/*! 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). 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 proton's velocity is approximately \(3.88 \times 10^{4} \mathrm{m/s}\).

Step by step solution

01

Understand the Problem

In this problem, we are dealing with a proton moving through a uniform electric field. The electric field will exert a force on the proton, causing it to accelerate. We need to find the final velocity of the proton after it has traveled a certain distance.
02

Calculate the Force on the Proton

The force exerted on the proton by the electric field can be calculated using the formula \( F = qE \), where \( q \) is the charge of the proton \( e = 1.6 \times 10^{-19} \mathrm{C} \) and \( E \) is the electric field strength. Thus, \( F = (1.6 \times 10^{-19} \mathrm{C})(2.3 \times 10^{3} \mathrm{N/C}) \). This gives us \( F = 3.68 \times 10^{-16} \mathrm{N} \).
03

Use Newton's Second Law

Apply Newton's second law \( F = ma \) to find the acceleration \( a \) of the proton. Rearrange to solve for \( a \): \( a = \frac{F}{m} \). Substitute in the force \( F = 3.68 \times 10^{-16} \mathrm{N} \) and mass of proton \( 1.67 \times 10^{-27} \mathrm{kg} \), \( a = \frac{3.68 \times 10^{-16}}{1.67 \times 10^{-27}} \approx 2.2 \times 10^{11} \mathrm{m/s^2} \).
04

Use Kinematic Equation to Find Final Velocity

With the displacement \( s = 2.0 \times 10^{-3} \mathrm{m} \) and initial velocity \( u = 2.5 \times 10^{4} \mathrm{m/s} \), apply the kinematic equation \( v^2 = u^2 + 2as \) to find the final velocity \( v \). Substituting in \( u = 2.5 \times 10^{4} \), \( a = 2.2 \times 10^{11} \), and \( s = 2.0 \times 10^{-3} \), we have \( v^2 = (2.5 \times 10^{4})^2 + 2 \times (2.2 \times 10^{11}) \times (2.0 \times 10^{-3}) \).
05

Calculate the Final Velocity

Calculate the values: \( (2.5 \times 10^{4})^2 = 6.25 \times 10^{8} \) and \( 2 \times (2.2 \times 10^{11}) \times (2.0 \times 10^{-3}) = 8.8 \times 10^{8} \). Add them: \( v^2 = 6.25 \times 10^{8} + 8.8 \times 10^{8} = 1.505 \times 10^{9} \). Take the square root: \( v = \sqrt{1.505 \times 10^{9}} \approx 3.88 \times 10^{4} \mathrm{m/s} \).

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Key Concepts

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

Proton Motion
Understanding the motion of a proton in an electric field is a critical concept in physics. Protons are positively charged particles that respond to forces exerted by electric fields. When a proton is placed in a uniform electric field, it experiences a force in the same direction as the field if the field is positively charged. This force causes the proton to accelerate, influencing its velocity and trajectory over time.
Proton motion is vital in understanding fundamental physics and plays a role in many technological applications, such as in particle accelerators. The dynamics of a proton in an electric field also serve as a good introductory example of how charged particles behave under the influence of forces.
Kinematic Equations
Kinematic equations are essential tools in physics that describe the motion of objects. They connect fundamental variables such as initial velocity, final velocity, acceleration, displacement, and time. In the context of a proton moving in an electric field, kinematic equations help determine the final velocity after the proton displaces over a distance.
The key equation used here is:
  • Final velocity squared: \( v^2 = u^2 + 2as \)
Where:
  • \( v \) is the final velocity.
  • \( u \) is the initial velocity.
  • \( a \) is the acceleration.
  • \( s \) is the displacement.
Using these variables appropriately can solve many motion-related problems efficiently and serve as a foundation for more complex physics concepts.
Newton's Second Law
Newton's Second Law is fundamental to understanding how forces influence motion. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration:
  • \( F = m \times a \)
In the context of a proton moving through an electric field, this law explains how the electric force causes the proton to accelerate. Here, the force (\( F \)) exerted on the proton in the electric field is calculated from the charge of the proton and the strength of the electric field. Then, by rearranging the equation to solve for acceleration \( a \), we gain insight into how the proton's motion changes:
  • \( a = \frac{F}{m} \)
By applying Newton’s second law, we can proceed to describe the subsequent motion of the proton using other motion equations.
Acceleration Calculation
Calculating acceleration is pivotal in understanding how objects speed up or slow down when a force is applied. Using Newton’s Second Law \( F = ma \), once we know the force exerted on the proton and its mass, we can calculate its acceleration. The force on a proton in an electric field is given by the product of its charge and the electric field strength:
  • \( F = qE \)
Given:
  • Proton's charge, \( q = 1.6 \times 10^{-19} \, \text{C} \)
  • Electric field strength, \( E = 2.3 \times 10^{3} \, \text{N/C} \)
The calculated force leads to acceleration by rearranging:
  • \( a = \frac{F}{m} \)
With these values plugged in, we understand not just the speed change, but the fundamental influence fields exert on particles. Such understanding is crucial, especially in fields like electromagnetism and particle physics.

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Most popular questions from this chapter

Two parallel plate capacitors have circular plates. The magnitude of the charge on these plates is the same. However, the electric field between the plates of the first capacitor is \(2.2 \times 10^{5} \mathrm{N} / \mathrm{C},\) whereas the field within the second capacitor is \(3.8 \times 10^{5} \mathrm{N} / \mathrm{C} .\) Determine the ratio \(r_{2} / r_{1}\) of the plate radius for the second capacitor to the plate radius for the first capacitor.

A particle of charge \(+12 \mu C\) and \(\operatorname{mass} 3.8 \times 10^{-5} \mathrm{kg}\) is released from rest in a region where there is a constant electric field of \(+480 \mathrm{N} / \mathrm{C}\). What is the displacement of the particle after a time of \(1.6 \times 10^{-2} \mathrm{s} ?\)

A plate carries a charge of \(-3.0 \mu \mathrm{C},\) while a rod carries a charge of \(+2.0 \mu \mathrm{C}\). How many electrons must be transferred from the plate to the rod, so that both objects have the same charge?

Four identical metallic objects carry the following charges: +1.6 \(+6.2,-4.8,\) and \(-9.4 \mu \mathrm{C} .\) The objects are brought simultaneously into contact, so that each touches the others. Then they are separated. (a) What is the final charge on each object? (b) How many electrons (or protons) make up the final charge on each object?

Conceptual Example 13 deals with the hollow spherical conductor in Figure \(18.30 .\) The conductor is initially electrically neutral, and then a charge \(+q\) is placed at the center of the hollow space. Suppose the conductor initially has a net charge of \(+2 q\) instead of being neutral. What is the total charge on the interior and on the exterior surface when the \(+q\) charge is placed at the center?

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