1. Concept used

The work required to bring a charge $Q$ from infinity to a point is

$W = QV$

where $V$ is the electric potential at that point due to the already existing charges.

2. Coordinates of the four existing charges

The four charges $Q$ are at:

$(0,2),\ (4,2),\ (4,-2),\ (0,-2)$

We need the potential at the origin $(0,0)$.

3. Distances of these charges from the origin

Using distance formula:

• For $(0,2)$: $r_1 = \sqrt{0^2+2^2}=2$

• For $(0,-2)$: $r_2 = \sqrt{0^2+(-2)^2}=2$

• For $(4,2)$: $r_3 = \sqrt{4^2+2^2}=\sqrt{20}=2\sqrt{5}$

• For $(4,-2)$: $r_4 = \sqrt{4^2+(-2)^2}=\sqrt{20}=2\sqrt{5}$

4. Potential at the origin due to all four charges

Potential due to one point charge is

$V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$

So total potential at the origin is

$V = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{2}+\frac{Q}{2}+\frac{Q}{2\sqrt{5}}+\frac{Q}{2\sqrt{5}}\right)$

Simplify:

$V = \frac{1}{4\pi\varepsilon_0}\left(Q+\frac{Q}{\sqrt{5}}\right)$

$V = \frac{Q}{4\pi\varepsilon_0}\left(1+\frac{1}{\sqrt{5}}\right)$

5. Work required to bring the fifth charge $Q$ to the origin

$W = QV$

Therefore,

$W = Q\cdot \frac{Q}{4\pi\varepsilon_0}\left(1+\frac{1}{\sqrt{5}}\right)$

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

6. Matching with options

This matches Option D:

$\boxed{\frac{Q^2}{4\pi\varepsilon_0}\left(1+\frac{1}{\sqrt{5}}\right)}$

7. Comparison with stored correct answer

Stored correct answer: D

My derived answer: D

So, they agree.