Animated Solution for Physics - Electrostatics: Two point charges $4q$ and $-q$ are fixed on the X-axis at $x=\frac{-d}{2}$ and $x=\frac{d}{2}$, respectively. If a third point charge $q$ is taken from the origin to $x=d$ along the semi-circle as shown in the figure, the energy of the charge will

Let's embark on a thrilling journey through the electrostatic landscape! Imagine you are standing on the x-axis, observing a battlefield of charges. We have two massive anchors: a positive charge $4q$ fixed at $x=-\frac{d}{2}$, and a negative charge $-q$ fixed at $x=\frac{d}{2}$.

Now, a third charge, $q$, is about to take a trip. It starts at the origin ($x=0$) and travels along a sweeping semi-circular path to land at $x=d$. Our mission is to find out exactly how its energy changes during this journey.

The energy of our moving charge $q$ is simply its electrostatic potential energy. The change in this energy, $\Delta U$, is given by the charge multiplied by the change in the electric potential it experiences between its final and initial positions.

Mathematically, we write this as:

Before we dive into messy calculations, let's take a step back and look for a beautiful symmetry in the problem. Notice the position of the $-q$ charge. It sits exactly at $x=\frac{d}{2}$.

At the start of the journey, our moving charge is at the origin ($x=0$). The distance between them is $\frac{d}{2}$. At the end of the journey, the moving charge is at $x=d$. The distance between $x=d$ and $x=\frac{d}{2}$ is also exactly $\frac{d}{2}$!

Because the distance to the $-q$ charge is identical at both the starting and ending points, the electric potential created by it at these two locations is exactly the same. When we subtract the initial potential from the final potential, the contribution from the $-q$ charge completely cancels out!

This is a massive time-saver. We can completely ignore the $-q$ charge and focus solely on the potential change caused by the $4q$ charge.

Let's measure the distances from the $4q$ charge, which is anchored at $x=-\frac{d}{2}$.

Initially, the moving charge is at the origin. The distance is simply $r_i=\frac{d}{2}$.

Finally, the moving charge is at $x=d$. The distance is the difference in their coordinates: $r_f=d-\left(-\frac{d}{2}\right)=\frac{3d}{2}$.

Now, we substitute these distances into our potential energy formula. The change in potential energy is solely due to the $4q$ charge:

Let's carefully simplify this algebraic expression. We can pull out a common factor of $\frac{8k{q}^2}{d}$:

Inside the bracket, we are left with $\frac{1}{3}-1$, which evaluates to $-\frac{2}{3}$. Multiplying this out, we get:

Finally, we replace the Coulomb constant $k$ with its standard form, $\frac{1}{4\pi {\epsilon }_0}$:

The negative sign is crucial here. It tells us that the potential energy of the charge has decreased by $\frac{4q^2}{3\pi {\epsilon }_{0}d}$.

As a final thought, notice how the semi-circular path was just a clever distraction! Because the electrostatic force is conservative, the change in energy depends entirely on the initial and final coordinates, not the path taken. You could have taken a zig-zag path through space, and the answer would remain beautifully unchanged.