Problem 86 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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:\[ L = 2.0 \text{ cm} = 0.02 \text{ m} \]\[ d = 0.30 \text{ mm} = 0.0003 \text{ m} \]

Step 2: 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:\[ t = \frac{L}{v} = \frac{0.02 \text{ m}}{20 \text{ m/s}} = 0.001 \text{ s} \]

Step 3: Find Vertical Acceleration

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

Step 4: Determine Force Required for Deflection

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

Step 5: 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 \( q \):\[ q = \frac{F}{E} = \frac{8.4 \times 10^{-9} \text{ N}}{8 \times 10^4 \text{ N/C}} = 1.05 \times 10^{-13} \text{ C} \]

Key Concepts

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’s 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.