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Step-by-Step Solution
1. Identify the Physical Situation
A uniformly charged ring of radius $3a$ carries a total charge $q$. Another point charge $q$ of mass $m$ is initially moving along the positive $z$-axis with speed $u$ at $z = 4a$. We want to find the minimum speed $u$ required for this charge to reach (cross) the origin, overcoming the electrostatic repulsion from the ring.
2. Use Conservation of Energy
The law of conservation of energy states:
$$
U_{\text{i}} + K_{\text{i}} \;=\; U_{\text{f}} + K_{\text{f}},
$$
where $U$ is the electrostatic potential energy and $K$ is the kinetic energy.
3. Write Expressions for Potential Energy
Let $k = \frac{1}{4\pi\varepsilon_0}$. The potential energy between charges $q$ and $q$ separated by a distance $r$ is
$$
U = \frac{k q^2}{r}.
$$
Below are the relevant distances in our problem:
Initial position of the point charge at $z = 4a$
The distance from this point to any point on the ring’s circumference is not needed directly. Instead, the potential at $z = 4a$ due to the entire uniformly charged ring is considered. However, from the solution approach provided:
The effective distance for potential energy is
$$
\sqrt{(3a)^2 + (4a)^2} \;=\; \sqrt{9a^2 + 16a^2} \;=\; 5a.
$$
Hence the initial electrostatic potential energy is
$$
U_{\text{i}} \;=\; \frac{k q^2}{5a}.
$$
Final position of the point charge at the origin $z = 0$
When the charge is at the origin, it is at a distance $3a$ from every point on the ring (because the ring has radius $3a$ and is centered at the origin in the $xy$-plane). The electrostatic potential energy at the origin is
$$
U_{\text{f}} \;=\; \frac{k q^2}{3a}.
$$
4. Write Expressions for Kinetic Energy
The kinetic energy at any instant is given by
$$
K = \tfrac12 m v^2.
$$
Here, $v$ is the speed of the charge at that instant.
Initial Kinetic Energy: $K_{\text{i}} = \tfrac12 m u^2$, where $u$ is the speed at $z = 4a$.
Final Kinetic Energy: $K_{\text{f}} = \tfrac12 m v_{f}^2$. But for the minimum speed required just to reach the origin, we consider $v_{f}$ (the speed at the origin) to be zero if it “just crosses” that point. Thus, $K_{\text{f}} = 0$ in the minimal case.
5. Apply Conservation of Energy
Putting all terms into $U_{\text{i}} + K_{\text{i}} = U_{\text{f}} + K_{\text{f}}$, we get:
$$
\frac{k q^2}{5a} + \tfrac12 m u^2 \;=\; \frac{k q^2}{3a} + 0.
$$
Therefore,
$$
\tfrac12 m u^2 \;=\; \frac{k q^2}{3a} \;-\; \frac{k q^2}{5a}
\;=\; k q^2 \left(\frac{1}{3a} \;-\; \frac{1}{5a}\right)
\;=\; k q^2 \left(\frac{5 - 3}{15a}\right)
\;=\; \frac{2 k q^2}{15a}.
$$
6. Solve for the Speed $u$
From
$$
\tfrac12 m u^2 = \frac{2 k q^2}{15 a},
$$
we have
$$
u^2 = \frac{4 k q^2}{15 m a}.
$$
Thus,
$$
u \;=\; \sqrt{\frac{4 k q^2}{15 m a}}
\;=\; \sqrt{\frac{2}{m}} \left(\frac{2q^2}{15 \cdot 4\pi\varepsilon_0 \, a}\right)^{\!\tfrac12}.
$$
We recognize that $k = \tfrac{1}{4\pi \varepsilon_0}$, so the final expression is typically written as:
$$
u \;=\; \sqrt{\frac{2}{m}} \,\Biggl(\frac{2}{15} \,\frac{q^2}{4\pi \varepsilon_0\,a}\Biggr)^{\!\tfrac12}.
$$
7. Final Answer
The minimum speed required for the point charge $q$ to cross the origin is:
$$
\boxed{
u = \sqrt{\frac{2}{m}}\;\Biggl(\frac{2}{15}\;\frac{q^2}{4\pi \varepsilon_0\,a}\Biggr)^{\tfrac12}.
}
$$
