INT Electrostatic cleaners remove small dust particles and pollen...

Question

INT Electrostatic cleaners remove small dust particles and pollen grains from air by first ionizing them, then flowing the air between the plates of a parallel-plate capacitor, parallel to the plates, where electric forces deposit charged particles on one of the electrodes. A typical pollen grain has a mass of 5.0×10^−10 g, the ionizer charges it with 750 extra electrons, and a fan moves the air at 3.0 m/s. Ignore air resistance and gravity.b. What minimum electric field strength is needed to deflect the grain by 3.0 mm before it leaves the electrodes?

Accepted Answer

Welcome back everyone in this problem. An electrostatic air purifier comprising two horizontal plates each 15 centimeters by 15 centimeters is employed to remove smoke particles from the air injected between the plates. The air purifier first ionizes the particles then passes the air through a region with a uniform electric field perpendicular to the airflow direction to deflect the charged particles towards a collection plate. If the smoke particles are charged with 100 extra electrons and the air flows at a speed of 2.5 m per second, what minimum electric field strength is required to deflect the smallest smoke particle by five millimeters before it reaches the collection plate, we're assuming air resistance and gravity are negligible and the mass of each smoke particle is four times 10 to the negative 11 g. Now, what are we talking about here? Let's try to make sense of what's going on. So we're talking about an air purifier with two horizontal plates. So let's do a quick sketch of that. So we have our horizontal plates here that probably look something like this. OK. And between our two plates where the length of each plate is or the width, rather depending on how you're looking at, it is 15 centimeters. We have e that's flowing. So I'm going to put that ear in blue. So this is the air that's flowing through the purifier and it's flowing at a speed of 2.5 m per second. So it's flowing with a velocity of 2.5 m per second. We also know that there's a force that's acting on the particles in the smoke in our airflow that's going perpendicular to the airflow direction perpendicular. So that means our force is acting in the y direction and it's pushing the particles toward a plate. So let me see if I can draw that plate here. It's pushing the particles toward this plate. OK. And it's expected to deflect it the smallest smoke particle by five millimeters. So we can also represent that on our diagram. OK. And let's just imagine that these are the smoke particles that are being defect deflected toward the plate. We also know that these smoke particles are charged with 100 extra electrons. So they have a negative charge. Is there anything else we know? Well, we don't know this force that's pushing the electrons to the electrostatic plate or the collection plate. But we know that we want to find the electric field strength that's required to deflect those smoke particles by that distance of five millimeters before it reaches the collection plate. So let's make a note of all the things we know from our problems so far. So first we know that let me make sure we can see our diagram. We know that our, our our particles are traveling a horizontal distance of 15 centimeters. So a change in the x direction is 15 centimeters which in meters would have been 0.15 m for our si units. We also know that our air flow is moving at a speed of 2.5 m per second. What else do we know? Well, we know that the mass of one of the smoke particles equals four times 10 to the negative 11 g. So in SI units, that's four times 10 to the negative 14 kg. OK. And we also know that the charge on each particle is 100 they are charged with 100 extra electrons. So if we think about it, since the charge of an electron is negative 1.6 times 10 to the negative 19 coulombs, then the charge for our extra 100 electrons would be 100 multiplied by negative 1.6 times 10 to the negative 19 coulombs, which would be equal to negative 1.6 times 10 to the negative 17 coulombs. Are we missing anything else? Well, let's see. We, we know that we want the electric field strength. That's what we don't know yet. And since our first is acting in a perpendicular direction to the airflow, our electric field is acting in the y direction. So we don't know what the strength of our electric field is. Neither do we know the force of our electric field? So how can we figure that out? Well, let's start with what we want to know, is there some way to relate our electric field strength to the force? Well, recall record that electric field strength, the magnitude of electric field strength equals the force of our field divided by the charge. And we know the charge of our particle, but we don't know the force. So how can we figure out the force? What do we know about force? Well, also recall that force equals mass multiplied by acceleration. And we're looking at the force in the y direction. So we're also going to be looking at the acceleration in the y direction, we don't know the force, but we know the mass of the particles. So if we can figure out the acceleration in the