Question 153 of 276

A charged particle (mass $m$ and charge $q$) moves along $X$ axis with velocity ${V}_{0}$. When it passes through the origin it enters a region having uniform electric field $\vec{E}=-E\hat{j}$ which extends upto $x=d$. Equation of path of electron in the region $x\gt d$ is:

A$y=\dfrac{qEd}{{mv}_{0}^{2}}\left(x-d\right)$B$y=\dfrac{qEd}{{mv}_{0}^{2}}\left(\dfrac{d}{2}-x\right)$C$y=\dfrac{qEd}{{mv}_{0}^{2}}$D$y=\dfrac{{qEd}^{2}}{{mv}_{0}^{2}}x$

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Answer: (B) $y=\dfrac{qEd}{{mv}_{0}^{2}}\left(\dfrac{d}{2}-x\right)$

$x\gt d$path is straight line
$\dfrac{-y-\dfrac{1}{2}a{t}^{2}}{x-d}=\dfrac{at}{{V}_{0}}$
$\dfrac{-y-\dfrac{1}{2}a{t}^{2}}{at}=\dfrac{x-d}{{V}_{0}}$
$\dfrac{-y}{at}-\dfrac{1}{2}\dfrac{d}{{V}_{0}}=\dfrac{x}{{V}_{0}}-\dfrac{d}{{V}_{0}}$
$\dfrac{-my{V}_{0}}{qEd}=\dfrac{x}{{V}_{0}}-\dfrac{d}{2{V}_{0}}$
$y=\dfrac{-qEd}{m{V}_{0}}\left(\dfrac{x}{{V}_{0}}-\dfrac{d}{2{V}_{0}}\right)$
$y=\dfrac{qEd}{m{V}_{0}^{2}}\left(\dfrac{d}{2}-x\right)$.