Question 147 of 276

An electron with kinetic energy ${K}_{1}$ enters between parallel plates of a capacitor at an angle $\alpha$ with the plates. It leaves the plates at angle $\beta$ with kinetic energy ${K}_{2}.$ Then the ratio of kinetic energies ${K}_{1}:{K}_{2}$ will be :

A$\dfrac{{\sin }^{2}\beta }{{\cos }^{2}\alpha }$B$\dfrac{\cos \beta }{\sin \alpha }$C$\dfrac{\cos \beta }{\cos \alpha }$D$\dfrac{{\cos }^{2}\beta }{{\cos }^{2}\alpha }$

Show answer and solution▾

Answer: (D) $\dfrac{{\cos }^{2}\beta }{{\cos }^{2}\alpha }$

velocity along the plate will not change.
$∴{v}_{1}\cos \alpha ={v}_{2}\cos \beta$
$\dfrac{{K}_{1}}{{K}_{2}}\Rightarrow \dfrac{{v}_{1}^{2}}{{v}_{2}^{2}}=\dfrac{{\cos }^{2}\beta }{{\cos }^{2}\alpha }$