Two thin wires 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

Let the magnitudes of charge on the two coaxial rings be $$Q$$ (one ring carries $$+Q$$, the other $$-Q$$). Both rings have the same radius $$R$$ and their centres are separated by a distance $$d$$ along the common axis.

Step 1: Electric potential on the axis of a uniformly charged ring
For a ring of total charge $$q$$ and radius $$R$$, the electric potential at a point on its axis at a distance $$x$$ from the centre is
$$V(x)=\frac{1}{4\pi\varepsilon_0}\,\frac{q}{\sqrt{R^{2}+x^{2}}}$$

Step 2: Potential at the centre of the +Q ring (point P)
(i) Due to its own charge $$+Q$$ (here $$x=0$$):
$$V_{P}^{(self)}=\frac{1}{4\pi\varepsilon_0}\,\frac{+Q}{R}$$
(ii) Due to the other ring (charge $$-Q$$, centre a distance $$d$$ away, so $$x=d$$):
$$V_{P}^{(other)}=\frac{1}{4\pi\varepsilon_0}\,\frac{-Q}{\sqrt{R^{2}+d^{2}}}$$

Hence the total potential at P is
$$V_P=\frac{1}{4\pi\varepsilon_0}\left[\frac{Q}{R}-\frac{Q}{\sqrt{R^{2}+d^{2}}}\right]$$

Step 3: Potential at the centre of the −Q ring (point Q)
(i) Due to its own charge $$-Q$$:
$$V_{Q}^{(self)}=\frac{1}{4\pi\varepsilon_0}\,\frac{-Q}{R}$$
(ii) Due to the other ring (charge $$+Q$$, now also at distance $$d$$):
$$V_{Q}^{(other)}=\frac{1}{4\pi\varepsilon_0}\,\frac{+Q}{\sqrt{R^{2}+d^{2}}}$$

Total potential at Q is
$$V_Q=\frac{1}{4\pi\varepsilon_0}\left[-\frac{Q}{R}+\frac{Q}{\sqrt{R^{2}+d^{2}}}\right]$$

Step 4: Required potential difference
The problem asks for the potential difference between the centres of the two rings, i.e. $$V_P-V_Q$$:
$$\begin{aligned} V_P-V_Q&=\frac{1}{4\pi\varepsilon_0}\Bigl[\frac{Q}{R}-\frac{Q}{\sqrt{R^{2}+d^{2}}}+\frac{Q}{R}-\frac{Q}{\sqrt{R^{2}+d^{2}}}\Bigr]\\[4pt] &=\frac{1}{4\pi\varepsilon_0}\,2Q\left[\frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}}\right]\\[4pt] &=\frac{Q}{2\pi\varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}}\right] \end{aligned}$$

Step 5: Matching with the options
The expression obtained matches exactly with Option B.

Hence, the potential difference between the centres of the two rings is:
$$\boxed{\displaystyle \frac{Q}{2\pi\varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}}\right]}$$
Option B.