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(D) The force acting on the particle in the electric field is $F = qE$. According to Newton's second law,the acceleration of the particle in the direc

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$\frac{qE}{2mv^{2}}$

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(D) The force acting on the particle in the electric field is $F = qE$. According to Newton's second law,the acceleration of the particle in the direction of the field is $a = \frac{F}{m} = \frac{qE}{m}$. Since the particle is thrown perpendicular to the field,the initial velocity component in the direction of the field is $0$. Along the $x$-direction (perpendicular to the field),there is no acceleration,so the velocity remains constant at $v$. Thus,the distance covered in time $t$ is $x = vt$,which gives $t = \frac{x}{v}$. Along the $y$-direction (parallel to the field),using the second equation of motion $y = u_y t + \frac{1}{2} a_y t^2$,where $u_y = 0$ and $a_y = a = \frac{qE}{m}$,we get: $y = 0 + \frac{1}{2} \left( \frac{qE}{m} \right) \left( \frac{x}{v} \right)^2$. $y = \frac{qE}{2mv^2} x^2$. Comparing this with the given equation $y = \alpha x^2$,we find $\alpha = \frac{qE}{2mv^2}$.

$A$ particle of mass $m$ and charge $q$ is thrown perpendicular to an electric field of intensity $E$ with an initial velocity $v$. The particle moves a distance $x$ perpendicular to the field and a distance $y$ along the direction of the field. If $y = \alpha x^{2}$,then $\alpha$ is given by:

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