/*! 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 92 Inkjet printers can be described... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\times \space 10^{-8}\) g. The drops will leave the nozzle and travel toward the paper at 50 m\(/\)s, 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 cm long, where there is a uniform vertical electric field with magnitude 8.0 \(\times \space 10^4 \space N/C\). 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 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 \(m/s\), what \(q\) value is needed to achieve the same 0.30-mm deflection?

Short Answer

Expert verified
(a) Charge needed is \(1.05 \times 10^{-13} C\), about 656 electrons removed. (b) With slower speed, charge needed is \(2.1 \times 10^{-13} C\).

Step by step solution

01

Understand the Forces

The force on the ink drop due to the electric field is given by the equation \( F = qE \), where \( q \) is the charge and \( E \) is the electric field. The deflection occurs because of this force.
02

Calculate the Time in the Field

The time \( t \) the drop spends between the plates can be found from the equation \( t = \frac{d}{v} \), where \( d = 0.02 \, \text{m} \) is the length of the plates and \( v = 50 \, \text{m/s} \) is the speed of the drops.
03

Determine Vertical Acceleration

The vertical deflection \( y \) of the drop is given by \( y = \frac{1}{2} a t^2 \), where \( a \) is the vertical acceleration. Solving for \( a \) gives \( a = \frac{2y}{t^2} \).
04

Relate Acceleration to Force

Since \( F = ma \), we can substitute for \( F \) using the electrostatic force formula \( F = qE \), hence \( ma = qE \) and solve for \( q \) as \( q = \frac{ma}{E} \).
05

Calculate Drop Mass and Required Charge

The mass of the drop \( m \) is given as \( 1.4 \times 10^{-11} \, \text{kg} \). Plugging this into the equation for \( q \), we obtain the required charge value.
06

Count Removed Electrons

The charge of an electron is \( 1.6 \times 10^{-19} C \). The number of electrons removed is \( \frac{q}{1.6 \times 10^{-19}} \).
07

Repeat for Redesign

Repeat Steps 2 to 5 using the new speed \( v = 25 \, \text{m/s} \). Calculate \( t \), \( a \), and finally \( q \) for the redesigned system.

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Key Concepts

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

Inkjet Printer Design
Inkjet printers use tiny droplets of ink to create high-quality prints on paper. There are two main types of inkjet printers: continuous and drop-on-demand.
Continuous inkjet printers, like the one in the exercise, continuously expel ink droplets from a nozzie toward the paper. Each droplet is formed and charged in sequence, passing through electric fields to create precise patterns of ink.
The ink drops travel at rapid speeds; in this example, at 50 m/s, and are precisely directed using electrostatic forces between deflecting plates. This precise control ensures accuracy in printing.
  • The speed of the ink droplets affects how long they interact with the deflecting plates.
  • A charging unit imparts electric charge onto the droplets by adding or removing electrons.
  • The uniform electric field between the plates moves the droplets into their final position on the paper.
The efficiency and quality of continuous inkjet printing are vital in applications such as product labeling and high-speed graphic printing.
Electrostatic Force
At the heart of inkjet printer operation is the concept of electrostatic force, a fundamental principle of physics. Electrostatic force in this context refers to the force exerted on the charged ink droplets by the electric field established between two plates. The calculation of this force is vital for understanding droplet movement.
This force is governed by the equation \( F = qE \), where:
  • \( F \) is the electrostatic force applied to the droplet.
  • \( q \) is the charge placed on the droplet by the charging unit.
  • \( E \) is the magnitude of the electric field between the plates.
This force is perpendicular to the direction of the ink drop's motion as it passes between the plates. It causes vertical acceleration, altering the drop's trajectory toward the paper to achieve the desired deflection.
The amount of charge \( q \), therefore, plays a crucial role, as a greater charge results in a stronger force, allowing for significant deflections even with a small electric field. This control allows designers to achieve various ink patterns.
Deflection Calculations
Deflection calculations determine how much an ink droplet moves from its original path as it passes through the electric field. To calculate the necessary charge and the corresponding droplet deflection, a series of physics principles come into play.
First, the time \( t \) the droplet spends under the influence of the electric field is determined using its speed and the length of the deflection plates:
\[ t = \frac{d}{v} \]
Where \( d \) is the length of the plates and \( v \) is the droplet speed. Knowing \( t \) is crucial because it affects how much the droplet can be deflected.
  • The vertical deflection \( y \) is given by \( y = \frac{1}{2}at^2 \), where \( a \) is the acceleration due to the electrostatic force.
  • The relationship between the force and acceleration is \( a = \frac{F}{m} \) with \( F = qE \).
Finally, using the mass of the ink droplet \( m \), you can connect the electrostatic force, electric charge, and field to find the required charge \( q \):
\[ q = \frac{ma}{E} \]
This calculation allows you to determine the exact charge needed for a specified deflection, ensuring that each drop lands precisely where intended on the paper. Adjustments to droplet speed or other parameters require reevaluating these calculations to maintain the quality and accuracy of printing.

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Most popular questions from this chapter

A \(+\)2.00-nC point charge is at the origin, and a second \(-\)5.00-nC point charge is on the \(x\)-axis at \(x = \)0.800 m. (a) Find the electric field (magnitude and direction) at each of the following points on the x-axis: (i) \(x =\) 0.200 m; (ii) \(x =\) 1.20 m; (iii) \(x = -\)0.200 m. (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).

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