Problem 27 - University Physics with Modern Physics

Question

CP A proton is traveling horizontally to the right at \(4.50 \times 10^{6} \mathrm{m} / \mathrm{s} .\) (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.20 \(\mathrm{cm} .\) (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

Step-by-Step Solution

Answer

(a) 3.30 kN/C reverse direction. (b) 1.42×10⁻⁸ s. (c) 17.97 N/C opposite direction.

Step 1: Understand the Problem

We need to stop a proton moving at \(4.50 \times 10^6 \; \text{m/s}\) over a distance of \(3.20\; \text{cm}\). We need to find the necessary electric field magnitude and direction, time taken to stop, and analogous information for an electron under the same conditions.

Step 2: Convert Units

The stopping distance is given in centimeters. Convert \(3.20 \; \text{cm}\) to meters: \[3.20 \; \text{cm} = 0.0320 \; \text{m}.\]

Step 3: Use Kinematic Equation for Acceleration

We use the kinematic equation \(v_f^2 = v_i^2 + 2a x\), where \(v_f = 0\), \(v_i = 4.50 \times 10^6 \; \text{m/s}\), and \(x = 0.0320 \; \text{m}\). Solve for \(a\): \[0 = (4.50 \times 10^6)^2 + 2a(0.0320).\]Simplifying gives:\[a = -\frac{(4.50 \times 10^6)^2}{2 \times 0.0320}.\]

Step 4: Calculate Acceleration

Plug in the numbers to calculate \(a\):\[a = -\frac{(4.50 \times 10^6)^2}{0.064} = -3.16 \times 10^{14} \; \text{m/s}^2.\]

Step 5: Relate Acceleration to Electric Field

Using \(F = ma\) and knowing \(F = qE\) for a proton, relate \(a\) to \(E\): \[qE = ma \Rightarrow E = \frac{ma}{q}.\]For a proton, \(m = 1.67 \times 10^{-27}\; \text{kg}\) and \(q = 1.60 \times 10^{-19}\; \text{C}\).

Step 6: Calculate the Electric Field Magnitude

Substitute the values into:\[E = \frac{1.67 \times 10^{-27} \times 3.16 \times 10^{14}}{1.60 \times 10^{-19}}\]This gives:\[E = 3.30 \times 10^3 \; \text{N/C}.\]The negative acceleration implies the electric field is directed opposite the proton's motion.

Step 7: Use Kinematic Equation for Time

Using \(v_f = v_i + at\), solve for \(t\): \[0 = 4.50 \times 10^6 + (-3.16 \times 10^{14})t\] \[t = \frac{4.50 \times 10^6}{3.16 \times 10^{14}} \approx 1.42 \times 10^{-8} \; \text{s}.\]

Step 8: Calculate Minimum Field for Electron

Repeat steps 5 and 6 with electron mass \(m_e = 9.11 \times 10^{-31} \; \text{kg}\) and charge of \(-1.60 \times 10^{-19} \; \text{C}\): \[E_{electron} = \frac{9.11 \times 10^{-31} \times 3.16 \times 10^{14}}{1.60 \times 10^{-19}} = 1.797 \times 10^1 \; \text{N/C}.\]Direction: opposite to the electron's initial velocity.

Key Concepts

Kinematics

Kinematics is the branch of physics that describes the motion of objects. It deals with parameters like displacement, velocity, and acceleration. In simpler terms, it's all about understanding how things move without worrying about the forces that cause the motion.
To analyze moving objects, kinematics uses equations that relate these parameters. For example, the equation \(v_f^2 = v_i^2 + 2ax\) is a kinematic formula where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(x\) is the displacement. This equation helps to calculate any one of these variables if the others are known. In our problem, kinematics helps us understand how a proton or electron can be stopped using an electric field by determining the necessary acceleration (or deceleration in this case).
Key kinematic concepts include:

• Displacement: The change in position of an object. Here it is given as \(3.20 \; \text{cm}\) which we convert to meters.
• Velocity: The rate of change of displacement. Initial velocity \(v_i\) for the proton is provided as \(4.50 \times 10^6 \; \text{m/s}\).
• Acceleration: The rate of change of velocity with respect to time, crucial in determining how quickly the proton and electron stop.

Acceleration

Acceleration is a fundamental concept in physics that describes the change in velocity over time. When an object speeds up, slows down, or changes direction, it's accelerating. In our problem scenario, we focus on deceleration (slowing down) of a proton and an electron to rest.
The acceleration can be calculated using the kinematic formula mentioned earlier. In this exercise, since the proton is to come to a rest, we assume the final velocity \(v_f\) to be zero and solve for acceleration \(a\): \[a = -\frac{(4.50 \times 10^6)^2}{2 \times 0.0320}.\]The negative sign in acceleration indicates that the proton is slowing down, i.e., deceleration.
Once acceleration is known, we use it to find the electric field needed to stop the proton, because the electric field exerts a force causing this acceleration. Understanding this relationship is key as it ties motion (kinematics) to forces (Newton's second law).
Factors that affect acceleration:

• Velocity: The initial speed and how quickly it needs to be stopped.
• Distance: Over which the stopping occurs—larger distances allow slower acceleration.

Proton and Electron Dynamics

In physics, dynamics is concerned with the forces and torques and their effect on motion. When dealing with charged particles like protons and electrons, dynamics involves understanding how these charges interact with electric fields.
Protons and electrons have opposite charges; protons carry a positive charge while electrons carry a negative one. This means they will move in opposite directions under the same electric field.
To stop a moving proton using an electric field, we need to know the force required for stopping. This force is related to acceleration by Newton's second law \(F = ma\).
The calculation for the electric field \(E\) uses:\[E = \frac{ma}{q},\]where \(m\) is the mass of the proton, \(a\) is the calculated acceleration, and \(q\) is the charge of the proton.
A similar calculation applies to the electron, but due to its much smaller mass, the electric field needed to stop it is different. The dynamics of charged particles under electric fields are crucial for applications ranging from particle accelerators to electronic devices.

Unit Conversion

Unit conversion is a critical skill in physics that ensures all quantities are expressed in consistent units. This facilitates accurate calculations and comparisons. In physics problems, units like meters, seconds, kilograms, and Coulombs are standard.
In this exercise, we converted the proton's stopping distance from centimeters to meters. This is because the kinematic equations demand consistency in units, typically in meters for distance, seconds for time, and meters per second (m/s) for velocity.
To convert centimeters to meters, remember that 1 centimeter is 0.01 meters. Thus, when converting 3.20 centimeters to meters, the calculation is:\[3.20 \; \text{cm} = 0.0320 \; \text{m}.\]Common conversions in physics include:

• Meters and kilometers: 1 km = 1000 m
• Seconds and minutes: 1 min = 60 s
• Grams and kilograms: 1 kg = 1000 g
• Coulombs: For electric charge, where 1 C is a standard unit

Ensuring units are consistent simplifies calculations and helps avoid errors in calculations, ensuring the final answers are accurate.