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Step-by-Step Solution
Step 1: Identify the Positions of the Charges
We have four point charges, each of magnitude $Q$, placed at:
(0, 2), (4, 2), (4, −2), and (0, −2). We want to bring a fifth charge $Q$ from infinity to the origin (0, 0).
Step 2: Determine the Distances of Each Charge from the Origin
Charge at (0, 2): Distance from (0,0) is $2$.
Charge at (0, −2): Distance from (0,0) is $2$.
Charge at (4, 2): Distance from (0,0) is $\sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}.$
Charge at (4, −2): Distance from (0,0) is also $2\sqrt{5}.$
Step 3: Calculate the Electric Potential at the Origin
The potential $V$ at a point due to a single point charge $Q$ at distance $r$ is given by:
$$
V = \frac{Q}{4\pi \varepsilon_0 \, r}.
$$
Therefore, we sum up the potentials from all four charges:
Potential due to charges at (0, 2) and (0, −2):
$$
V_1 + V_2 = \frac{Q}{4 \pi \varepsilon_0 \times 2} + \frac{Q}{4 \pi \varepsilon_0 \times 2}
= \frac{Q}{8 \pi \varepsilon_0} + \frac{Q}{8 \pi \varepsilon_0}
= \frac{Q}{4 \pi \varepsilon_0}.
$$
Potential due to charges at (4, 2) and (4, −2):
$$
V_3 + V_4
= \frac{Q}{4 \pi \varepsilon_0 \times 2\sqrt{5}} + \frac{Q}{4 \pi \varepsilon_0 \times 2\sqrt{5}}
= \frac{Q}{8 \pi \varepsilon_0 \sqrt{5}} + \frac{Q}{8 \pi \varepsilon_0 \sqrt{5}}
= \frac{Q}{4 \pi \varepsilon_0 \sqrt{5}}.
$$
Summing all these contributions, the total potential at the origin becomes:
$$
V_{\text{total}} = \frac{Q}{4 \pi \varepsilon_0} + \frac{Q}{4 \pi \varepsilon_0 \sqrt{5}}
= \frac{Q}{4 \pi \varepsilon_0} \Bigl(1 + \frac{1}{\sqrt{5}}\Bigr).
$$
Step 4: Compute the Work Required
To bring the fifth charge $Q$ from infinity to the origin, the work done is
$$
W = Q \times V_{\text{total}}
= Q \times \frac{Q}{4 \pi \varepsilon_0}\Bigl(1 + \frac{1}{\sqrt{5}}\Bigr)
= \frac{Q^2}{4 \pi \varepsilon_0}\Bigl(1 + \frac{1}{\sqrt{5}}\Bigr).
$$
Final Answer
$$
\boxed{\frac{Q^2}{4 \pi \varepsilon_0}\Bigl(1 + \frac{1}{\sqrt{5}}\Bigr)}
$$
