$(a)\;\large\frac{3 m_{e}v^2}{2ea}\qquad(b)\;-\large\frac{3 m_{e}v^2}{2ea}\qquad(c)\;\large\frac{ m_{e}v^2}{2ea}\qquad(d)\;-\large\frac{ m_{e}v^2}{2ea}$

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Answer : (b) $\;-\large\frac{3 m_{e}v^2}{2ea}$

Explanation :

Work done on particle in x axis :

Final velocity of particle along x - axis = $\;\sqrt{3} v$

$\bigtriangleup K.E = \;along\;x-axis=\;\large\frac{1}{2}\;m_{e}\;(\sqrt{3}v)^2-\large\frac{1}{2}\;m_{e}\;(0)^2$

$=\large\frac{3\;m_{e}\;v^2}{2}$

Let the $\;\overrightarrow{E}\;$ be = $\;E_{x} \hat{i} + E_{y} \hat{j}$

Work done by electric field along x-axis

$=E_{x} (-e) \times(2a-a)$

$=-eE_{x}a$

Now

$-eE_{x}a=\large\frac{3\;m_{e}\;v^2}{2}$

$E_{x}=-\large\frac{3\;m_{e}\;v^2}{2\;e\;a}$

$\bigtriangleup K.E \;along \;y - axis =\;\large\frac{1}{2}\;mv^2-\large\frac{1}{2}\;mv^2=0$

$-E_{y} e d =0$

$E_{y}=0$

$\overrightarrow{E}=-\large\frac{3m_{e}v^2}{2ea} \;\hat{i}$

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