JEE PYQ: Electrostatics Question 23
Question 23 - 2020 (04 Sep Shift 2)
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:
(1) $a$
(2) $\sqrt{\frac{2}{a}}$
(3) $\sqrt{\frac{3}{a}}$
(4) $\frac{1}{\sqrt{a}}$
Show Answer
Answer: (3)
Solution
$W = \int_0^x qE_0(1 - ax^2)dx = qE_0\left(x - \frac{ax^3}{3}\right) = 0$. This gives $x - \frac{ax^3}{3} = 0$, i.e., $x^2 = \frac{3}{a}$, so $x = \sqrt{\frac{3}{a}}$.
BETA
