Q.Figure shows a narrow beam of electrons entering with a velocity of $3\times10^{7}\ \text{m/s}$, symmetrically through the space between two parallel horizontal plates $P_1P_1'$ and $P_2P_2'$ kept $2\ \text{cm}$ apart. If each plate is $3\ \text{cm}$ long, calculate the potential difference $V$ applied between the plates so that the beam just strikes the end $P_2'$.

Concept: Projectile motion of a charged particle in a uniform electric field — the electron moves like a horizontal projectile under constant vertical acceleration.

Reasoning:

1. Time of flight: The electron travels the length of the plates $L = 3\ \text{cm} = 0.03\ \text{m}$ at constant horizontal speed $v = 3\times10^7\ \text{m/s}$.

2. Required vertical deflection: The beam enters midway (1 cm from each plate) and must just graze the lower plate end $P_2'$. So the downward displacement is half the gap:

3. Vertical acceleration: Using $y = \frac12 a t^2$, $a = \frac{2y}{t^2} = \frac{2 \times 0.01}{(1\times10^{-9})^2} = 2\times10^{16}\ \text{m/s}^2$ …

The key here is to recognise that the electron beam, once it enters the region between the plates, experiences a constant vertical force due to the uniform electric field. There is no horizontal force, so the horizontal velocity stays constant. This is exactly the same physics as a projectile thrown horizontally under gravity — except here the "gravity" is electric, and the acceleration is $a = \frac{eE}{m} = \frac{eV}{md}$, where $d$ is the plate separation.

The problem says the beam "just strikes the end $P_2'$". That means the electron enters exactly midway between the plates (1 cm from each plate) and, after travelling the full length of the plates, just grazes the lower plate at its far end. So the vertical deflection needed is exactly half the gap: $y = 1\ \text{cm} = 0.01\ \text{m}$.

Let's work through the numbers step by step.

1. Find the time the electron spends between the plates. The horizontal velocity is constant at $v_x = 3 \times 10^7\ \text{m/s}$, and the plate length is $L = 3\ \text{cm} = 0.03\ \text{m}$. Time of flight:

2. Relate vertical deflection to acceleration. Starting from rest vertically, the displacement in time $t$ under constant acceleration $a$ is:

Here $y = 0.01\ \text{m}$, so:

3. Connect acceleration to the electric field and potential difference. The electric field between parallel plates is $E = V/d$, where $d = 2\ \text{cm} = 0.02\ \text{m}$. The force on an electron is $F = eE$, so acceleration is:

Rearranging for $V$:

4. Substitute the numbers. We have $a = 2 \times 10^{16}\ \text{m/s}^2$, $d = 0.02\ \text{m}$, and the charge-to-mass ratio for an electron is $e/m = 1.76 \times 10^{11}\ \text{C/kg}$, so $m/e = 1/(1.76 \times 10^{11}) \approx 5.68 \times 10^{-12}\ \text{kg/C}$. …