A particle of charge q and mass m is subjected to an electric field E = E _{0}left(1- ax ^{2}right)

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2. Electric Potential and Capacitance

hard

A particle of charge $q$ and mass $m$ is subjected to an electric field $E = E _{0}\left(1- ax ^{2}\right)$ 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

A

$\sqrt{\frac{2}{a}}$

B

$\sqrt{\frac{1}{a}}$

C

$a$

D

$\sqrt{\frac{3}{a}}$

(JEE MAIN-2020)

Solution

$E = E _{0}\left(1- ax ^{2}\right)$

$W =\int q E\, d x = q E _{0} \int_{0}^{ x _{0}}\left(1- ax ^{2}\right) dx$

$=q E_{0}\left[x_{0}-\frac{a x_{0}^{3}}{3}\right]$

For $\Delta KE =0, \quad W =0$

Hence $x _{0}=\sqrt{\frac{3}{ a }}$

Standard 12

Physics

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