1. Use the potential due to a charged ring on its axis

For a ring of radius $a$ and total charge $q$, the potential at a point on its axis at distance $x$ from its centre is

$V=\frac{1}{4\pi\varepsilon_0}\frac{q}{\sqrt{a^2+x^2}}.$

At the centre of the ring, $x=0$, so

$V_{\text{centre}}=\frac{1}{4\pi\varepsilon_0}\frac{q}{a}.$

2. Potential at the centre of the first ring

Let ring 1 carry charge $+Q$ and ring 2 carry charge $-Q$. The distance between their centres is $s$.

At the centre of the $+Q$ ring:

• Potential due to its own ring: $V_1^{(self)}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{a}$

• Potential due to the other ring ($-Q$), whose centre is at distance $s$: $V_1^{(other)}=\frac{1}{4\pi\varepsilon_0}\frac{-Q}{\sqrt{a^2+s^2}}$

So total potential there is

$V_1=\frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{\sqrt{a^2+s^2}}\right).$

3. Potential at the centre of the second ring

At the centre of the $-Q$ ring:

• Potential due to its own ring: $V_2^{(self)}=\frac{1}{4\pi\varepsilon_0}\frac{-Q}{a}$

• Potential due to the $+Q$ ring at distance $s$: $V_2^{(other)}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{\sqrt{a^2+s^2}}$

Hence

$V_2=\frac{1}{4\pi\varepsilon_0}\left(-\frac{Q}{a}+\frac{Q}{\sqrt{a^2+s^2}}\right).$

4. Potential difference between the centres

Taking potential difference as $V_1-V_2$,

$V_1-V_2=\frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{\sqrt{a^2+s^2}}+\frac{Q}{a}-\frac{Q}{\sqrt{a^2+s^2}}\right).$

So,

$V_1-V_2=\frac{Q}{2\pi\varepsilon_0}\left(\frac{1}{a}-\frac{1}{\sqrt{a^2+s^2}}\right).$

This matches option D.

5. Check options systematically

• A: has wrong sign and wrong factor.
• B: has wrong sign and smaller by factor $2$.
• C: correct sign but smaller by factor $2$.
• D: exactly matches the derived expression.

Final Answer: $\boxed{\text{D}}$