed that it will be at infinity and there is no potential at any finite point due to this charge.
In the given condition there are four charges already placed in the plane. There are four equal point charges $Q$ each are placed in $XY$ plane $(0,2),(4,2),(4,-2)$ and $(0,-2)$ .

Let's assume there are four charges A, B, C, and D at $(0,2),(4,2),(4,-2)$ and $(0,-2)$ respectively.
So, potential at any point is by a $Q$ is given by $\dfrac{KQ}{r}$ where, $K=\dfrac{Q}{4\pi {{\varepsilon }_{0}}}$ , $r$ is distance from charge, $Q$ is charge by whom the potential at that point is calculating. ${{\varepsilon }_{0}}$ is permittivity of free space.
So, the distance of A, B, C, D from the origin is $2,\sqrt{20},\sqrt{20},2$ respectively. We determine the distance of B and C from origin by using Pythagoras theorem.
So the total potential energy at origin is equal to the sum of potential energies by A, B, C, D charges to origin.
So, potential energy $ = \dfrac{{KQ}}{2} + \dfrac{{KQ}}{{\sqrt {20} }} + \dfrac{{KQ}}{{\sqrt {20} }} + \dfrac{{KQ}}{2}$
So, $P.E.=\dfrac{KQ}{\sqrt{5}}(\sqrt{5}+1)$
So, the work done by a $Q$ charge is equal to potential energy i.e. $\dfrac{KQ}{\sqrt{5}}(\sqrt{5}+1)$
So put the value of $K$.
So, work done is equals to $\dfrac{{{Q}^{2}}}{4\pi {{\varepsilon }_{0}}}(1+\dfrac{1}{\sqrt{5}})$ .

Hence, option (B) is correct.

Note: The initial position of the charge has to be assumed at infinity because we assume that if there is any charge is at infinity then it will not show any effect like potential energy at any point. We assume infinity as a datum in which the potential is zero in every condition.