Inkjet printers can be described as either continuous or...

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?

Key Concepts

Charge Quantization and Electron Charge

Charge quantization refers to the fact that electric charge exists in discrete amounts, with the electron charge being a fundamental unit. Calculating the number of electrons that must be removed or added to a particle to achieve a specific net charge is an application of this concept. Understanding this helps in practical engineering tasks where exact charge values are required, ensuring that particles receive the correct amount of charge for the intended deflection behavior.

Time of Flight and Its Impact on Deflection

The time a particle spends in the electric field (also known as the time of flight) is crucial because it determines how long the particle is subject to the electric force. A longer exposure time, which can occur at lower speeds, results in a greater displacement for the same acceleration. This concept illustrates the interplay between the initial velocity of a particle and the resulting deflection, which is an important design consideration in systems requiring precise control of particle trajectories.

Electric Force on Charged Particles

This concept describes the force that an electric field exerts on a charged particle. The force is given by the product of the charge and the electric field (F = qE). This fundamental relationship is essential for understanding how charged particles can be manipulated, such as being deflected by electric fields, which is crucial in applications like inkjet printing deflection systems and many other areas in electromagnetism.

Newton's Second Law of Motion

Newton's Second Law (F = ma) links the force acting on an object to its mass and the resulting acceleration. In the context of charged particles in an electric field, the electric force leads to an acceleration that can be calculated using this relationship. This law is key to understanding how a known force (like that from an electric field) affects the motion of a particle, allowing the prediction of displacement over a given time.

Kinematics of Constant Acceleration

Kinematics under constant acceleration provides the mathematical framework to relate the time, acceleration, and displacement of an object. For a particle accelerating in a uniform electric field, equations such as x = (1/2)at^2 are used to determine how far the particle is deflected during a set time period. This concept is vital for designing systems where precise control over the particle's position is needed, for example, ensuring a specific deflection in an inkjet printer's operation.

Transcript

0:00 Hello everyone.

00:01 In this problem, we're asked to find out various details about an inkjet printer.

00:06 Namely, in the first part, we're asked to find how much charge we need to put on the droplets of ink in order to deflect it by the right amount to get it onto a paper.

00:17 So the situation is that we have a droplet of mass 1 .4 times the term is 11 kilograms, i .e.

00:26 1 .4 times that i .m .a.

00:28 Grams.

00:28 It's traveling with a horizontal velocity of 50 meters per second, and it enters a uniform vertical electric field of 8 times 10 to 4 units per cool arms between two parallel plates that are two centimeters long.

00:45 And so the question is, how much charge do we need to give this droplet in order for it to be deflected by three times centimeter minus 4 meters, easier with 30 millimeters? so first of all, we need to calculate how much time the droplet actually spends between the plates.

01:03 In order to do that, we can use its horizontal velocity because basically what this is is just a projectile motion for this ink droplet.

01:11 So in order to find what distance or how much time it takes to cover the horizontal distance of the length of the plates, we just use the horizontal velocity and the horizontal distance.

01:23 So we find t to be l, so the plates divided by the velocity, the horizontal velocity of the droplets.

01:31 And so that is 0 .020 divided by 50.

01:35 And so this equals 4 .0 times nymphom minus 4 seconds.

01:39 So that's a really, really short time.

01:42 And since we wanted to be, we wanted the droplet to be deflected by 30, 0 .30 millimeters, we can check how much deflection, just gravity on its own is going to be.

01:53 Give us.

01:54 So the deflection due to gravity, right, now we're kind of assuming downwards to be positive.

02:00 So the deflection due to be due to gravity is going to be a half times g times t squared.

02:06 Right.

02:07 So this is the deflection that gravity gives us.

02:09 And we can calculate what this is.

02:10 So this is minus 8.

02:26 So if you punch that into the calculator, then that gives you 8 times 16 times 10 to the minus 8.

02:36 So that's 7 .84 times 10 to the minus 7 meters.

02:46 Right.

02:46 So you can see that this is orders of magnitude like a thousand smaller than what we need, which is 3 .0 times 10 to the minus 4 meters.

02:56 So this tells us that the electric field is going to be pointing down.

03:03 So e must point down.

03:09 So if we go back to our diagram, we know the electric field is going to be pointing this way.

03:17 Now we know that it's going to be pointing this way because the way that we put the charges on the droplets is by removing the electrons.

03:24 So we know that the electrons removal means that we're going to give positive charge to these droplets, right? so these are going to be positively charged droplets.

03:35 Okay, so since the electric field is down, we can then calculate, we can then, you know, use the full power of union's second law.

03:44 So we're going to say that this is our droplet.

03:47 It's got the weight, so it's mass pulling it down, but it's also going to have the electric fields force pushing it down.

03:56 So if we analyze this diagram and we take down to be positive, as before, then we're going to say that the weight plus fe is f -natt in the y direction.

04:14 And this is not equal to zero.

04:19 It is equal to m -a -y.

04:24 So then we can put in the expressions that we know for these various forces.

04:28 So the weight is m times g, while the electric field gives a force or produces a force in the chart that is q times e.

04:37 And this is going to be equal to m times a y so this allows us to find the acceleration a y to be g plus q e over m so this is the expression that we get for the acceleration in the wide direction all right so what we want to do after that is use this acceleration to find the deflection so we're going to say that the deflection y is equal to a half times a y times t squared where a y is now the full acceleration due to electric and gravitational fields so this means that this is equal to half times g plus q e over m times t squared and we want to rearrange this for q since charge is what we're looking for that so charge is our target variable so we're going to multiply this by two and divide by t squared so then we get two y over t squared is equal to g plus q e over m you're going to subtract g so we're going to get 2y over t squared minus g put that in a bracket and we're going to multiply by m over e and that is going to give us the charge so if you're putting in all the values here so the mass of the droplet is 1 .4 times 10 to minus 11 is 1 .4 times 10 to minus 11 divided by the electric field, which was 8 times 10 to 4, neutas per coulomb, so 8 times 10 to the 4.

06:35 And then 2 times 3 .0 times 10 to the minus 4 as well, divided by t squared, which was 4 .0 times 10 .3.

06:51 I want to say minus 4 as well.

06:55 Today, yeah, exactly.

06:55 It's 4 times 10 to minus 4, but this time squared, minus 9 .8.

07:06 So if you kind of think about the numbers here a little bit, this one of the powers of 10 to the minus 4 cancel in the first term, but you still have a 1 over 10 to minus 4 on the bottom.

07:16 So that means that it's about three orders of magnitude bigger.

07:20 Than 9 .8...

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