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Question

(D) The potential at a distance $z$ on the axis of a ring of radius $R$ and charge $q$ is $V = \frac{1}{4\pi \epsilon_0} \frac{q}{\sqrt{R^2 + z^2}}$.H

Vedclass · Accepted Answer

$\sqrt {\frac{2}{m}} {\left( {\frac{2}{15}\frac{{{q^2}}}{{4\pi { \in _0}a}}} \right)^{1/2}}$

Vedclass · Answer

(D) The potential at a distance $z$ on the axis of a ring of radius $R$ and charge $q$ is $V = \frac{1}{4\pi \epsilon_0} \frac{q}{\sqrt{R^2 + z^2}}$.Here,$R = 3a$. At $z = 4a$,the potential energy is $U_i = qV = \frac{q^2}{4\pi \epsilon_0 \sqrt{(3a)^2 + (4a)^2}} = \frac{q^2}{4\pi \epsilon_0 (5a)}$.At the origin $(z = 0)$,the potential energy is $U_f = qV = \frac{q^2}{4\pi \epsilon_0 (3a)}$.By the law of conservation of energy,$K_i + U_i = K_f + U_f$. For the charge to just cross the origin,$K_f = 0$.$\frac{1}{2}mv^2 + \frac{q^2}{4\pi \epsilon_0 (5a)} = 0 + \frac{q^2}{4\pi \epsilon_0 (3a)}$.$\frac{1}{2}mv^2 = \frac{q^2}{4\pi \epsilon_0 a} (\frac{1}{3} - \frac{1}{5}) = \frac{q^2}{4\pi \epsilon_0 a} (\frac{2}{15})$.$v^2 = \frac{2}{m} \cdot \frac{2}{15} \cdot \frac{q^2}{4\pi \epsilon_0 a}$.$v = \sqrt{\frac{2}{m}} \left( \frac{2}{15} \frac{q^2}{4\pi \epsilon_0 a} \right)^{1/2}$.

$A$ uniformly charged ring of radius $3a$ and total charge $q$ is placed in the $xy$-plane centered at the 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

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