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Question

(B) Let $v$ be the velocity at the highest point. The forces acting on the particle at the highest point are the gravitational force $Mg$ (downwards),

Vedclass · Accepted Answer

$2 \sqrt{g R}$

Vedclass · Answer

(B) Let $v$ be the velocity at the highest point. The forces acting on the particle at the highest point are the gravitational force $Mg$ (downwards), the electrostatic repulsive force $F_e = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{R^2}$ (upwards), and the tension $T$ (downwards). Given $Mg = \frac{Q^2}{4 \pi \epsilon_0 R^2}$, the net force towards the center is $F_{net} = Mg - F_e + T = \frac{M v^2}{R}$. At the highest point, $T = 0$, so $Mg - \frac{Q^2}{4 \pi \epsilon_0 R^2} = \frac{M v^2}{R}$. Since $Mg = \frac{Q^2}{4 \pi \epsilon_0 R^2}$, we have $0 = \frac{M v^2}{R}$, which implies $v = 0$. Now, applying the law of conservation of energy between the lowest point (velocity $v_0$) and the highest point (velocity $v = 0$): Total energy at lowest point = Total energy at highest point. $\frac{1}{2} M v_0^2 + U_{lowest} = \frac{1}{2} M v^2 + U_{highest}$. The potential energy includes both gravitational and electrostatic components. $U_{grav} = Mgh$, $U_{elec} = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{r}$. Change in energy: $\frac{1}{2} M v_0^2 = Mg(2R) + \Delta U_{elec}$. Since the distance $R$ from the center is constant, the electrostatic potential energy remains constant throughout the circular path. Thus, $\frac{1}{2} M v_0^2 = Mg(2R)$. $v_0^2 = 4gR$. $v_0 = 2 \sqrt{gR}$.

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