Question

Consider a ring of radius 3 m having charge 10 mC fixed at a point. A point charge $-1\mu C$ having mass 1 mg is given a velocity v towards the ring from a point (4, 0) meter. Find the minimum velocity (in km/s) for which the particle will cross the ring only once.

Answer 1

6

Answer 2

The potential energy of the point charge $q$ at a distance $x$ from the center of a ring of radius $R$ and charge $Q$ on its axis is given by $U(x) = \frac{kQq}{\sqrt{R^2 + x^2}}$.

Given $R = 3$ m, $Q = 10$ mC $= 10 \times 10^{-3}$ C, $q = -1 \mu C = -1 \times 10^{-6}$ C, $m = 1$ mg $= 1 \times 10^{-6}$ kg. The initial position is $x_i = 4$ m. The initial velocity is $v$ towards the ring (in the negative x direction).

The initial potential energy is $U_i = U(4) = \frac{kQ q}{\sqrt{3^2 + 4^2}} = \frac{kQ q}{\sqrt{9 + 16}} = \frac{kQ q}{5}$. The initial kinetic energy is $K_i = \frac{1}{2}mv^2$. The total initial energy is $E_i = K_i + U_i = \frac{1}{2}mv^2 + \frac{kQ q}{5}$.

The particle moves along the x-axis under the influence of the electric force from the ring. The force is conservative, so the total mechanical energy is conserved.

To cross the ring only once, the particle must pass through the plane of the ring (x=0) and then move to infinity (or a point far away) without returning. Since the force is attractive towards the center for both $x>0$ and $x<0$, if the particle stops at any finite point $x_s < 0$, it will be pulled back towards the origin and cross the ring again. Therefore, to cross the ring only once, the particle must have enough energy to reach infinity after crossing the ring. The minimum velocity required for the particle to reach infinity is when its kinetic energy at infinity is zero.

The potential energy at infinity ($x \to \infty$) is $U(\infty) = 0$. Let the velocity at infinity be $v_\infty$. The kinetic energy at infinity is $K_\infty = \frac{1}{2}mv_\infty^2$. By conservation of energy, $E_i = E_\infty$. $\frac{1}{2}mv^2 + U_i = \frac{1}{2}mv_\infty^2 + U(\infty)$. $\frac{1}{2}mv^2 + \frac{kQ q}{5} = \frac{1}{2}mv_\infty^2 + 0$.

For the minimum velocity to reach infinity, we set $v_\infty = 0$. $\frac{1}{2}mv_{min}^2 + \frac{kQ q}{5} = 0$. $\frac{1}{2}mv_{min}^2 = -\frac{kQ q}{5}$.

We are given $Q = 10 \times 10^{-3}$ C and $q = -1 \times 10^{-6}$ C, so $Qq = (10 \times 10^{-3}) \times (-1 \times 10^{-6}) = -10^{-8}$ C$^2$. Also, $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9$ N m$^2$/C$^2$. $m = 1 \times 10^{-6}$ kg.

Substitute these values into the equation: $\frac{1}{2}(1 \times 10^{-6})v_{min}^2 = -(9 \times 10^9)(-10^{-8})/5$. $\frac{1}{2} \times 10^{-6} v_{min}^2 = \frac{9 \times 10^1}{5} = \frac{90}{5} = 18$. $v_{min}^2 = \frac{2 \times 18}{10^{-6}} = 36 \times 10^6$. $v_{min} = \sqrt{36 \times 10^6} = 6 \times 10^3$ m/s.

The question asks for the velocity in km/s. $v_{min} = 6 \times 10^3$ m/s $= 6$ km/s.

If the initial velocity is $v_{min} = 6$ km/s, the particle will reach infinity with zero velocity. It will cross the ring at x=0 with some positive velocity and continue moving towards negative x, slowing down but never stopping, eventually reaching infinity. Thus, it crosses the ring only once. If the initial velocity is greater than $6$ km/s, it will reach infinity with positive velocity and also cross the ring only once. If the initial velocity is less than $6$ km/s, the total energy $E_i$ will be negative. Since $U(x) \to 0$ as $|x| \to \infty$, the particle cannot reach infinity because $K(x) = E_i - U(x)$ would need to be non-negative, and $U(x)$ approaches 0 from negative values for $x>0$ and from values more negative than $U(4)$ for $x>4$, and approaches 0 from values less negative than $U(0)$ for $x<0$. Specifically, for $x<0$, $U(x)$ is negative and approaches 0 from negative values as $x \to -\infty$. If $E_i < 0$, then $K(x) = E_i - U(x) < -U(x)$. As $x \to -\infty$, $U(x) \to 0$, so $K(x) \to E_i < 0$, which is impossible. Thus, if $E_i < 0$, the particle cannot reach infinity and must turn back at some point. If it turns back at $x_s < 0$, it will cross the ring again. So, the minimum velocity to cross the ring only once is the minimum velocity required to reach infinity, which is 6 km/s.