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# A uniform electric field of strength $\;\overrightarrow{E}\;$ exists in a region . An electron enters a point $A$ with velocity $v$ as shown . It moves through the electric field and reaches at point $B$ . Velocity of particle at $B$ is $2v$ at $30^{\circ}\;$ with x - axis . Then electric field $\;\overrightarrow{E}\;$ is

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

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
$-eE_{x}a=\large\frac{3\;m_{e}\;v^2}{2}$
$E_{x}=-\large\frac{3\;m_{e}\;v^2}{2\;e\;a}$
$\bigtriangleup K.E \;along \;y - axis =\;\large\frac{1}{2}\;mv^2-\large\frac{1}{2}\;mv^2=0$
$-E_{y} e d =0$
$E_{y}=0$
$\overrightarrow{E}=-\large\frac{3m_{e}v^2}{2ea} \;\hat{i}$