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A charged particle (mass m and charge q) moves along X axis with velocity V0. 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 > d is :

(1) $y={qEd\over mV_0^2}x$

(2) $y={qEd\over mV_0^2}(x-d)$

(3) $y={qEd\over mV_0^2}\left({d\over2}-x\right)$

(4) $y={qEd^2\over mV_0^2}x$

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Electrostatics, electric charges and Fields

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Ans. (3) $y={qEd\over mV_0^2}\left({d\over2}-x\right)$

Sol.

x > d path is straight line

${-y-{1\over2}at^2\over x-d}={at\over V_0}$

${-y-{1\over2}at^2\over at}={x-d\over V_0}$

${-y\over at}-{1\over2}{d\over V_0}={x\over V_0}-{d\over V_0}$

${-myV_0\over qEd}={x\over V_0}-{d\over2V_0}$

$y={-qEd\over mV_0}\left({x\over V_0}-{d\over2V_0}\right)$

$y={qEd\over mV_0^2}\left({d\over2}-x\right)$

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