Proton Motion Between Two Charged Parallel Metal Plates

To determine the potential difference $X$ between the two vertical parallel metal plates, we analyze the forces acting on the proton immediately after it is released from rest.

1. Forces Acting on the Proton

When the proton is released, it experiences two mutually perpendicular forces:

• Vertical Force (Gravity): $F_y = mg \quad \text{(downwards)}$
• Horizontal Force (Electrostatic): $F_x = e E = e \left(\frac{X}{d}\right) \quad \text{(horizontally towards the negative plate)}$

where:

• $m \approx 1.67 \times 10^{-27} \text{ kg}$ is the mass of the proton,
• $g \approx 9.8 \text{ m/s}^2$ is the acceleration due to gravity,
• $e \approx 1.6 \times 10^{-19} \text{ C}$ is the elementary charge,
• $d = 1 \text{ cm} = 10^{-2} \text{ m}$ is the separation between the plates,
• $X$ is the potential difference between the plates.

---

2. Condition for Motion at $45^\circ$ to the Vertical

Since the proton moves at an angle of $\theta = 45^\circ$ relative to the vertical immediately following its release, the ratio of the horizontal force to the vertical force must satisfy:

$\tan 45^\circ = \frac{F_x}{F_y}$

Since $\tan 45^\circ = 1$, we have:

$F_x = F_y$

Substituting the expressions for $F_x$ and $F_y$:

$e \left(\frac{X}{d}\right) = mg$

Solving for the potential difference $X$:

$X = \frac{m g d}{e}$

---

3. Calculation

Substitute the standard numerical values into the equation:

$X = \frac{(1.67 \times 10^{-27} \text{ kg}) \times (9.8 \text{ m/s}^2) \times (10^{-2} \text{ m})}{1.6 \times 10^{-19} \text{ C}}$

$X = \frac{1.6366 \times 10^{-28}}{1.6 \times 10^{-19}} \text{ V} \approx 1.02 \times 10^{-9} \text{ V} \approx 1 \times 10^{-9} \text{ V}$

Thus, the value of $X$ is approximately $1 \times 10^{-9} \text{ V}$.

Correct Option: (C)