CP A proton is traveling horizontally to the right at 4.50 * 10^6 m/s. (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

Video transcript

Hey, everyone in this problem, we have a hydrogen ion which we're told is essentially a proton has a horizontal velocity of 2.4 times 10 to the six m per second to the east. If the ion should stop within four cm inside an electric field, were asked to determine the minimum electric field strength and direction of an electric field that slows the ion uniformly. So we have our eye on moving to the right. Okay, because the right is the eastward direction. We're gonna take that as our positive direction and it is moving 2. times 10 to the 6m/s. Now, we're asked to find an electric field, OK. The strength and direction and recall that the electric field E can be given by the electric force F divided by the charge. Cute. Now, we know the charge Q okay. We're told that we can treat this as essentially a proton. We know the charge of a proton. But we don't know the electric force f recall through Newton Law though that we can relate the force to the mass times acceleration. Okay. So we can write this as mass times acceleration divided by Q. Now again, we have essentially a proton. So we know the mass just like we know the charge were given information about velocity and distance. And so let's use that to try to calculate the acceleration. If we can calculate the acceleration, then we'll be able to calculate E okay, we'll be able to find the electric field strength and direction. So let's write what we know, we know that the initial speed of this proton or of the hydrogen ion is 2.4 times 10 to the six m per second. It's going in the positive direction, the direction we've chosen to be positive. And so the velocity is going to be positive 2.4 times 10 to the six m per second. We know that the final velocity is going to be 0m per second because we want it to come to a stop. We know that the change in the exposition is going to be 4cm. Okay. We wanted to come to a stop within 4cm. Okay. That means that the minimum electric electric field strength would stop it at exactly 4cm. Okay. So we're gonna say that the change in position is four centimeters okay. We want to convert this into meters or standard unit. So we multiply by one m per 100 centimeters, the unit of centimeter cancels. Okay. We're essentially dividing by 100 and we get 0.4 m. The acceleration is what we're trying to find. Okay, we don't know the acceleration, but we're trying to find that. So we can use that in our equation for the electric force and then the electric field, we don't have information about the time. Um And we don't really care about the time. So we're going to choose the U A M equation that ignores T okay. Sometimes students get a little confused here. We're dealing with electrons, we're dealing with electric fields and electric forces. Um And so they kind of forget about the um equations but this electron moving, it's just like a ball moving or anything else. We're told that it slows uniformly okay. So we have uniform acceleration which means we can use our you am or Kinnah Matic equations. So we're going to choose the one without time and that's gonna be V F squared is equal to V not squared Plus two A Delta X. Okay. Substituting the information we know we have zero. On the left hand side, We have 2.4 Times 10 to the six m/s, all squared Plus two A times 0.04 m. Okay. If we move the two A times 0.4 m to the left hand side, in order to try to isolate A, we get negative 0.8 m times the acceleration A is equal to 2.4 times 10 to the six m per second squared gives us 5.76 times to the 12 meters squared per second squared. And then we divide by negative 0.8 m to isolate a and we get an acceleration of negative 7.2 times 10 to the 13. And the unit we have meters squared per second squared divided by meters. So we get a unit of meters per second squared, which is what we would expect for acceleration. Now, we have our acceleration and we can get back to looking at our electric field. And recall from above, we wrote that the electric field E can be written as the mass times the acceleration divided by the charge. Q We're told that this is essentially a proton, this ion that we're looking at. And so we have the mass of a proton is 1. times 10 To the -27 kg. The acceleration we just found negative 7.2 times to the 13 m/s squared. And all of this divided by the charge Q again, we're talking about a proton. So we have 1.6 times 10 to the negative 19 coolio's. Now, if we work this out on our calculators, we're going to get negative 7. times 10 to the five newtons per cool. Oh okay. We have kilogram meter per second squared. That's a newton over a cooler. Alright. And so when we're talking about the electric field strength. Okay. That's gonna be the magnitude of this, which is gonna be 7.515 times 10 to the five newtons per kilo. And the negative indicates the direction. So this is a negative value we've chosen east or to the right as our positive direction, which means that the electric field direction is going to be west or to the left. All right. And that's what our calculation shows. Okay. That's what that negative indicates. But we can also think about this conceptually if we wanted to just double check that, that sign is right, that that direction is correct. And we have a positive charge. In this case, what that means is that our electric force F and the electric field E point in the same direction, we know that the electric force F points in the same direction as our acceleration, acceleration is negative indicating that it points to the left. Okay. It's opposing the motion, which means that our force F also points to the left. Now, like we just said, the electric force and the electric field pointing the same direction because we have a positive charge, which means that our electric field is also pointing to the left or in the west direction. So that checks out. OK. So we can match up our calculations and the sign of our calculations with our conceptual understanding of the problem. All right, if we look at our answer choices here, we found that we had a uniform or sorry, a minimum electric field strength of approximately 7. times 10 to the five newtons per cool. Um And that the direction was west and so we have answer choice. F thanks everyone for watching. I hope this video helped you in the next one.