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(B) Let $u$ be the velocity of the electron while entering the field and $v$ be the velocity when it leaves the plates. The electric field between the

Vedclass · Accepted Answer

$(\cos \beta / \cos \alpha)^2$

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(B) Let $u$ be the velocity of the electron while entering the field and $v$ be the velocity when it leaves the plates. The electric field between the plates is perpendicular to the plates. Therefore,the component of velocity parallel to the plates remains unchanged throughout the motion.Equating the parallel components: $u \cos \alpha = v \cos \beta$.From this,we get the ratio of velocities: $\frac{u}{v} = \frac{\cos \beta}{\cos \alpha}$.The kinetic energy $K$ is given by $K = \frac{1}{2} m v^2$. The ratio of initial kinetic energy $K_i$ to final kinetic energy $K_f$ is:$\frac{K_i}{K_f} = \frac{\frac{1}{2} m u^2}{\frac{1}{2} m v^2} = \left(\frac{u}{v}\right)^2 = \left(\frac{\cos \beta}{\cos \alpha}\right)^2$.

If an electron enters into a space between the plates of a parallel plate capacitor at an angle $\alpha$ with the plates and leaves at an angle $\beta$ to the plates,the ratio of its kinetic energy while entering the capacitor to that while leaving will be

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