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

Hint:

In projectile motion under a constant force (like gravity or a uniform electric field), the component of velocity perpendicular to the force remains constant. This is a key principle for solving such problems.

Answer 1

The Correct Option is D

Let the initial velocity of the electron be $v_1$ and the final velocity be $v_2$.
The electric field inside the capacitor is perpendicular to the plates. Let's assume the plates are horizontal.
The force on the electron is vertical, so there is no horizontal acceleration. The horizontal component of the velocity remains constant.
The initial horizontal component of velocity is $v_{1x} = v_1 \cos\alpha$.
The final horizontal component of velocity is $v_{2x} = v_2 \cos\beta$.
Since the horizontal velocity is constant, we have $v_{1x} = v_{2x}$.
$v_1 \cos\alpha = v_2 \cos\beta$.
We can find the ratio of the final velocity to the initial velocity: $\frac{v_2}{v_1} = \frac{\cos\alpha}{\cos\beta}$.
The kinetic energies are $K_1 = \frac{1}{2}mv_1^2$ and $K_2 = \frac{1}{2}mv_2^2$.
The ratio of the kinetic energies is $\frac{K_2}{K_1} = \frac{\frac{1}{2}mv_2^2}{\frac{1}{2}mv_1^2} = \left(\frac{v_2}{v_1}\right)^2$.
Substituting the velocity ratio we found:
$\frac{K_2}{K_1} = \left(\frac{\cos\alpha}{\cos\beta}\right)^2 = \frac{\cos^2\alpha}{\cos^2\beta}$.
The question asks for the ratio $K_1 : K_2$, which is the inverse:
$\frac{K_1}{K_2} = \frac{\cos^2\beta}{\cos^2\alpha}$.