Animated Solution for Physics - Electrostatics: 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

The journey of an electron through a uniform electric field is a beautiful dance between constant velocity and constant acceleration. This problem is a classic example of how a seemingly complex electrostatics question can be elegantly unraveled using basic 2D kinematics. Let's dive into the physics of this electron's flight!

Imagine the parallel plates of the capacitor oriented vertically. When the electron enters the space between these plates, it finds itself in a uniform electric field. Because the plates are vertical, this electric field is strictly horizontal.

What does this mean for the forces acting on the electron? The electric force ${\vec{F}}_e=-e\vec{E}$ will also be purely horizontal. There is absolutely no force pushing or pulling the electron in the vertical direction. This is the crucial insight that cracks the problem wide open!

Conservation of Vertical Velocity

According to Newton's First Law, an object in motion will stay in motion with a constant velocity unless acted upon by a net external force. Since the vertical force $F_y$ is zero, the vertical component of the electron's velocity must remain perfectly constant throughout its entire journey between the plates.

Let's break down the initial and final velocities into their components. At the entry point, the electron has a velocity $v_1$ at an angle $\alpha$ with the vertical plates. Using basic trigonometry, the vertical component is:

At the exit point, the electron has a new velocity $v_2$ at an angle $\beta$ with the plates. Its vertical component is:

Because the vertical velocity is conserved, we can equate these two expressions:

From Velocity to Kinetic Energy

We are asked to find the ratio of the initial and final kinetic energies, $K_1/K_2$. We know that kinetic energy is given by $K=\frac{1}{2}m{v}^2$. Since the mass of the electron $m$ is constant, the kinetic energy is directly proportional to the square of the velocity ($K\propto v^2$).

Therefore, the ratio of the kinetic energies is simply the square of the ratio of the velocities:

From our master equation, we can easily find the velocity ratio:

Substituting this back into our kinetic energy ratio, we arrive at our final, elegant result:

And there we have it! By simply recognizing that the vertical velocity is unaffected by the horizontal electric field, we bypassed complex work-energy calculations and arrived straight at the answer.