Problem 86 - University Physics with Modern Physics

Question

CP 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 \(1.4 \times 10^{-8}\) 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 needed is \(1.05 \times 10^{-13}\) C.

Step 1: Convert Mass to kg

The mass of each ink drop is given as \(1.4 \times 10^{-8}\) grams. To convert this to kilograms, use the conversion factor \(1 \mathrm{g} = 10^{-3} \mathrm{kg}\). This gives \(1.4 \times 10^{-8} \mathrm{g} = 1.4 \times 10^{-11}\) kg.

Step 2: Determine the Drop's Travel Time

The nozzle speed is \(20 \mathrm{m/s}\) and the length of the deflecting plates is \(2.0 \mathrm{cm} = 0.02 \mathrm{m}\). The time \(t\) for the drop to travel through the plates is calculated by dividing the length of the plates by the speed: \[ t = \frac{0.02}{20} = 0.001 \text{ s} \].

Step 3: Calculate the Required Vertical Acceleration

To be deflected \(0.30 \mathrm{mm} = 0.0003 \mathrm{m}\), the drop must accelerate vertically. Use the kinematic equation \[ y = \frac{1}{2} a t^2 \]Solve for acceleration \(a\): \[ 0.0003 = \frac{1}{2} a (0.001)^2 \] \[ a = \frac{2 \times 0.0003}{0.000001} = 600 \mathrm{m/s}^2 \].

Step 4: Relate Acceleration to Electric Force

The force \(F\) acting on the drop due to the electric field is given by \(F = m a\). Now substitute the known values:\[ F = (1.4 \times 10^{-11}) \times 600 = 8.4 \times 10^{-9} \mathrm{N} \].

Step 5: Find Charge Using Electric Force

The force due to the electric field is also given by \(F = qE\), where \(E = 8.0 \times 10^4 \mathrm{N/C}\). Solving for \(q\):\[ q = \frac{F}{E} = \frac{8.4 \times 10^{-9}}{8.0 \times 10^4} = 1.05 \times 10^{-13} \mathrm{C} \].

Key Concepts

Kinematics

In this exercise, kinematics plays a crucial role in understanding the ink drop's motion through the deflection plates. Kinematics is all about describing the motion of objects without considering the forces that cause the motion. Here, it's used to calculate how quickly the ink drop travels through the plates, which is determined by the speed of 20 m/s, and how far it needs to be deflected vertically by the time it exits the plates.

We start by determining the time it takes for the drop to travel through the plates. Since the plates are 2.0 cm long (which is 0.02 meters), and the ink droplets move at a constant velocity, the travel time can be found using the formula:\[ t = \frac{\text{Distance}}{\text{Speed}}\]Thus, the time is 0.001 seconds. This brief time interval is crucial for the ink drop to achieve the needed vertical deflection using the electric field. Next, kinematics helps us determine the required vertical acceleration for the ink drop to be deflected by 0.30 mm upon exiting the plates, using the equation:\[ y = \frac{1}{2} a t^2\]Here, \(a\) represents acceleration, and \(y\) is the deflection distance needed.

Electric Force

The electric force experienced by the ink drop is necessary to calculate to find the required charge. The force is related to the electric field and the charge on the drop. Electric force can be visualized as a push or pull within an electric field. It provides the acceleration that kinematics calculates.

In the case of the inkjet printer, the ink droplets are subjected to a vertical electric field of magnitude \(8.0 \times 10^4 \text{ N/C}\). This field exerts a force on the charged ink droplet, which can be calculated using:\[ F = m \times a\]This equation tells us the product of mass and the acceleration of the drop gives the force exerted on it. For this exercise, the vertical acceleration needed was previously calculated as \(600 \text{ m/s}^2\). By multiplying with the mass of the ink drop (converted to kilograms, \(1.4 \times 10^{-11} \text{ kg}\)), we find the force required to achieve this acceleration: \(8.4 \times 10^{-9} \text{ N}\).

By knowing this force, you can now relate it to the charge and the electric field through the formula \(F = qE\).

Electric Charge

Electric charge delineates how much an electric field can exert force on an object. In this problem, we need to find the magnitude of charge \(q\) necessary to cause the vertical deflection of the ink drop. An electric charge is required for the electric field to generate a force on the ink droplets, and this charge is what ultimately leads to the deflection.

We have already calculated the necessary force \(F = 8.4 \times 10^{-9} \text{ N}\). Given the electric field \(E = 8.0 \times 10^4 \text{ N/C}\), the magnitude of charge is calculated using the relation:\[ q = \frac{F}{E}\]Substituting the known values into the formula provides \(q = 1.05 \times 10^{-13} \text{ C}\).

This charge is essential for the proper functioning of the inkjet printer, as it ensures that each ink drop is accurately directed onto the paper at the correct location, enabling printers to produce clear, precise images and text.