/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 In a vacuum, a proton (charge \(... [FREE SOLUTION] | 91Ó°ÊÓ

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In a vacuum, a proton (charge \(=+e,\) mass \(=1.67 \times\) \(10^{-27} \mathrm{kg}\) ) is moving parallel to a uniform electric field that is directed along the \(+x\) axis (see the figure below). The proton starts with a velocity of \(+2.5 \times\) \(10^{4} \mathrm{m} / \mathrm{s}\) and accelerates in the same direction as the electric field, which has a value of \(+2.3 \times 10^{3} \mathrm{N} / \mathrm{C}\). Find the velocity of the proton when its displacement is \(+2.0 \mathrm{mm}\) from the starting point.

Short Answer

Expert verified
The final velocity of the proton is \(3.88 \times 10^4 \text{ m/s}\).

Step by step solution

01

Identify the Forces Involved

The electric force on a proton in an electric field is given by the equation \( F = qE \). The charge of the proton \( q \) is equal to \( +e \), where \( e = 1.6 \times 10^{-19} \text{ C} \). The electric field \( E \) is given as \( +2.3 \times 10^3 \text{ N/C} \). Substituting these values, the force on the proton is \( F = (1.6 \times 10^{-19})(2.3 \times 10^3) \text{ N} \).
02

Calculate the Acceleration

Use Newton's second law \( F = ma \) to find the acceleration \( a \) of the proton. The force \( F \) calculated previously is \( 3.68 \times 10^{-16} \text{ N} \). The mass \( m \) of the proton is \( 1.67 \times 10^{-27} \text{ kg} \). Thus, the acceleration is \( a = \frac{3.68 \times 10^{-16} \text{ N}}{1.67 \times 10^{-27} \text{ kg}} \approx 2.2 \times 10^{11} \text{ m/s}^2 \).
03

Apply the Kinematic Equation

Use the kinematic equation \( v^2 = u^2 + 2a s \) to find the final velocity \( v \). The initial velocity \( u \) is \( 2.5 \times 10^4 \text{ m/s} \), the acceleration \( a \) is \( 2.2 \times 10^{11} \text{ m/s}^2 \), and the displacement \( s \) is \( 2.0 \text{ mm} = 2.0 \times 10^{-3} \text{ m} \). Plug the values into the equation: \( v^2 = (2.5 \times 10^4)^2 + 2 (2.2 \times 10^{11}) (2.0 \times 10^{-3}) \).
04

Solve for the Final Velocity

Calculate \( v^2 = 6.25 \times 10^8 + 8.8 \times 10^8 = 1.505 \times 10^9 \). Solve for \( v \) by taking the square root: \( v \approx \sqrt{1.505 \times 10^9} \approx 3.88 \times 10^4 \text{ m/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proton Dynamics
Protons are fascinating subatomic particles with a positive charge, symbolized by the charge of an electron, denoted as \(+e\). Understanding how protons move in an electric field helps us see the beautiful symphony of physics in action.

When placed in a uniform electric field, a proton experiences a force that propels it in the direction of the field. This movement is dictated by its charge interacting with the field. The equation \( F = qE \), where \( F \) is the force on the proton, \( q \) is its charge, and \( E \) is the electric field strength, tells us how strong this force is.

For our exercise, with \( q = 1.6 \times 10^{-19} \text{ C} \) and \( E = 2.3 \times 10^3 \text{ N/C} \), the dynamic force calculation aids in mapping the proton’s journey through space.
Kinematics
Kinematics involves studying the motion of objects without considering the forces that cause the motion. It plays a crucial role in physics to predict how objects like protons move through space.

In our problem, the proton starts with a velocity of \(+2.5 \times 10^4 \text{ m/s}\) and travels a displacement of \(+2.0 \text{ mm} \).

Using the kinematic equation \( v^2 = u^2 + 2as \), where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the displacement, we can solve for the final velocity. This approach lets us describe the proton's swift journey precisely.
Newton's Second Law
Newton's Second Law is a fundamental principle that relates the motion of an object to the forces acting on it. It states that \( F = ma \), which means the force exerted on an object is equal to the mass \( m \) of the object times its acceleration \( a \).

For the moving proton, having calculated the force \( F = 3.68 \times 10^{-16} \text{ N} \), and knowing its mass is \( 1.67 \times 10^{-27} \text{ kg} \), we can find its acceleration by rearranging the formula to \( a = \frac{F}{m} \).

This acceleration \( 2.2 \times 10^{11} \text{ m/s}^2 \) helps us further appreciate how protons respond to electric forces in their environment, enabling us to predict their behavior with stunning accuracy.
Electric Charge
Electric charge is a basic property of matter that causes it to experience a force when placed in an electric field. It comes in two types: positive and negative. Here, we focus on the positive charge of a proton.

Each proton has a charge of \(+e\), where \( e = 1.6 \times 10^{-19} \text{ C} \). This positive charge is fundamental to interactions in electric fields. It determines how the proton will respond, moving in the direction of the field lines due to its compatible charge.

In an electric field, the charge creates a clear path for the force equation \( F = qE \), a pivotal part of understanding the proton's motion, as it ultimately influences the dynamics and kinematics of the proton within the field.

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Most popular questions from this chapter

A charge of \(-3.00 \mu \mathrm{C}\) is fixed at the center of a compass. Two additional charges are fixed on the circle of the compass, which has a radius of \(0.100 \mathrm{m} .\) The charges on the circle are \(-4.00 \mu \mathrm{C}\) at the position due north and \(+5.00 \mu \mathrm{C}\) at the position due east. What are the magnitude and direction of the net electrostatic force acting on the charge at the center? Specify the direction relative to due east.

A tiny ball (mass \(=0.012 \mathrm{kg}\) ) carries a charge of \(-18 \mu \mathrm{C}\). What electric field (magnitude and direction) is needed to cause the ball to float above the ground?

Two point charges are fixed on the \(y\) axis: a negative point charge \(q_{1}=-25 \mu \mathrm{C}\) at \(y_{1}=+0.22 \mathrm{m}\) and a positive point charge \(q_{2}\) at \(y_{2}=+0.34 \mathrm{m}\) A third point charge \(q=+8.4 \mu \mathrm{C}\) is fixed at the origin. The net electrostatic force exerted on the charge \(q\) by the other two charges has a magnitude of \(27 \mathrm{N}\) and points in the \(+y\) direction. Determine the magnitude of \(q_{2}\)

Two identical small insulating balls are suspended by separate \(0.25-\mathrm{m}\) threads that are attached to a common point on the ceiling. Each ball has a mass of \(8.0 \times 10^{-4} \mathrm{kg} .\) Initially the balls are uncharged and hang straight down. They are then given identical positive charges and, as a result, spread apart with an angle of \(36^{\circ}\) between the threads. Determine (a) the charge on each ball and (b) the tension in the threads.

A charge \(Q\) is located inside a rectangular box. The electric flux through each of the six surfaces of the box is: \(\Phi_{1}=+1500 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) \(\Phi_{2}=+2200 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}, \Phi_{3}=+4600 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}, \Phi_{4}=-1800 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}, \Phi_{5}=\) \(-3500 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) and \(\Phi_{6}=-5400 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} .\) What is \(Q ?\)

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