1. Potential due to a charged ring on its axis

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

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

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

$V_{\text{centre}} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}.$

2. Let the centres of the two rings be $C_1$ and $C_2$

• Ring 1 has charge $+q$
• Ring 2 has charge $-q$
• Distance between centres $=d$

We need the potential difference between the centres.

3. Potential at centre $C_1$

Potential at $C_1$ is the sum of contributions from both rings.

• Due to ring 1 itself:

$V_{11}=\frac{1}{4\pi\varepsilon_0}\frac{q}{R}$

• Due to ring 2, whose centre is at distance $d$ from $C_1$:

$V_{12}=\frac{1}{4\pi\varepsilon_0}\frac{-q}{\sqrt{R^2+d^2}}$

So,

$V_{C_1}=\frac{1}{4\pi\varepsilon_0}\left(\frac{q}{R}-\frac{q}{\sqrt{R^2+d^2}}\right).$

4. Potential at centre $C_2$

Similarly,

• Due to ring 2 itself:

$V_{22}=\frac{1}{4\pi\varepsilon_0}\frac{-q}{R}$

• Due to ring 1 at distance $d$:

$V_{21}=\frac{1}{4\pi\varepsilon_0}\frac{q}{\sqrt{R^2+d^2}}$

Hence,

$V_{C_2}=\frac{1}{4\pi\varepsilon_0}\left(-\frac{q}{R}+\frac{q}{\sqrt{R^2+d^2}}\right).$

5. Potential difference between the centres

Taking

$\Delta V = V_{C_1}-V_{C_2},$

we get

$\Delta V=\frac{1}{4\pi\varepsilon_0}\left(\frac{q}{R}-\frac{q}{\sqrt{R^2+d^2}}+\frac{q}{R}-\frac{q}{\sqrt{R^2+d^2}}\right).$

So,

$\Delta V=\frac{1}{4\pi\varepsilon_0}\,2q\left(\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right).$

Therefore,

$\Delta V=\frac{q}{2\pi\varepsilon_0}\left(\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right).$

6. Matching with options

This matches Option A.

So the correct answer is:

$\boxed{\text{A}}$

7. Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

They agree.