Question

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. $\frac{\sin ^{2} \beta}{\cos ^{2} \alpha}$
• B. $\frac{\cos ^{2} \beta}{\cos ^{2} \alpha}$
• C. $\frac{\cos \beta}{\cos \alpha}$
• D. $\frac{\cos \beta}{\sin \alpha}$

Answer 1

Answer: (B)

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