y direction of the particles, then we should be able to solve. No, let's think about it. We have our distance in the X direction, our speed in the X direction, we have our charge and we want to figure out our acceleration in the y direction. We're talking about the movement of a particle. Can you think of any equations that relate the movement of a particle to its acceleration and its its its direction or the distance it has traveled its displacement? Well, yes, we can think about the basic kinematic equation. So applying, let me write this in a different color, apply the basic kinematic equations kinematic equations in the Y direction. OK? Because we want the acceleration in the Y direction recall that it says or change in Y OK. Let me put that in red or change in Y equals our initial velocity multiplied by T plus half of A T squared which in this case are acceleration in the Y direction. Now remember there's no initial speed in the Y direction. OK. So our initial speed in the Y direction would have been equal to zero. So since let me write that a little bit more clearly, since our initial speed in the Y direction equals zero, that means our equation now becomes our change in Y equals half A T squared. Now we're almost there, we don't know a, we already know why neither do we know the time. So if we can figure out the time, then we should be able to figure out our acceleration in the Y direction. So how can we do that? In other words, how long does it take our particle to pass through the purifier? Well, if we think about it, when we look on our diagram, our time is independent of the X and Y directions, so whatever time we get for it to travel in the X direction would be the same in the Y direction. So if we can apply our kinematic equation in the X direction, we should be able to solve for a time and thus be able to solve for our acceleration. So in the X direction, let me put that in blue in the X direction, our kinematic equation would read our change in X equals our initial velocity in the X direction multiplied by T plus half A T squared. Now, what do we know about our acceleration in the X direction? Well, if we think about it, our force is acting in the Y direction. So then that tells us since there's no acceleration, no movement from the force in the X direction, then the acceleration in the X direction equals zero. And that's pretty important because if that equals zero, if we go back to our kinematic equation in the X direction, then our term over here half A T squared would all be equal to zero. So now that means our change in X would be equal to our initial velocity multiplied by T where we know that it's traveling zero point 1 5 m OK across our purifier. And our initial velocity is our air flow. The air comes in at a speed of 2.5 m per second. So that means our time would be equal to 2.5 m per second, divided by 0.15 m, which would be equal to 0.06 seconds. So now we have a value for our time since we have a value for our time, we can substitute it into our expression because no, that means our acceleration, OK. Our acceleration would be equal to two multiplied by our change. In why? In the Y direction all divided by our time squared? No, we're not going to work it out yet because we want an exact value. So we're going to substitute that acceleration in the Y direction to find a force in the Y direction. So now this tells us that our force in the Y direction will be equal to our mass multiplied by our acceleration, which is two multiplied by the distance, the change of distance in Y all divided by T squared. And if this is our force, then since our electric field equals force divided by charge, then the magnitude of our electric field, the magnitude of our, let me write it below it. The magnitude of our electric field equals our force which is two multiplied by and multiplied by the change in Y all divided by T squared divided by Q. So it would be that divided by Q. So our Q multiplied by T squared. Now we know all of these variables, we know the mass, the change in distance in the Y direction, the charge and our time. So we can substitute to solve. So that will be two multiplied by our mass which was four multiply by 10 to the negative 14 kg multiplied by our change in the Y direction which was 50 millimeters was 50 millimeters. We didn't make or five millimeters. My apologies, we didn't make a note of that. So that would have been five times 10 to the negative 3 m. And all of that is being divided by our charge which is negative 1.6 multiplied by 10 to the negative 17 Coombs multiplied by our time, which is 0.06 seconds all squared. Now we can put these values into our calculator. And when we work these values out, you will find that your answer would equate to negative 6.9 multiplied by 10 to the three newtons per coo. But now remember we only want the magnitude of our field strength. So we would write it as 6.9 times 10 to the three neurons per column, which we can also rewrite that as six, which would all which when we look at our answer choices would be equal to answer choice. A thanks a lot for watching everyone. I hope this video helped.

Struggling with Physics?

Watch next

Electric Fields in Capacitors practice set