Problem 77 - 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 \(h_{max}\) that the proton descends vertically below its initial elevation. 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 \(h_{max}\) and \(d\) if \(E =\) 500 N\(/\)C, \(v_0 =\) 4.00 \(\times 10^5\) m\(/\)s, and \(\alpha =\) 30.0\(^\circ\).

Step-by-Step Solution

Answer

The maximum descent is \(h_{max}\) downwards, occurring before returning to the initial position after horizontal distance \(d\). Compute values using kinetic equations and known constants.

Step 1: Understanding the Problem

We are analyzing the motion of a proton in a uniform electric field. The electric field affects only vertical motion. We need to determine the maximum vertical distance downward (part (a)), the horizontal distance to return to initial elevation (part (b)), and provide a sketch of the trajectory (part (c)).

Step 2: Vertical and Horizontal Components of Initial Velocity

The proton is launched with an initial velocity of magnitude \(v_0\) at an angle \(\alpha\) below the horizontal. We decompose this velocity into vertical and horizontal components using trigonometric functions:- Vertical component: \(v_{0y} = -v_0 \sin(\alpha)\)- Horizontal component: \(v_{0x} = v_0 \cos(\alpha)\).

Step 3: Vertical Motion in an Electric Field

The uniform electric field exerts a force on the proton, causing it to undergo constant acceleration \(a = \frac{qE}{m}\) in the vertical direction. The proton's charge \(q\) is the elementary charge, and \(m\) is its mass. Using equations of motion, the maximum vertical distance \(h_{max}\) from its original position can be found using the formula:\[v_{y}^2 = v_{0y}^2 + 2ah\]where final vertical velocity \(v_y = 0\) at maximum descent. Solve for \(h\).

Step 4: Horizontal Motion and Time of Flight

The proton moves horizontally with constant velocity \(v_{0x}\). The time \(t\) taken to reach maximum descent and return to its original elevation can be found by solving the equation for vertical displacement;\[0 = v_{0y}t + \frac{1}{2}at^2\].The horizontal distance \(d = v_{0x}t_{total}\), where \(t_{total}\) is the full time of flight (i.e., twice the time to reach maximum descent).

Step 5: Numeric Calculation for h_max and d

Given the values:- Electric field \(E = 500\, \text{N/C}\)- Initial speed \(v_0 = \text{4.00} \times 10^{5} \, \text{m/s}\)- Angle \(\alpha = 30^\circ\)We compute:1. Decompose \(v_0 = \, \text{4.00} \times 10^{5} \, \text{m/s}\) into components: - \(v_{0y} = -v_0 \sin(30^\circ) = -2.00 \times 10^5\, \text{m/s}\) - \(v_{0x} = v_0 \cos(30^\circ) = 3.46 \times 10^5\, \text{m/s}\)2. Calculate acceleration \(a = \frac{qE}{m}\) with \(q = 1.6 \times 10^{-19}\, \text{C}\) and \(m = 1.67 \times 10^{-27}\, \text{kg}\).3. Solve for \(h_{max}\) using the derived formula.4. Solve for time and calculate \(d\).

Step 6: Sketching the Trajectory

In a sketch of the trajectory, the proton starts at its initial elevation with initial velocity at angle \(\alpha\). It travels downward due to the electric field, reaching its maximum vertical descent before returning to its original elevation, tracing a parabolic path when observed in projection.

Key Concepts

Proton Trajectory

In this exercise, we delve into the trajectory of a proton moving in an electric field. The proton is projected into a uniform electric field, which influences its path by exerting force only in the vertical direction. Initially, the proton is propelled with a velocity that forms an angle below the horizontal, setting it on a unique path. This trajectory represents a parabola when viewed in a plane, showcasing the influence of the electric field on the proton's movement.

• Inspired by projectile motion, the proton's trajectory showcases the simultaneous action of its horizontal velocity and the vertical force from the electric field.
• Ignoring gravitational forces allows us to focus solely on the electric field's impact, simplifying the analysis.

Understanding this path is crucial, as it reveals the balance between initial momentum and external forces, a key concept in electric field motion.

Uniform Electric Field

Uniform electric fields are regions where electric forces remain constant in magnitude and direction. In this scenario, the field points vertically upward, impacting the proton's vertical motion. This uniformity simplifies calculations, as the force exerted on the proton is constant, leading to a consistent vertical acceleration.

• The electric field's magnitude is denoted by \(E\), serving as a central value in our calculations.
• In this field, a positively charged proton experiences a force \(F = qE\), where \(q\) is the charge of the proton.

This force acts vertically, altering the proton's vertical velocity, which is crucial in determining its downward maximum mid-journey and the time it takes to return to its original height. A uniform field thereby provides a straightforward framework for understanding electric force impacts on motion.

Motion Equations

Motion equations play a pivotal role in analyzing the proton's journey through an electric field. Decomposing the motion into vertical and horizontal components allows for a clear understanding. Initially, the proton has a velocity \(v_0\), divided into horizontal \(v_{0x} = v_0 \cos(\alpha)\) and vertical \(v_{0y} = -v_0 \sin(\alpha)\) components.

• Vertical motion undergoes constant acceleration \(a = \frac{qE}{m}\), derived from the proton's charge and mass.
• The maximum vertical displacement \(h_{max}\) is calculated using: \[ v_{y}^2 = v_{0y}^2 + 2ah \] where the final vertical velocity \(v_{y} = 0\) at peak descent.

These equations demonstrate the direct relationship between the proton's charged properties and its motion through a uniform electric field, highlighting how theoretical physics underpins practical problem-solving.

Projectile Motion Analysis

Analyzing the proton's motion as projectile motion offers insights into its path and behavior in the electric field. Though typically associated with gravity, here it’s analogous due to the electric field's uniform influence. This motion analysis involves dissecting the time of flight, horizontal distance covered, and peak height descended in a given electric field.

• The proton's horizontal distance \(d\) can be obtained by calculating the total time of its journey (twice the descent time) and using horizontal velocity \(v_{0x}\).
• The time \(t_{total}\) involves the vertical motion equation, solving \[ 0 = v_{0y}t + \frac{1}{2}at^2 \] resulting in the requisite projectile calculations.

By viewing the proton's travel as projectile motion, students can apply familiar concepts to electric field-induced motion, reinforcing the versatility and interconnectedness of physics principles.