Problem 87 - University Physics with Modern Physics

Question

A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\) . The initial velocity of the proton has a magnitude \(v_{0}\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{\text { max }}\) that the proton descends vertically below its initial elevation. You can ignore gravitational forces. ( \(b\) ) After what horizontal distance \(d\) does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{\text { max }}\) and \(d\) if \(E=500 \mathrm{N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{m} / \mathrm{s},\) and \(\alpha=30.0^{\circ} .\)

Step-by-Step Solution

Answer

(a) Determine \( h_{\text{max}} \) using \( \frac{-v_{0y}^2}{2a} \); (b) Calculate \( d \) using \( v_{0x} \times t_{\text{total}} \); (c) See sketch. Use given values for exact calculations.

Step 1: Identify Forces and Analyze Motion

The only force acting on the proton in the vertical direction is the electric force. Since gravitational forces are ignored, use Newton's second law, where the electric force is given by \( F = qE \), where \( q \) is the charge of the proton and \( E \) is the electric field. The acceleration \( a \) of the proton can be described by \( a = \frac{qE}{m} \), where \( m \) is the mass of the proton.

Step 2: Find Initial Velocity Components

Decompose the initial velocity \( v_0 \) into horizontal and vertical components using the angle \( \alpha = 30.0^\circ \). The horizontal component is \( v_{0x} = v_0 \cos \alpha \) and the vertical component is \( v_{0y} = -v_0 \sin \alpha \), since \( \alpha \) is below the horizontal.

Step 3: Determine Time to Maximum Descent

At the maximum descent, the vertical velocity component is zero. Use the equation \( v_{y} = v_{0y} + at \), where the calculated vertical velocity is zero. Solve for the time \( t = \frac{-v_{0y}}{a} \).

Step 4: Calculate Maximum Descent Distance \( h_{\text{max}} \)

Use the equation for vertical displacement: \( 0 = v_{0y}^2 + 2a h_{\text{max}} \) and solve for \( h_{\text{max}} \) as \( h_{\text{max}} = \frac{-v_{0y}^2}{2a} \). Insert \( v_{0y} \) and \( a \) from previous steps.

Step 5: Calculate Time to Return to Original Elevation

Double the time to maximum descent to obtain the total time the proton takes to return to its original elevation: \( t_{\text{total}} = 2t_{\text{max}} \).

Step 6: Compute Horizontal Distance \( d \) to Return to Original Elevation

The horizontal velocity is constant. Compute the horizontal distance \( d \) using \( d = v_{0x} \times t_{\text{total}} \).

Step 7: Plug in Numerical Values

Use the given values: \( E = 500 \; \text{N/C} \),\( v_0 = 4.00 \times 10^5 \; \text{m/s} \), \( \alpha = 30.0^\circ \), \( q = 1.6 \times 10^{-19} \; \text{C} \), and \( m = 1.67 \times 10^{-27} \; \text{kg} \). Calculate \( v_{0x} = v_0 \cos(30^\circ) \) and \( v_{0y} = -v_0 \sin(30^\circ) \), then substitute into \( h_{\text{max}} \) and \( d \).

Step 8: Sketch the Trajectory

The trajectory is parabolic. Initially, the proton descends until its velocity is zero, then ascends back to the original elevation. Sketch a downward-facing parabola beginning at the origin.

Key Concepts

Electric Fields

Electric fields are essential in understanding how charged particles like protons behave. An electric field exerts a force on charged particles, influencing their motion and trajectory.
In this exercise, the electric field is uniform and points upwards. The strength of this field is denoted by \( E \), measured in newtons per coulomb (N/C). A proton, having a positive charge \( q = 1.6 \times 10^{-19} \text{ C} \), will thus experience an upward force \( F = qE \).
This force acts on the proton, opposing its initial downward velocity, and is crucial in determining the proton's motion, as observed in the following sections.

Kinematics

Kinematics allows us to describe the motion of particles without considering the forces that cause this motion. It's essential for determining the path of the proton in this electric field.
The initial velocity \( v_0 \) of the proton can be split into two components:

• The horizontal component \( v_{0x} = v_0 \cos(\alpha) \)
• The vertical component \( v_{0y} = -v_0 \sin(\alpha) \)

Both components are pivotal for calculating the trajectory and maximum descent of the proton. The horizontal component remains constant as no forces act in this direction, while the vertical component changes due to the electric force acting on the proton.

Newton's Second Law

Newton’s Second Law is central to predicting how forces affect the motion of a particle. It states \( F = ma \), meaning force equals mass times acceleration.
For our proton in the electric field, the net force is provided solely by the electric field, and thus:
\[ a = \frac{qE}{m} \]
where

• \( q \) is the charge of the proton
• \( E \) is the electric field strength
• \( m \) is the mass of the proton

This acceleration acts in the positive vertical direction and is crucial to computing both the time it takes for the proton to reach maximum descent and the subsequent return to its original elevation.

Trajectory Analysis

Analyzing the trajectory of the proton involves understanding its motion in both horizontal and vertical directions. The trajectory describes how the proton's position changes over time.
Initially, the proton is projected downward, and its motion is influenced by the vertical electric field. At maximum descent, the vertical velocity component becomes zero as the proton stops descending and starts moving upwards.
Determining the path involves calculating:

• Maximum descent using the kinematic equation for vertical motion:
  \[ h_{\text{max}} = \frac{-v_{0y}^2}{2a} \]
• The full horizontal distance \( d \) it travels by doubling the time to maximum descent and using
  \[ d = v_{0x} \times t_{\text{total}} \]

This results in a parabolic path, determined by continuously changing vertical motion and constant horizontal motion.