A uniformly charged ring of radius 3a and total charge q is placed in x-y plane centred at origin. A point charge q is moving towards the ring along the z-axis and has speed v at z = 4a. The minimum value of v such that it crosses the origin is:

We are dealing with two identical positive charges. One charge $$q$$ is smeared uniformly on a ring of radius $$R = 3a$$ that lies in the $$x\text{-}y$$ plane and is centred at the origin. The second charge, also of magnitude $$q$$ and mass $$m$$, starts on the axis of the ring (the $$z$$-axis) at the point $$z_0 = 4a$$ with speed $$v$$ directed towards the origin. Only electrostatic forces act, so the mechanical energy of the moving charge is conserved.

First we need the electric potential due to the ring on its axis. For a ring of total charge $$q$$ and radius $$R$$, the standard formula for the potential at an axial distance $$z$$ from the centre is

$$ V(z) \;=\; \frac{1}{4\pi\varepsilon_0}\; \frac{q}{\sqrt{R^{2}+z^{2}}}. $$

The electrostatic potential energy of our point charge $$q$$ at a position $$z$$ is therefore

$$ U(z) \;=\; q\,V(z) \;=\; \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{\sqrt{R^{2}+z^{2}}}. $$

We now evaluate this energy at the two positions of interest.

Initial position $$z_0 = 4a$$:

$$ U_{\text{initial}} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{\sqrt{R^{2}+z_0^{2}}} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{\sqrt{(3a)^{2} + (4a)^{2}}} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{\sqrt{9a^{2} + 16a^{2}}} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{5a}. $$

Final position (the origin) $$z = 0$$:

$$ U_{\text{final}} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{\sqrt{R^{2}+0}} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{R} = \frac{1}{4\pi\varepsilon_0}\; \frac{q^{2}}{3a}. $$

Because the ring and the charge have the same sign, the potential energy nearer the ring is higher: $$U_{\text{final}} > U_{\text{initial}}.$$ To reach the origin, the moving charge must supply exactly the additional energy difference out of its initial kinetic energy. We therefore employ the principle of conservation of mechanical energy:

$$ \text{Initial kinetic} + \text{Initial potential} \;=\; \text{Final kinetic} + \text{Final potential}. $$

For the minimum speed required to just cross the origin, the charge will momentarily come to rest there, so the final kinetic energy is zero. Writing this out gives

$$ \frac{1}{2}\,m\,v^{2} + U_{\text{initial}} \;=\; U_{\text{final}}. $$

Substituting the two potential energies derived above, we have

$$ \frac{1}{2}\,m\,v^{2} \;=\; U_{\text{final}} - U_{\text{initial}} = \frac{1}{4\pi\varepsilon_0}\,q^{2} \left( \frac{1}{3a} - \frac{1}{5a} \right). $$

Now we simplify the bracket:

$$ \frac{1}{3a} - \frac{1}{5a} = \frac{5 - 3}{15a} = \frac{2}{15a}. $$

Substituting this back,

$$ \frac{1}{2}\,m\,v^{2} = \frac{1}{4\pi\varepsilon_0}\,q^{2}\; \frac{2}{15a}. $$

Multiplying both sides by 2 and dividing by $$m$$ to isolate $$v^{2}$$ gives

$$ v^{2} = \frac{2}{m}\; \frac{1}{4\pi\varepsilon_0}\,q^{2}\; \frac{2}{15a} = \frac{4}{15m}\; \frac{q^{2}}{4\pi\varepsilon_0\,a}. $$

Taking the square root to obtain the minimum speed, we find

$$ v = \sqrt{\frac{4}{15m}\; \frac{q^{2}}{4\pi\varepsilon_0\,a}} = \sqrt{\frac{2}{m}}\; \sqrt{\frac{2}{15}\; \frac{q^{2}}{4\pi\varepsilon_0\,a}}. $$

This matches option D exactly.

Hence, the correct answer is Option D.