Q 11-05-174JEE MainJEE Main 2020 (4 Sep, Shift 2)Medium
A particle of charge $q$ and mass $m$ is subjected to an electric field $E = E_0(1 - ax^{2})$ in the $x$-direction, where $a$ and $E_0$ are constants. Initially the particle was at rest at $x = 0$. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is:
Answer: (C) $\sqrt{\dfrac3a}$
By the work-energy theorem, the kinetic energy at $x$ equals the work done by the field:
$$K = \int_0^{x} qE_0(1 - ax^{2})\,dx = qE_0\left(x - \frac{ax^{3}}{3}\right)$$
$K = 0$ for $x \neq 0$ when $x^{2} = \dfrac3a$, i.e. $x = \sqrt{\dfrac3a}$.
Solution by Sreeraj P, M.Sc Physics