Animated Solution for Physics - Electrostatics: Two identical thin rings, each of radius $R$, are coaxially placed a distance $R$ apart. If $Q_1$ and $Q_2$ are respectively the charges uniformly spread on the two rings, the work done in moving a charge $q$ from the centre of one ring to that of the other is

Imagine two identical thin rings, each with a radius of $R$, placed coaxially at a distance of $R$ from each other. The first ring carries a uniformly distributed charge $Q_1$, and the second ring carries a uniformly distributed charge $Q_2$. Our goal is to find the work done in moving a test charge $q$ from the center of the first ring ($C_1$) to the center of the second ring ($C_2$).

To solve this, we need to rely on the principle of superposition for electric potential. The work done by an external agent in moving a charge between two points in an electrostatic field is simply the product of the charge and the potential difference between those two points: $W=q\Delta V$.

Let's focus on the center of the first ring, $C_1$. The total electric potential at this point is not just due to its own charge $Q_1$; it also experiences a potential due to the charge $Q_2$ on the adjacent ring.

The potential at $C_1$ due to its own ring is straightforward: $\frac{1}{4\pi {\epsilon }_0}\frac{Q_1}{R}$.

But what about the potential from the second ring? Every point on the circumference of the second ring is at a perpendicular radial distance $R$ from the central axis, and the two rings are separated by an axial distance $R$. Using the Pythagorean theorem, the straight-line distance from $C_1$ to any point on the second ring is $\sqrt{R^2+R^2}=R\sqrt{2}$.

Therefore, the total potential at $C_1$ is:

By applying the exact same logic and symmetry, the potential at the center of the second ring, $C_2$, is:

Now, we need to find the potential difference between the two centers, $\Delta V=V_{C_1}-V_{C_2}$. Let's carefully subtract the two expressions:

Grouping the terms for $Q_1$ and $Q_2$ together, we get:

Factoring out the common term:

To match the options provided in the question, we can write $1-\frac{1}{\sqrt{2}}$ as $\frac{\sqrt{2}-1}{\sqrt{2}}$:

Finally, the work done in moving the charge $q$ from $C_1$ to $C_2$ is given by $W=q\Delta V$. Substituting our potential difference into this formula yields:

It is crucial to remember that the electrostatic field is a conservative field. This means the work done depends exclusively on the initial and final potentials, regardless of the specific path the charge $q$ takes between the two centers. This is a classic and highly scoring concept in JEE Physics!