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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
${\left( {\cos \,\alpha /\cos \,\beta } \right)^2}$
${\left( {\cos \,\beta /\cos \,\alpha } \right)^2}$
${\left( {\sin \,\alpha /\sin \,\beta } \right)^2}$
${\left( {\sin \,\beta /\sin \,\alpha } \right)^2}$
Solution
Let $\mathrm{u}$ be the velocity of electron while entering the field and $\mathrm{v}$ the velocity when it leaves the plates. Component of velocity parallel to plates will remain unchanged.
$u \cos \alpha=v \cos \beta \quad \therefore \frac{u}{v}=\frac{\cos \beta}{\cos \alpha}$
The desired ratio is.
$\frac{\frac{1}{2} m u^{2}}{\frac{1}{2} m v^{2}}=\left(\frac{u}{v}\right)^{2}=\left(\frac{\cos \beta}{\cos \alpha}\right)^{2}$
