An electron with kinetic energy K_{1} enters | vedclass

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Question

(B) When an electron enters the electric field between parallel plates,the electric force acts only in the direction perpendicular to the plates (vert

Vedclass · Accepted Answer

$\frac{\cos^{2} \beta}{\cos^{2} \alpha}$

Vedclass · Answer

(B) When an electron enters the electric field between parallel plates,the electric force acts only in the direction perpendicular to the plates (vertical direction).Therefore,there is no force acting in the direction parallel to the plates (horizontal direction).As a result,the component of velocity parallel to the plates remains constant throughout the motion.Let $v_{1}$ be the initial velocity and $v_{2}$ be the final velocity.The horizontal component of velocity at entry is $v_{1} \cos \alpha$.The horizontal component of velocity at exit is $v_{2} \cos \beta$.Since the horizontal component is constant,we have: $v_{1} \cos \alpha = v_{2} \cos \beta$.This implies: $\frac{v_{1}}{v_{2}} = \frac{\cos \beta}{\cos \alpha}$.The kinetic energy $K$ is given by $K = \frac{1}{2}mv^{2}$.Therefore,the ratio of kinetic energies is: $\frac{K_{1}}{K_{2}} = \frac{\frac{1}{2}mv_{1}^{2}}{\frac{1}{2}mv_{2}^{2}} = \left(\frac{v_{1}}{v_{2}}\right)^{2}$.Substituting the velocity ratio: $\frac{K_{1}}{K_{2}} = \left(\frac{\cos \beta}{\cos \alpha}\right)^{2} = \frac{\cos^{2} \beta}{\cos^{2} \alpha}$.

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