One type of ink-jet printer, called an electrostatic ink-jet prin...

Question

One type of ink-jet printer, called an electrostatic ink-jet printer, forms the letters by using deflecting electrodes to steer charged ink drops up and down vertically as the ink jet sweeps horizontally across the page. The ink jet forms 30-μm-diameter drops of ink, charges them by spraying 800,000 electrons on the surface, and shoots them toward the page at a speed of 20 m/s . Along the way, the drops pass through two horizontal, parallel electrodes that are 6.0 mm long, 4.0 mm wide, and spaced 1.0 mm apart. The distance from the center of the electrodes to the paper is 2.0 cm. To form the tallest letters, which have a height of 6.0 mm, the drops need to be deflected upward (or downward) by 3.0 mm. What electric field strength is needed between the electrodes to achieve this deflection? Ink, which consists of dye particles suspended in alcohol, has a density of 800 kg/m^3 .

Accepted Answer

Hey, everyone. Let's go through this practice problem. An electrostatic spray painting system forms a uniform coating by directing charged paint droplets vertically with deflecting electrodes. While the nozzle moves horizontally, it forms 25 micro radius droplets charges them with 1 million electrons and propels them at 25 m per second. These droplets traverse between two parallel electrodes, eight millimeters long and 5 mL wide spaced 1.5 mL apart, the electrodes are three centimeters away from the surface to achieve the desired eight millimeter coating height. The droplets must be deflected four millimeters. What electric field strength is needed for this deflection? The paint a mix of pigment and a solvent has a 1200 kg per cubic meter density. All right. So we have four multiple choice options to choose from. Option A 3.82 multiplied by 10 to the power of seven nuance per Coolum. Option B 7.6 multiplied by 10 to the power of six nuance per Coolum. Option C 5.0 multiplied by 10 to the power of six nuance per Coolum. And option D 1.9 multiplied by 10 raised to the power of six nuance per Coolum. OK. So there is a lot going on in this problem. So the very first thing I'm going to do is draw a diagram to hopefully clarify what's happening here. So first off, there is some surface onto which these paint droplets are being projected. Uh This little rectangle here represents the surface. And at some distance away, we have two electrodes and these electrodes are chain are deflecting the charged paint droplets according to their electric charge. So let's say that in a normal case, the paint droplets would be just kind of projected in a straight line. But due to the deflection from the electrodes, they're actually going to travel in more of an angle that's some angle theta. So this is the new direction they go in. And of course, this goes both ways as well since it also is going to have the same deflections towards the top part of my crude diagram drawing here. The problem mentions that the electrodes are positioned at a distance of three centimeters away from the surface and that the electrodes themselves each have a length of 8 mL millimeters. We're also told that it is expected for there to be an eight millimeter coating height, which means that this full length here, the full length that the paint is hitting should be 8 mL, which means that each individual deflection should have a height of 4 mL since four plus 4 mL millimeters is eight millimeters So that's basically what's happening in this problem. And for the sake of our equations, we're just going to use a standard xy coordinate system. So the problem is just basically asking considering all this information, find what the electric field strength must be, what must be needed for all these variables to be true. So first off recall that the strength of an electric field is defined as the force on a particle in that field affected by the electric field divided by the charge of that particle, the electric charge or more mechanically. Another alternative way to represent this, the mass of each particle multiplied by the acceleration divided by the charge. And keep in mind, there are two different potential accelerations. There's one in the X direction and one in the y direction. The one that's going to be relevant here is a sub y the acceleration in the y direction because the electric field is only causing there to be a deflection in the along the y axis. In order to solve for the electric field strength, we're going to need to solve for all three of these variables individually and then put them into this equation. So we'll want the mass of each particle, the charge of each particle and the acceleration in the y direction. Let's start with mass. And to do this, let's assume that each paint particle is a perfect sphere. So recall that the mass of an object is equal to its density multiplied by its volume. So if we assume that each particle is a perfect sphere, then recall that the volume of the sphere is equal to four thirds multiplied by pi multiplied by the cube of its radius. So now since the problem gives us both the radius of each paint particle and the density of the paint, so we'll just plug those values into this equation. So the problem tells us that the density of the paint pigment is kg per cubic meter to 1200 kg per cubic meter multiplied by four thirds multiplied by pi multiplied by the cube of the radius of each paint particle. And the problem tells us that each droplet has a radius of microns. So that's 25 multiplied by 10 to the power of negative 6 m that's cubed. And we put that into a calculator. Then we find a mass for each paint droplet of 7.85 multiplied by 10, raised the power of negative 11 kg. So that was easy enough. Now, let's find the charge on each paint droplet. Now, the problem tells us that each droplet of paint has 1 million electrons. That