Animated Solution for Physics - Electrostatics: Two thin wire rings each having a radius $R$ are placed at a distance $d$ apart with their axes coinciding. The charges on the two rings are $+q$ and $-q$. The potential difference between the centres of the two rings is

When dealing with electric fields, we often find ourselves tangled in a web of vectors, resolving components along the x, y, and z axes. It can get messy. But electric potential? That's where the magic happens. Potential is a scalar quantity, meaning it only has magnitude. To find the total potential at any point due to multiple charges, we simply add them up algebraically. No vectors, no resolving components. Just pure, elegant addition.

In this problem, we are presented with two thin wire rings, let's call them Ring A and Ring B. They both have the same radius $R$ and are separated by a distance $d$ along their common axis. Ring A carries a positive charge $+q$, while Ring B carries a negative charge $-q$. Our goal is to find the potential difference between their centers.

Let's focus on the center of Ring A, which we'll call point $A$. The potential at this point is influenced by two sources: 1. The charge on Ring A itself: Since every point on Ring A is at a distance $R$ from its center, the potential due to its own charge $+q$ is simply $\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}$. 2. The charge on Ring B: This is where it gets interesting. Every point on Ring B is at a specific distance from the center of Ring A. If we draw a line from the center of Ring A to any point on the circumference of Ring B, we form a right-angled triangle with base $d$ and height $R$. By the Pythagorean theorem, this distance is $\sqrt{R^2+d^2}$. Therefore, the potential at $A$ due to the charge $-q$ on Ring B is $\frac{1}{4\pi {\epsilon }_0}\frac{-q}{\sqrt{R^2+d^2}}$.

Using the principle of superposition, the total potential at center $A$ ($V_A$) is the sum of these two contributions:

Now, let's shift our focus to the center of Ring B, point $B$. By symmetry, the situation is identical but with reversed charges. The potential at $B$ due to its own charge $-q$ is $\frac{1}{4\pi {\epsilon }_0}\frac{-q}{R}$, and the potential due to Ring A is $\frac{1}{4\pi {\epsilon }_0}\frac{q}{\sqrt{R^2+d^2}}$.

So, the total potential at center $B$ ($V_B$) is:

Notice that $V_B$ is exactly the negative of $V_A$. This makes perfect sense given the anti-symmetric nature of the charge distribution.

The question asks for the potential difference between the centers of the two rings, which is $\Delta V=V_A-V_B$.

Substituting our expressions for $V_A$ and $V_B$:

Simplifying the fraction by canceling the 2, we arrive at our final, elegant result:

This problem beautifully illustrates the power of scalar addition in electrostatics. By breaking down the system into individual components and applying the superposition principle, a seemingly complex 3D geometry problem reduces to a straightforward algebraic calculation.