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 centers of the two rings is $\ldots .$ (A) 0 (B) $\left.\left[\mathrm{q} /\left(2 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{(}^{2}+\mathrm{d}^{2}\right)\right\\}\right]$ (C) $\left[\mathrm{q} /\left(4 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{\left. \left.\left(\mathrm{R}^{2}+\mathrm{d}^{2}\right)\right\\}\right]}\right.\right.$ (D) \(\left[(q R) /\left(4 \pi \epsilon_{0} d^{2}\right)\right]\)

To find the potential difference, we first need to find the electric potential at the center of each ring due to the charges on the rings themselves. The formula for electric potential V due to charged ring at its center is given by: \[V = \frac{kq}{R}\] Where: - k is the Coulomb's constant, - q is the charge on the ring (positive for the first ring, negative for the second), - R is the radius of the ring.

Using the formula from Step 1, we can find the electric potential for both rings: For the first ring with charge +q: \(V_{1} = \frac{kq}{R}\) For the second ring with charge -q: \(V_{2} = \frac{-kq}{R}\)

Since the electric potential is a scalar quantity, we can calculate the potential difference between the centers of the two rings by finding the difference between their potentials, which is given by: \(V_{difference} = V_{1} - V_{2}\) Substitute the values from Step 2: \(V_{difference} = \left(\frac{kq}{R}\right) - \left(\frac{-kq}{R}\right)\) \(V_{difference} = 2\frac{kq}{R}\) Finally, replace k with its expression in terms of the vacuum permittivity constant \(\epsilon_0\), which is \(k = \frac{1}{4\pi\epsilon_0}\): \(V_{difference} = 2\frac{\frac{q}{4\pi\epsilon_0R}}{1}\) \(V_{difference} = \frac{q}{2\pi\epsilon_0R}\) The correct answer is option (B): \[\frac{q}{2 \pi \epsilon_0}\left(\frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}}\right)\]