means that the charge for each individual pain droplet is going to be the number of those electrons multiplied by the charge of an electron, the elementary charge. So that's 1 million or one multiplied by 10 rate. The power of six multiplied by the elementary charge, which is, which recall that the charge on an electron is 1.6 multiplied by 10, raise the power of negative 19 columns, put that into a calculator. And we find that the charge on each page droplet is 1. multiplied by 10 rates to the power of negative columns. So we're two thirds of the way there. Now, all that's left is we need to determine the acceleration on each page droplet in the Y direction. This is probably the trickiest part. However, since it's going to involve some kinematics, one way we could do this is to first find a relationship between the final speeds and both axes. Basically taking advantage of what we know about the maximum deflection. So let's say we're looking at the final Y speed of each droplet divided by the final X speed of each droplet. The Y speed is the distance that it travels in the Y axis per unit time divided by the X speed, which is the distance it travels in the X direction per unit time. The T is cancel out and the ratio is just the same as the distance it travels in each direction. Recall that according to the problem, the maximum deflection for each particle is going to be four millimeters or four multiplied by rate. The power of negative 3 m. This is because of the condition. The problem tells us about how there is going to be an eight millimeter coating height, which means that the droplets can get as far away as four millimeters from the central axis. And then the dis the verdict, the horizontal distance H sub X is the same for all of them. It's just the distance from the electrodes to the surface which is three multiplied by 10, raise the power of negative 2 m because it's three centimeters put this into a calculator. We find that this ratio is equal to about 0.13. Since this ratio of the maximum distances is the same as the ratio of the maximum speeds. One way to interpret this result is that the maximum speed in the vertical direction is equal to 0.13 multiplied by the maximum speed in the horizontal direction. And the problem actually tells us what that horizontal speed is going to be because remember that according to the problem, there's nothing that should be accelerating the particles horizontally only vertically. So when the problem tells us that the pain droplets are being propelled at a speed of 25 m per second, that speed is going to remain constant throughout the entire flight of the pain droplet's motion. So the maximum speed of each droplet in the vertical direction is 0. multiplied by 25 m per second. Or if you put it into a calculator about 3. m per second. This should help us find the acceleration in the y direction because the definition of acceleration is that it's the change in the velocity. So the final Y velocity minus the initial Y velocity divided by the time interval. And so we now know that in the maximum case, the final speed in the Y direction is 3.25 m per second and then minus the initial speed, which is going to be zero since it's being accelerated by the electrodes divided by the time interval. But we don't know what the time interval is yet. So we're going to have to find that time interval. Using another means, the easiest way we could do that is to take advantage of the fact that the particles that drop its horizontal speeds are going to be constant throughout their motion. So one way we could find that time is to look at the horizontal direction and take the direction, the distance that the droplet travels horizontally. In other words, the distance between the electrodes and the surface and divide that by what we know to be the horizontal speed of those droplets. So the distance between the electrodes and the surface is given in the problem as eight millimeters or eight multiplied by 10 to the power of negative m divided by that speed, the horizontal speed which is 25 m per second. And putting this into a calculator gives us a time interval of about 3.2 multiplied by 10 to the power of negative four seconds. So going back to our acceleration formula, the acceleration is 3.25 m per second, divided by 3.2 multiplied by 10, raise the power of negative four seconds. And putting this into a calculator, we find an acceleration of about 1. multiplied by 10 rate the power of m per second squared. And that is our acceleration. So now that we have that acceleration, we have everything we need to plug into the earlier equation we established for the electric field strength. So our earlier equation recall is that the electric field strength is equal to the mass of each particle multiplied by the vertical component of acceleration divided by the charge on each particle. And we found all three of those things already. The mass is 7.85 multiplied by 10 to the power of negative 11 kg. The vertical acceleration is 1.02 multiplied by 10 to the power of 4 m per second divided by the charge which is 1.60 multiplied by 10 to the power of negative 13 coupons. And if you put this into a calculator, we find a final value for the, the the field strength of about 5. multiplied by 10 of the power of six newtons per coon. And that is our answer to the problem. And so let's see here. This is our answer. And if we look at our multiple choice options we can see if this agrees with. Option C So option C is our answer to this problem and that is it for this video. I hope this video helped you out if it did and you'd like more experience, more practice. Please check out some of our other tutoring videos which will give you more experience with these types of problems, but that's all for now and I hope you all have a lovely day. Bye bye.

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