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 a fifth charge $$Q$$ at the origin of the coordinate system will be:

We have four identical point charges of magnitude $$Q$$ lying in the $$xy$$-plane at the Cartesian coordinates $$(0,\,2)\,,\;(4,\,2)\,,\;(4,\,-2)\;$$and$$(0,\,-2)$$. We wish to calculate the external work that must be done to slowly bring a fifth charge, also of magnitude $$Q$$, from infinity to the origin $$(0,\,0)$$.

The work required equals the increase in the electrostatic potential energy of the system, and that is given by the product of the fifth charge and the electric potential already present at the origin because of the four fixed charges. Symbolically,

$$W = q\,V_{\text{origin}},$$

where here $$q = Q$$ (the charge we are bringing in) and $$V_{\text{origin}}$$ is the net electric potential at the origin due to the other four charges.

The electric potential due to a single point charge is, by definition,

$$V = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{q}{r},$$

where $$r$$ is the straight-line distance from the charge to the point where the potential is being evaluated. The superposition principle tells us that potentials simply add algebraically. Therefore we first find the distance of each of the four charges from the origin, then add the individual contributions.

For the charge at $$(0,\,2)$$ we note that it lies on the $$y$$-axis 2 units above the origin, so

$$r_1 = \sqrt{0^2 + 2^2} = 2.$$

The charge at $$(0,\,-2)$$ is 2 units below the origin, giving

$$r_2 = \sqrt{0^2 + (-2)^2} = 2.$$

The charge at $$(4,\,2)$$ has coordinates that produce

$$r_3 = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}.$$

Finally, for the charge at $$(4,\,-2)$$ we obtain the identical distance

$$r_4 = 2\sqrt{5}.$$

Now we compute each potential contribution. Using $$k = \dfrac{1}{4\pi\varepsilon_0}$$ for brevity, we write

$$V_1 = k\,\dfrac{Q}{r_1} = k\,\dfrac{Q}{2},$$

$$V_2 = k\,\dfrac{Q}{r_2} = k\,\dfrac{Q}{2},$$

$$V_3 = k\,\dfrac{Q}{r_3} = k\,\dfrac{Q}{2\sqrt{5}},$$

$$V_4 = k\,\dfrac{Q}{r_4} = k\,\dfrac{Q}{2\sqrt{5}}.$$

Adding them, the total potential at the origin becomes

$$\begin{aligned} V_{\text{origin}} & = V_1 + V_2 + V_3 + V_4 \\[4pt] & = kQ\!\left(\dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2\sqrt{5}} + \dfrac{1}{2\sqrt{5}}\right) \\[6pt] & = kQ\!\left(1 + \dfrac{1}{\sqrt{5}}\right). \end{aligned}$$

We now substitute this potential into the work formula:

$$W = Q \, V_{\text{origin}} = Q \, \left[kQ\left(1 + \dfrac{1}{\sqrt{5}}\right)\right] = kQ^2\left(1 + \dfrac{1}{\sqrt{5}}\right).$$

Replacing $$k$$ by its explicit form $$\dfrac{1}{4\pi\varepsilon_0}$$, we obtain

$$W = \dfrac{Q^2}{4\pi\varepsilon_0}\left(1 + \dfrac{1}{\sqrt{5}}\right).$$

This matches Option C.

Hence, the correct answer is Option C.