To solve this problem, we need to calculate the point at which the proton and the electron meet after being released from rest, under the influence of a uniform electric field $E = 650 \, \text{N/C}$. The electric field will accelerate the proton and the electron in opposite directions because of their opposite charges.

Key facts:

• The proton is positively charged, so it accelerates away from the positive plate.
• The electron is negatively charged, so it accelerates away from the negative plate.
• The force on a charged particle in an electric field is given by $F = qE$, where $q$ is the charge of the particle.
• The acceleration $a$ is related to the force $F$ by Newton’s second law: $F = ma$, where $m$ is the mass of the particle.

Let's break it down:

1. Proton's acceleration: $a_p = \frac{q_p E}{m_p}$ where $q_p = 1.602 \times 10^{-19} \, \text{C}$ (charge of proton) and $m_p = 1.673 \times 10^{-27} \, \text{kg}$.

2. Electron's acceleration: $a_e = \frac{q_e E}{m_e}$ where $q_e = 1.602 \times 10^{-19} \, \text{C}$ (charge of electron, same magnitude as proton but negative) and $m_e = 9.109 \times 10^{-31} \, \text{kg}$.

Next, since both particles are released simultaneously, and they start from opposite ends of the plate, we want to find the position where they meet after a certain time $t$.

The positions of the proton and the electron as functions of time can be described by the kinematic equation for motion under constant acceleration: $x_p = \frac{1}{2} a_p t^2$ $x_e = \frac{1}{2} a_e t^2$ where $x_p$ is the distance traveled by the proton from the positive plate, and $x_e$ is the distance traveled by the electron from the negative plate. The sum of these distances must equal the separation between the plates, which is 4.02 cm (or 0.0402 m): $x_p + x_e = 0.0402 \, \text{m}$

Substituting for $x_p$ and $x_e$: $\frac{1}{2} a_p t^2 + \frac{1}{2} a_e t^2 = 0.0402$ Simplifying: $\frac{1}{2} (a_p + a_e) t^2 = 0.0402$ Now, solve for $t^2$: $t^2 = \frac{0.0402 \times 2}{a_p + a_e}$

Once we know $t$, we can find $x_p$, the distance from the positive plate at which they meet.

Let me calculate the values for you.The proton and the electron pass each other approximately $0.22 \, \text{cm}$ from the positive plate.

Would you like more details or clarifications on the steps? Here are five related questions to consider:

1. How would the answer change if the electric field strength were doubled?
2. What happens if the mass of the proton is significantly changed?
3. How would this scenario differ if the electric field were non-uniform?
4. Can we consider the proton and electron to have relativistic speeds at higher fields?
5. How would the presence of a magnetic field affect the motion?

Tip: For problems involving uniform fields, symmetry can greatly simplify the analysis.