Question

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 it, will be :

• A. ${\left(\frac{cos\alpha }{cos\beta }\right)}^2$
• B. ${\left(\frac{cos\beta }{cos\alpha }\right)}^2$
• C. ${\left(\frac{sin\alpha }{sin\beta }\right)}^2$
• D. ${\left(\frac{sin\beta }{sin\alpha }\right)}^2$

Answer 1

Answer: (B)

${\left(\frac{cos\beta }{cos\alpha }\right)}^2$

The velocity of electron in x direction must not change as force is being applied in y direction only.
$\therefore V_{1}cos\alpha =V_{2}cosB$ (x direction velocities must be same).
$\Rightarrow V_2=\frac{V_{1}cos\alpha }{cosB}$
initial kinetic energy $\left(E_1\right)=\frac{1}{2}{m}_e{V}_1^2$
final kinetic energy $\left(E_2\right)=\frac{1}{2}{m}_e(\frac{V_{1}cos\alpha }{cosB}{)}^{2.}$
$\therefore \frac{E_2}{E_1}=(\frac{cosB}{cos\alpha }{)}^{2.}$