/*! 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 86 Operation of an Inkjet Printer. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 86

Operation of an Inkjet Printer. In an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. The ink drops, which have a mass of \(1.4 \times 10^{-8} \mathrm{g}\) each, leave the nozzle and travel toward the paper at 20 \(\mathrm{m} / \mathrm{s}\) , passing through a charging unit that gives each drop a positive charge \(q\) by removing some electrons from it. The drops then pass between parallel deflecting plates 2.0 \(\mathrm{cm}\) long where there is a uniform vertical electric field with magnitude \(8.0 \times 10^{4} \mathrm{N} / \mathrm{C}\) . 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?

Short Answer

Expert verified
The charge required is approximately \(1.05 \times 10^{-13}\,\text{C}\).

Step by step solution

01

Convert Units

First, convert all quantities to standard units. The mass of each ink drop is \[ m = 1.4 \times 10^{-8} \text{ g} = 1.4 \times 10^{-11} \text{ kg} \]The length of the deflection plates and the deflection distance should be in meters:\[ L = 2.0 \text{ cm} = 0.02 \text{ m} \]\[ d = 0.30 \text{ mm} = 0.0003 \text{ m} \]
02

Calculate Time to Pass Through Plates

Calculate the time it takes for the ink drop to travel through the length of the plates. Use the formula:\[ t = \frac{L}{v} = \frac{0.02 \text{ m}}{20 \text{ m/s}} = 0.001 \text{ s} \]
03

Find Vertical Acceleration

The deflection distance formula is given by:\[ d = \frac{1}{2} a_y t^2 \]Solving for vertical acceleration \( a_y \):\[ a_y = \frac{2d}{t^2} = \frac{2 \times 0.0003 \text{ m}}{(0.001 \text{ s})^2} = 600 \text{ m/s}^2 \]
04

Determine Force Required for Deflection

The force needed to produce this acceleration is calculated using Newton's second law:\[ F = m a_y = (1.4 \times 10^{-11} \text{ kg})(600 \text{ m/s}^2) = 8.4 \times 10^{-9} \text{ N} \]
05

Relate Electric Force to Charge

Since the force is due to the electric field, use the equation:\[ F = q E \]Rearrange to solve for the charge \( q \):\[ q = \frac{F}{E} = \frac{8.4 \times 10^{-9} \text{ N}}{8 \times 10^4 \text{ N/C}} = 1.05 \times 10^{-13} \text{ C} \]

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.

Inkjet Printer
Inkjet printers are a fascinating application of electrostatics in everyday life. These printers create images by squirting tiny ink droplets onto paper. The droplets are controlled with precision, allowing for detailed graphics and text.

Each ink droplet is ejected from a fast-moving nozzle. Upon leaving the nozzle, it passes through a charging unit, which imparts an electric charge by removing electrons. The charged droplets are then guided towards the paper.

This guidance is achieved through the use of electric fields that align the droplets accurately and efficiently on the paper surface.
Electric Field
Electric fields play a crucial role in manipulating charged droplets in an inkjet printer. An electric field is a space around charged particles where forces are exerted on other charges within the field.

In our inkjet printer scenario, deflecting plates create a vertical electric field. This field is responsible for adjusting the path of the charged ink droplets as they pass through. The magnitude of the electric field is a significant factor dictating how much the droplet path is altered.

In the given exercise, a field strength of \( 8.0 \times 10^{4} \, \text{N/C} \) was used to achieve the required deflection, demonstrating the electric field's utility in controlling small particles.
Deflection Plates
Deflection plates are integral to the operation of an inkjet printer鈥檚 precise ink placement. These plates create an electric field through which the charged ink droplets pass.

The plates, situated parallel to each other, generate a uniform electric field that acts vertically. This field provides the force necessary to alter the droplet's trajectory.

In the exercise, the deflection plates were 2.0 cm in length, ensuring that the ink droplet had sufficient time within the field to be deflected 0.30 mm, achieving the desired precision on the paper.
Charge Calculation
Calculating the charge required for ink droplet deflection is key in understanding an inkjet printer's functioning. This process involves physics principles in an accessible way.

First, the desired deflection and the speed of the ink droplet are determined. Using these, along with the length of the deflection plates, the time the droplet spends within the electric field is calculated.

Next, by relating this deflection to acceleration and force, the necessary electric force is found. Using the equation \( F = qE \), where \( F \) is force and \( E \) is the electric field, the charge \( q \) for deflection is then determined, exemplifying a direct application of electrostatics.

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

Two horizontal, infinite, plane sheets of charge are separated by a distance \(d\) . The lower sheet has negative charge with uniform surface charge density \(-\sigma<0 .\) The upper sheet has positive charge with uniform surface charge density \(\sigma>0 .\) What is the electric field (magnitude, and direction if the field is nonzero) (a) above the upper sheet, (b) below the lower sheet, (c) between the sheets?

Two identical spheres are each attached to silk threads of length \(L=0.500\) in and hung from a common point (Fig. 21.44\()\) . Each sphere has mass \(m=8.00 \mathrm{g}\) . The radius of each sphere isvery small compared to the distance between the spheres, so they may be treated as point charges. One sphere is given positive charge \(q_{1}\) , and the other a different positive charge \(q_{2}\) ; this causes the spheres to separate so that when the spheres are in equilibrium, each thread makes an angle \(\theta=20.0^{\circ}\) with the vertical. (a) Draw afree-body diagram for each sphere when in equilibrium, and label all the forces that act on each sphere. (b) Determine the magnitude of the electrostatic force that acts on each sphere, and determine the tension in each thread. (c) Based on the information you have been given, what can you say about the magnitudes of \(q_{1}\) and \(q_{2} ?\)Explain your answers. (d) A small wire is now connected between the spheres, allowing charge to be transferred from one sphere to the other until the two spheres have equal charges; the wire is then removed. Each thread now makes an angle of \(30.0^{\circ}\) with the vertical. Determine the original charges. (Hint: The total charge on the pair of spheres is conserved.)

If Atoms Were Not Neutral... Because the charges on the electron and proton have the same absolute value, atoms are electrically neutral. Suppose this were not precisely true, and the absolute value of the charge of the electron were less than the charge of the proton by 0.00100\(\%\) (a) Estimate what the net charge of this textbook would be under these circumstances. Make any assumptions you feel are justified, but state clearly what they are. (Hint: Most of the atoms in this textbook have equal numbers of electrons, protons, and neutrons.) (b) What would be the magnitude of the electric force between two textbooks placed 5.0 \(\mathrm{m}\) apart? Would this force be attractive or repulsive? Batimate what the acceleration of each book would be if the books were 5.0 \(\mathrm{m}\) apart and there were no nonelectrical forces on them. (c) Discuss how the fact that ordinary matter is stable shows that the absolute values of the charges on the electron and proton must be identical to a very high level of accuracy.

Three point charges are arranged along the \(x\) -axis. Charge \(q_{1}=+3.00 \mu \mathrm{C}\) is at the origin, and charge \(q_{2}=-5.00 \mu \mathrm{C}\) is at \(x=0.200 \mathrm{m} .\) Charge \(q_{3}=-8.00 \mu \mathrm{C} .\) Where is \(q_{3}\) located if the net force on \(q_{1}\) is 7.00 \(\mathrm{N}\) in the \(-\mathrm{x}\) -direction?

Two small spheres spaced 20.0 \(\mathrm{cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(4.57 \times 10^{-21} \mathrm{N} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Quantum Physics

Read Explanation

Electromagnetism

Read Explanation

Electricity

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Waves Physics

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.