/*! 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 20 A uniform electric field exists ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 20

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, \(1.60 \mathrm{~cm}\) distant from the first, in a time interval of \(3.20 \times 10^{-6} \mathrm{~s} .\) (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

Short Answer

Expert verified
The magnitude of the electric field is \(6.56 \times 10^4 \, \mathrm{N/C}\) and the speed of the proton when it hits the negatively charged plate is \(200 \, \mathrm{m/s}\).

Step by step solution

01

Identify the given values

From the exercise, we can identify the given values. The distance between the plates \(d = 1.60 \, \mathrm{cm} = 1.60 \times 10^{-2} \, \mathrm{m}\), the time it takes for the proton to travel \(t = 3.20 \times 10^{-6} \, \mathrm{s}\), and the charge and the mass of the proton \(q = 1.60 \times 10^{-19} \, \mathrm{C}\) and \(m = 1.67 \times 10^{-27} \, \mathrm{kg}\).
02

Find the acceleration of the proton

We know from kinematics that the distance travelled is given by \(d = \frac{1}{2} a t^2\), where a is the acceleration. We can solve the equation for the acceleration: \(a = \frac{2d}{t^2}\). Substituting the given values, we find \(a = \frac{2 \cdot 1.60 \times 10^{-2} \, \mathrm{m}}{(3.20 \times 10^{-6} \, \mathrm{s})^2} = 6.25 \times 10^{7} \, \mathrm{m/s^2}\).
03

Calculate the magnitude of the electric field

The electric field \(E\) is related to the force on the proton and its charge by \(E = \frac{F}{q}\), where \(F\) is the force. Since the force is related to the mass and acceleration of the proton by \(F = ma\), we can substitute this into the previous equation to find \(E = \frac{ma}{q} = \frac{1.67 \times 10^{-27} \, \mathrm{kg} \cdot 6.25 \times 10^{7} \, \mathrm{m/s^2}}{1.60 \times 10^{-19} \, \mathrm{C}} = 6.56 \times 10^4 \, \mathrm{N/C}\).
04

Calculate the speed of the proton

The speed \(v\) of the proton can be found using the equation \(v = at\), where \(a\) is the acceleration and \(t\) is the time. Substituting the given values, we find \(v = 6.25 \times 10^{7} \, \mathrm{m/s^2} \cdot 3.20 \times 10^{-6} \, \mathrm{s} = 200 \, \mathrm{m/s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Proton Motion
Protons are positively charged particles found in the nucleus of an atom. In this exercise, the proton is set into motion under the influence of an electric field. When a proton is placed in an electric field, it experiences a force along the direction of the field. This is due to the interaction between the proton's positive charge and the electric field's force lines.
The motion is straightforward: the proton starts from rest and accelerates towards the opposite charge, in this case, towards the negatively charged plate. This type of motion is predictable and is a great example of how charged particles interact with electric fields to yield observable behavior.
Newton's Laws
Newton's laws of motion are fundamental to understanding how forces affect motion. Here, Newton's second law is particularly relevant. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, given by the formula:
  • \( F = ma \)
For the proton, we know that the force acting on it is due to the electric field. Therefore, we express this force in terms of the electric field (\( E \)) and the proton's charge (\( q \)):
  • \( F = qE \)
Combining these facts gives us a way to relate the electric field to the proton's acceleration through the equation \( E = \frac{ma}{q} \). This relationship is pivotal in calculating how quickly the proton will speed up in response to the electric field.
Kinematics
Kinematics is the study of motion without considering the forces that cause the motion. To find how far and how fast the proton travels between the plates, we employ kinematics equations.

The formula for the distance \( d \) that an object covers when starting from rest under constant acceleration is:
  • \( d = \frac{1}{2} a t^2 \)
This equation allows us to compute the acceleration \( a \) when we have values for \( d \) and the time \( t \). Furthermore, once the acceleration is determined, the final speed \( v \) of the proton can be calculated using:
  • \( v = at \)
Both these equations illustrate how motion parameters are connected and how kinematics plays a critical role in understanding particle motion under uniform acceleration.
Uniform Electric Field
A uniform electric field is one where the electric force is consistent throughout. This means that a charged particle, such as a proton, will experience the same force magnitude and direction regardless of where it is in the field.

In the scenario provided, the uniform electric field exists between two plane parallel plates, with one positively charged and the other negatively charged. When released from rest at the positively charged plate, the proton moves towards the negatively charged plate under the influence of this electric field. The field provides constant acceleration, creating a steady increase in the proton's speed as it traverses the space between the plates.

The magnitude of the uniform electric field can be established by using the relationship between force, charge, and electric field (\( E = \frac{F}{q} \)). This enables us to quantify the field's influence on the proton, showcasing how a uniform electric field can precisely control the motion of charged particles.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Point charge \(A\) is on the \(x\) -axis at \(x=-3.00 \mathrm{~cm}\). At \(x=1.00 \mathrm{~cm}\) on the \(x\) -axis its electric field is \(2700 \mathrm{~N} / \mathrm{C}\). Point charge \(B\) is also on the \(x\) -axis, at \(x=5.00 \mathrm{~cm}\). The absolute magnitude of charge \(B\) is twice that of \(A .\) Find the magnitude and direction of the total electric field at the origin if (a) both \(A\) and \(B\) are positive; (b) both are negative; (c) \(A\) is positive and \(B\) is negative; (d) \(A\) is negative and \(B\) is positive.

(a) What must the charge (sign and magnitude) of a \(1.45 \mathrm{~g}\) particle be for it to remain stationary when placed in a downwarddirected electric field of magnitude \(650 \mathrm{~N} / \mathrm{C} ?\) (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

Inkjet printers can be described as either continuous or drop-on-demand. In a continuous inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. You are part of an engineering group working on the design of such a printer. Each ink drop will have a mass of \(1.4 \times 10^{-8} \mathrm{~g}\). The drops will leave the nozzle and travel toward the paper at \(50 \mathrm{~m} / \mathrm{s}\) in a horizontal direction, passing through a charging unit that gives each drop a positive charge \(q\) by removing some electrons from it. The drops will then pass between parallel deflecting plates, \(2.0 \mathrm{~cm}\) long, where there is a uniform vertical electric field with magnitude \(8.0 \times 10^{4} \mathrm{~N} / \mathrm{C}\). Your team is working on the design of the charging unit that places the charge on the drops. (a) If a drop is to be deflected \(0.30 \mathrm{~mm}\) by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop? How many electrons must be removed from the drop to give it this charge? (b) If the unit that produces the stream of drops is redesigned so that it produces drops with a speed of \(25 \mathrm{~m} / \mathrm{s},\) what \(q\) value is needed to achieve the same \(0.30 \mathrm{~mm}\) deflection?

Negative charge \(-Q\) is distributed uniformly around a quarter-circle of radius \(a\) that lies in the first quadrant, with the center of curvature at the origin. Find the \(x\) - and \(y\) -components of the net electric field at the origin.

A very long, straight wire has charge per unit length \(3.20 \times 10^{-10} \mathrm{C} / \mathrm{m} .\) At what distance from the wire is the electric- field magnitude equal to \(2.50 \mathrm{~N} / \mathrm{C} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Electric Charge Field and Potential

Read Explanation

Energy Physics

Read Explanation

Magnetism

Read Explanation

Electricity and Magnetism

Read Explanation

Force

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.