Question

An insulated sphere of radius R with uniform distributed charge Q is fixed at distance 3R from a point charge -q. The point charge is projected with velocity $V_0$ as shown and in subsequent motion charge -q remains bounded is the field of Q and minimum distance of point charge from centre of sphere is 2R then possible value of angle $\theta$ satisfy the relation.

Answer 1

$\sin^2\theta > \frac{2}{3}$

Answer 2

The initial position of the point charge $-q$ is at a distance $3R$ from the center of the sphere, which is at the origin. Let's set the initial position at $(-3R, 0)$. The initial velocity is $\vec{V}_0$ at an angle $\theta$ with the positive x-axis. So, the initial velocity components are $V_{0x} = V_0 \cos \theta$ and $V_{0y} = V_0 \sin \theta$. The initial position vector is $\vec{r}_i = -3R \hat{i}$. The initial velocity vector is $\vec{V}_0 = V_0 \cos \theta \hat{i} + V_0 \sin \theta \hat{j}$.

Conservation of angular momentum about the center of the sphere: Initial angular momentum $\vec{L}_i = \vec{r}_i \times (m \vec{V}_0) = (-3R \hat{i}) \times (m (V_0 \cos \theta \hat{i} + V_0 \sin \theta \hat{j})) = -3R m V_0 \sin \theta (\hat{i} \times \hat{j}) = -3R m V_0 \sin \theta \hat{k}$. The magnitude of the initial angular momentum is $L_i = | -3R m V_0 \sin \theta | = 3R m V_0 |\sin \theta|$.

Conservation of mechanical energy: The potential energy of a point charge $-q$ at a distance $r$ from the center of a uniformly charged sphere of charge $Q$ and radius $R$ is $U(r) = -\frac{kQq}{r}$ for $r \ge R$, where $k = \frac{1}{4\pi\epsilon_0}$. Initial potential energy $U_i = -\frac{kQq}{3R}$. Initial kinetic energy $K_i = \frac{1}{2} m V_0^2$. Total initial energy $E_i = K_i + U_i = \frac{1}{2} m V_0^2 - \frac{kQq}{3R}$.

Let the minimum distance of the point charge from the center of the sphere be $r_{min} = 2R$. At the point of minimum distance, the velocity is perpendicular to the position vector. Let the velocity at this point be $v_{min}$. Potential energy at minimum distance $U_{min} = -\frac{kQq}{r_{min}} = -\frac{kQq}{2R}$. At minimum distance, the velocity is purely tangential. The magnitude of the angular momentum is $L_{min} = r_{min} m v_{min} = 2R m v_{min}$. By conservation of angular momentum, $L_i = L_{min}$. $3R m V_0 |\sin \theta| = 2R m v_{min}$. $v_{min} = \frac{3}{2} V_0 |\sin \theta|$. Kinetic energy at minimum distance $K_{min} = \frac{1}{2} m v_{min}^2 = \frac{1}{2} m (\frac{3}{2} V_0 |\sin \theta|)^2 = \frac{9}{8} m V_0^2 \sin^2 \theta$. Total energy at minimum distance $E_{min} = K_{min} + U_{min} = \frac{9}{8} m V_0^2 \sin^2 \theta - \frac{kQq}{2R}$.

By conservation of energy, $E_i = E_{min}$. $\frac{1}{2} m V_0^2 - \frac{kQq}{3R} = \frac{9}{8} m V_0^2 \sin^2 \theta - \frac{kQq}{2R}$. $\frac{kQq}{2R} - \frac{kQq}{3R} = \frac{9}{8} m V_0^2 \sin^2 \theta - \frac{1}{2} m V_0^2$. $\frac{3kQq - 2kQq}{6R} = m V_0^2 (\frac{9}{8} \sin^2 \theta - \frac{4}{8})$. $\frac{kQq}{6R} = \frac{m V_0^2}{8} (9 \sin^2 \theta - 4)$. $9 \sin^2 \theta - 4 = \frac{8kQq}{6RmV_0^2} = \frac{4kQq}{3RmV_0^2}$. $9 \sin^2 \theta = 4 + \frac{4kQq}{3RmV_0^2}$.

The condition that the charge $-q$ remains bounded in the field of $Q$ means that the total energy $E$ is less than 0 (for attractive potential). $E = \frac{1}{2} m V_0^2 - \frac{kQq}{3R} < 0$. $\frac{1}{2} m V_0^2 < \frac{kQq}{3R}$. $m V_0^2 < \frac{2kQq}{3R}$. $\frac{1}{m V_0^2} > \frac{3R}{2kQq}$.

From $9 \sin^2 \theta = 4 + \frac{4kQq}{3RmV_0^2}$, we have $\frac{4kQq}{3RmV_0^2} = 9 \sin^2 \theta - 4$. Using the boundedness condition, $\frac{4kQq}{3RmV_0^2} = \frac{4kQq}{3R} \cdot \frac{1}{mV_0^2} > \frac{4kQq}{3R} \cdot \frac{3R}{2kQq} = 2$. So, $9 \sin^2 \theta - 4 > 2$. $9 \sin^2 \theta > 6$. $\sin^2 \theta > \frac{6}{9} = \frac{2}{3}$. $|\sin \theta| > \sqrt{\frac{2}{3}}$.

Also, we know that $0 \le \sin^2 \theta \le 1$. So, the possible values of $\sin^2 \theta$ are in the range $(\frac{2}{3}, 1]$. Therefore, $\frac{2}{3} < \sin^2 \theta \le 1$.

The question asks for the relation satisfied by the possible value of angle $\theta$. We have found that $\sin^2 \theta > \frac{2}{3}$.