JEE PYQ: Electrostatics Question 45
Question 45 - 2019 (10 Apr Shift 1)
A uniformly charged ring of radius $3a$ and total charge $q$ is placed in $xy$-plane centred at 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:
(1) $\sqrt{\frac{2}{m}\left(\frac{4}{15}\frac{q^2}{4\pi\varepsilon_0 a}\right)^{1/2}}$
(2) $\sqrt{\frac{2}{m}\left(\frac{1}{5}\frac{q^2}{4\pi\varepsilon_0 a}\right)^{1/2}}$
(3) $\sqrt{\frac{2}{m}\left(\frac{2}{15}\frac{q^2}{4\pi\varepsilon_0 a}\right)^{1/2}}$
(4) $\sqrt{\frac{2}{m}\left(\frac{1}{15}\frac{q^2}{4\pi\varepsilon_0 a}\right)^{1/2}}$
Show Answer
Answer: (3)
Solution
At $z = 4a$: distance from ring $= \sqrt{9a^2 + 16a^2} = 5a$. $V_i = \frac{kq}{5a}$. At origin: $V_f = \frac{kq}{3a}$. By energy conservation: $\frac{1}{2}mv^2 = q(V_f - V_i) = kq^2\left(\frac{1}{3a} - \frac{1}{5a}\right) = \frac{2kq^2}{15a}$. $v = \sqrt{\frac{2}{m} \cdot \frac{2q^2}{15 \cdot 4\pi\varepsilon_0 a}}$.
BETA
