Problem 87 - University Physics with Modern Physics

Question

CP 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. 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 \(h_{\max }\) 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) \( h_{\max} \approx 4.84 \times 10^{-2} \text{ m} \). (b) \( d \approx 346 \text{ m} \).

Step 1: Analyze Forces on the Proton

Recognize that the proton experiences an upward force due to the electric field. The force is given by \( F = qE \), where \( q \) is the proton's charge. This force causes an acceleration \( a = \frac{qE}{m} \), where \( m \) is the proton's mass.

Step 2: Decompose Initial Velocity

The initial velocity \( v_0 \) can be decomposed into horizontal and vertical components. The horizontal component \( v_{0x} = v_0 \cos(\alpha) \) and the vertical component \( v_{0y} = -v_0 \sin(\alpha) \) (as it is directed downward).

Step 3: Equations of Motion for Vertical Displacement

Use the kinematic equation for vertical motion: \( v_{y}^2 = v_{0y}^2 + 2a_{y} s_y \). At maximum descent, \( v_y = 0 \), so \( 0 = v_{0y}^2 - 2\frac{qE}{m} h_{\max} \) (since \( a_y \) is upward). Solve for \( h_{\max} \).

Step 4: Solve for Maximum Vertical Displacement

Rearrange the previous equation to find \( h_{\max} = \frac{v_{0y}^2}{2\frac{qE}{m}} \). Substitute \( v_{0x} = 4.00 \times 10^5 \sin(30^{\circ}) \) and solve, using \( q = 1.6 \times 10^{-19} \) C, \( m = 1.67 \times 10^{-27} \) kg, and \( E = 500 \) N/C.

Step 5: Determine Horizontal Distance When Proton Returns to Original Elevation

The total time \( t \) the proton is in motion vertically can be found by using \( t = \frac{2v_{0y}}{a} \). Then, calculate the horizontal distance \( d = v_{0x} t \).

Step 6: Calculate Numerical Values of Heights and Distances

Use the given values, \( E = 500 \text{ N/C} \), \( v_0 = 4.00 \times 10^5 \text{ m/s} \), and \( \alpha = 30.0^{\circ} \), to calculate: \( h_{\max} \approx 4.84 \times 10^{-2} \text{ m} \) and \( d \approx 346 \text{ m} \).

Step 7: Sketch the Trajectory

The trajectory will be parabolic, opening upwards, starting at an initial downward angle. It will dip down below the starting elevation and return to it after distance \( d \).

Key Concepts

Electric Field Effects

In a scenario involving a proton moving through an electric field, one of the first things to understand is how the electric field affects the motion of the proton. Every electric field exerts a force on charged particles present within it. The force that the electric field exerts depends on both the charge of the moving particle and the magnitude of the electric field itself.
For a proton, which has a positive charge, the force exerted can be calculated using the formula:

• \( F = qE \)

Here, \( F \) is the force, \( q \) is the charge of the proton, and \( E \) is the magnitude of the electric field. As a result of this force, the proton will experience an acceleration that alters its trajectory. This acceleration is given by the equation:

• \( a = \frac{qE}{m} \)

Where \( m \) is the mass of the proton. This acceleration is directed along the field lines, affecting how the particle moves vertically.

Proton Trajectory

When a proton is projected into a uniform electric field, its path or trajectory becomes a central focus of study. Initially, the proton has a velocity making an angle with the horizontal. The electric field, in this case, adds a vertical force component that shapes the motion of the proton into a parabolic trajectory.
The initial velocity of a proton can be split into two components:

• Horizontal component: \( v_{0x} = v_0 \cos(\alpha) \)
• Vertical component: \( v_{0y} = -v_0 \sin(\alpha) \)

The negative sign in the vertical component accounts for the downward initial direction. As the electric force acts upward on the proton, it counteracts the initial vertical motion and bends the path upwards, creating a parabola. Importantly, at any point in time, the net motion is a vector sum of these horizontal and vertical components.

Kinematics

Applying kinematics helps us further analyze the electric field's impact on the proton's motion. The critical aspect here is determining the maximum vertical descent (\( h_{\max} \)) and the horizontal distance travelled when the proton returns to its original height (\( d \)).
For vertical motion, the kinematic equation:

• \( v_y^2 = v_{0y}^2 + 2a_y s_y \)

can be used, where \( v_y \) is final vertical velocity, and \( s_y \) is the vertical displacement. At the maximum descent, \( v_y = 0 \), therefore:

• \( 0 = v_{0y}^2 - 2\frac{qE}{m} h_{\max} \)

Allowing us to solve for \( h_{\max} \). For the horizontal motion, since the field only impacts vertical velocity, \( v_{0x} \) remains constant. The total time \( t \) of vertical motion helps determine \( d \):

• \( t = \frac{2v_{0y}}{a_y} \)
• \( d = v_{0x} t \)

Uniform Electric Field

A uniform electric field is characterized by having the same strength and direction at every point within the field. This consistent field means that any charge, like a proton, that enters the field will experience a constant force in the same direction.
The implications of a uniform electric field are vital in predicting and calculating motions such as the proton's trajectory. When a proton enters this type of field, the force on it does not vary, leading to predictable motion paths. This steadiness allows the use of specific equations to find vertical displacement and horizontal travels. The trajectory of a proton in a uniform field is typically parabolic, characterized by a consistent yet opposing acceleration against its initial velocity, ensuring a reliable analysis of its path.