The two thin coaxial rings, each of radius $$a$$ and having charges $$+Q$$ and $$-Q$$ respectively are separated by a distance of $$s$$. The potential difference between the centres of the two rings is:

We have two identically sized, thin, coaxial rings of radius $$a$$ separated along their common axis by a distance $$s$$. The upper ring (call it Ring 1) carries charge $$+Q$$, while the lower ring (Ring 2) carries charge $$-Q$$. We wish to find the potential difference between the centres of the two rings; let these centres be the points $$O_1$$ (on Ring 1) and $$O_2$$ (on Ring 2).

First, we recall the standard result for the electric potential on the axis of a uniformly charged ring. If a ring of total charge $$q$$ and radius $$a$$ is observed from a point on its axis at a distance $$x$$ from its centre, the potential there is

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

Now we evaluate the potentials at the two centres one by one, adding the contributions from both rings in each case.

Potential at point $$O_1$$ (centre of Ring 1):

• Contribution of Ring 1 itself: here the axial distance is $$x=0$$, so

$$V_{(1\to O_1)}=\frac{1}{4\pi\varepsilon_0}\frac{+Q}{\sqrt{a^2+0^2}}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{a}\;.$$

• Contribution of Ring 2: its charge is $$-Q$$ and the distance from $$O_1$$ to the centre of Ring 2 is $$x=s$$, hence

$$V_{(2\to O_1)}=\frac{1}{4\pi\varepsilon_0}\frac{-Q}{\sqrt{a^2+s^2}}\;.$$

Adding these two gives the net potential at $$O_1$$:

$$$\begin{aligned} V_1&=V_{(1\to O_1)}+V_{(2\to O_1)}\\[4pt] &=\frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{\sqrt{a^2+s^2}}\right). \end{aligned}$$$

Potential at point $$O_2$$ (centre of Ring 2):

• Contribution of Ring 2 itself: here $$x=0$$ and the charge is $$-Q$$, so

$$V_{(2\to O_2)}=\frac{1}{4\pi\varepsilon_0}\frac{-Q}{a}\;.$$

• Contribution of Ring 1: its charge is $$+Q$$ and the distance from $$O_2$$ to Ring 1 is again $$x=s$$, giving

$$V_{(1\to O_2)}=\frac{1}{4\pi\varepsilon_0}\frac{+Q}{\sqrt{a^2+s^2}}\;.$$

Thus, the net potential at $$O_2$$ is

$$$\begin{aligned} V_2&=V_{(2\to O_2)}+V_{(1\to O_2)}\\[4pt] &=\frac{1}{4\pi\varepsilon_0}\left(-\frac{Q}{a}+\frac{Q}{\sqrt{a^2+s^2}}\right). \end{aligned}$$$

We are asked for the potential difference between the two centres. Choosing the order $$V_1-V_2$$ (the magnitude will be the same even if the sign is reversed), we obtain

$$$\begin{aligned} V_1-V_2 &=\frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{\sqrt{a^2+s^2}}\right) -\frac{1}{4\pi\varepsilon_0}\left(-\frac{Q}{a}+\frac{Q}{\sqrt{a^2+s^2}}\right)\\[6pt] &=\frac{1}{4\pi\varepsilon_0}\left[\frac{Q}{a}-\frac{Q}{\sqrt{a^2+s^2}}+\frac{Q}{a}-\frac{Q}{\sqrt{a^2+s^2}}\right]\\[6pt] &=\frac{1}{4\pi\varepsilon_0}\left(\frac{2Q}{a}-\frac{2Q}{\sqrt{a^2+s^2}}\right). \end{aligned}$$$

Extracting the common factor $$2Q$$ and dividing by $$4$$ in the denominator gives

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

This expression matches Option A exactly. Hence, the correct answer is Option A.