Animated Solution for Physics - Electrostatics: 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 (in joule) will be

Imagine a blank canvas, the $xy$-plane, where we are about to place four identical point charges, each carrying a charge $Q$. We pin them down at specific coordinates: $(0,2)$, $(0,-2)$, $(4,2)$, and $(4,-2)$. These charges create an invisible landscape of electric potential around them.

Our mission is to find out how much work it takes to bring a fifth charge, also $Q$, from infinitely far away and place it exactly at the origin $(0,0)$.

Work and Potential

In electrostatics, the work done $W$ by an external agent to bring a charge $q$ from infinity to a point is directly proportional to the electric potential $V$ at that point. The relationship is beautifully simple:

Here, our test charge $q$ is $Q$. So, our entire problem boils down to finding the net electric potential $V$ at the origin due to the four existing charges.

Adding it All Up

The principle of superposition tells us that the net potential at any point is simply the algebraic sum of the potentials due to each individual charge. Unlike electric fields, potential is a scalar quantity, which means we don't have to worry about vector components or angles—we just add the numbers!

The potential $V$ due to a single point charge $Q$ at a distance $r$ is given by:

where $K=\frac{1}{4\pi {\epsilon }_0}$.

Let's find the distance of each of our four charges from the origin $(0,0)$: 1. For the charge at $(0,2)$, the distance is $r_1=\sqrt{0^2+2^2}=2$. 2. For the charge at $(0,-2)$, the distance is $r_2=\sqrt{0^2+(-2{)}^2}=2$. 3. For the charge at $(4,2)$, the distance is $r_3=\sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20}$. 4. For the charge at $(4,-2)$, the distance is $r_4=\sqrt{4^2+(-2{)}^2}=\sqrt{16+4}=\sqrt{20}$.

Now, we plug these distances into our superposition equation:

Let's factor out $KQ$ to make the algebra cleaner:

We know that $\frac{1}{2}+\frac{1}{2}=1$. Also, $\sqrt{20}$ can be simplified to $\sqrt{4\times 5}=2\sqrt{5}$.

Finally, we calculate the work done by multiplying this net potential by our fifth charge $Q$:

Substituting $K=\frac{1}{4\pi {\epsilon }_0}$, we arrive at our final, elegant result:

This matches option (d) perfectly. The beauty of this problem lies in recognizing that scalar addition makes complex geometric arrangements surprisingly easy to handle!