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. $\raisebox{1ex}{${\sin }^2\beta $}\!\left/ \!\raisebox{-1ex}{${\cos }^2\alpha $}\right.$
• B. $\raisebox{1ex}{${\cos }^2\beta $}\!\left/ \!\raisebox{-1ex}{${\cos }^2\alpha $}\right.$
• C. $\raisebox{1ex}{$\cos \beta $}\!\left/ \!\raisebox{-1ex}{$\sin \alpha $}\right.$
• D. $\raisebox{1ex}{$\cos \beta $}\!\left/ \!\raisebox{-1ex}{$\cos \alpha $}\right.$

Answer 1

The correct option is B

$\raisebox{1ex}{${\cos }^2\beta $}\!\left/ \!\raisebox{-1ex}{${\cos }^2\alpha $}\right.$

Step 1. Electron doesn't experience a force in x-direction thus momentum in x-direction is conserved:

$<span\; style="font-family:"Cambria\; Math","serif";mso-bidi-font-family:"Cambria\; Math";">\because </span>v_1\cos \alpha =v_2\cos \beta$ [electric field inside a parallel plate capacitor is perpendicular to the plates, hence, there will be no change in parallel component of velocity]

$\Rightarrow \frac{v_1}{v_2}=\frac{\cos \beta }{\cos \alpha }$

Step 2. Then the ratio of kinetic energies:

$$\begin{gathered} \frac{k_1}{k_2}=\raisebox{1ex}{$\frac{1}{2}m{v_1}^2$}\!\left/ \!\raisebox{-1ex}{$\frac{1}{2}m{v_2}^2$}\right. \\ \Rightarrow \frac{k_1}{k_2}={\left(\frac{v_1}{v_2}\right)}^2 \\ \Rightarrow \frac{k_1}{k_2}={\left(\frac{\cos \beta }{\cos \alpha }\right)}^2 \end{gathered}$$

$\therefore \frac{k_1}{k_2}=\frac{{\cos }^2\beta }{{\cos }^2\alpha }$

Hence, the correct option is (B).