Question

An electron with kinetic energy $K_1$ enters between parallel plates of a capacitor at an angle ${}^{\prime }{\alpha }^{\prime }$ with the plates . It leaves the plates at an angle ${}^{\prime }{\beta }^{\prime }$ 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. $\frac{\cos \beta }{\sin \alpha }$
• D. $\frac{\cos \beta }{\cos \alpha }$

Answer 1

The correct option is B $\frac{{\cos }^2\beta }{{\cos }^2\alpha }$


$V_1$$\cos \alpha$ $=$ $V_2\cos \beta$
[Electric field inside the plates of a parallel plate capacitor is perpendicular to the plates , hence there will be no change in parallel component of velocity ]

$\frac{V_1}{V_2}$ $=$ $\frac{\cos \beta }{\cos \alpha }$

$\frac{K_1}{K_2}=\left(\frac{\frac{1}{2}m{V}_1^2}{\frac{1}{2}m{V}_2}\right)={\left(\frac{V_1}{V_2}\right)}^2={\left(\frac{\cos \beta }{\cos \alpha }\right)}^2$

$\frac{K_1}{K_2}$ $=$ $\frac{{\cos }^2\beta }{{\cos }^2\alpha }$