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 $x > d$ is

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$y = \frac{q E d}{m V_{0}^{2}} (\frac{d}{2} - x)$

$y = \frac{q E d^{2}}{m V_{0}^{2}} x$

$y = \frac{q E d}{m V_{0}^{2}} (x - d)$

$y = \frac{q E d}{m V_{0}^{2}} x$

answer is C.

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Detailed Solution

$\begin{array}{l} N o a c c e l e r a t i o n i n x d i r e c t i o n . \\ V_{x} = V_{0} \\ T i m e t a k e n t o t r a v e l d i s \tan c e d = \frac{d}{V_{0}} \\ A c c e l e r a t i o n i n n e g a t i v e y d i r e c t i o n = \frac{q E}{m} \\ M a g n i t u d e o f f i n a l v e l o c i t y i n y d i r e c t i o n w h e n i t e x i t t h e r e g i o n = V_{y} = (\frac{q E}{m}) . \frac{d}{V_{0}} \\ when x = d, \\ y = - \frac{1}{2} (\frac{q E}{m}) {(\frac{d}{v_{0}})}^{2} = y_{0} (a s s u m e) \\ Now for (x > d) no acceleration in y direction \\ y - y_{0} = - V_{y} t \\ x - d = V_{x} t = V_{0} t \\ eliminating t, \\ \Rightarrow - (\frac{y - y_{0}}{V_{y}}) = \frac{x - d}{V_{0}} \\ \Rightarrow y = \frac{d - x}{V_{0}} V_{y} + y_{0} = \frac{d - x}{V_{0}^{2}} . \frac{q E d}{m} - \frac{q E d^{2}}{2 m V_{0}^{2}} = \frac{q E d}{m V_{0}^{2}} (\frac{d}{2} - x) \end{array}$