CP A proton is traveling horizontally to the right at 4.50 * 10^6 m/s. (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.20 cm.

Video transcript

Hey, everyone. Welcome back in this problem, we have an electron that moves horizontally eastward at 7. times 10 to the 6m/s, Were asked to determine the least electric field strength and direction of an electric field that slows the electron at a constant rate to a complete stop within 3.6 cm. Alright. So we're looking at an electric field and we want to figure out the electric field. Now let's recall that an electric field E can be given as a force divided by a charge F divided by Q. Now we know the charge Q because this is an electron. Okay. So we know Q because this is an electron. What about the force F? Well, let's recall that the force F can be written as mass times acceleration from Newton's loss. So we know the mass again, it's an electron. So that's a value that we can look up in a table in our textbook. So if we're able to find the acceleration of the electron, then we'd be able to calculate the electric field through F over Q. All right. So just because this is an electron and it seems kind of like an electricity problem doesn't mean we can't use our cinematic equations. Okay. This is gonna slow at a constant rate. So we have uniform acceleration so we can use our U A M or Kinnah Matic equations. So let's write out the information we know We know that the initial speed is going to be 7.2 m Whoops Times 10 to the six meters per second. Now, the electron is moving eastward horizontally. So that's to the right and we're going to take that eastward direction to be our positive direction. And so this speed or velocity is positive, the final velocity is going to be zero m per second because we want the electron to come to a complete stop delta X It's gonna be 3.6 cm that change in the horizontal position because we want this to come to rest within 3.6 cm. Now let's convert this to our standard unit of meters. So we multiply by one m per 100 centimeters, the unit of centimeter cancels. So we get 3.6 divided by 100 which gives us 0.36 m. We don't know the acceleration and we don't know the time t okay, but we want to find the acceleration because that's going to help us in our equation for the electric field. E Alright. So we have three things that are known one that we want to find. So let's choose an equation that doesn't have time in it. And the equation we're going to use is V F squared is equal to be not squared Plus two a delta x okay. Substituting in our values we have zero on the left hand side, we have 7.2 times 10 to the six m per second. All squared plus two times a times 0.36 m on the right hand side. All right, if we move our acceleration term to the left hand side, we're gonna have negative two times 0.36 m gives us 0.72 m. So negative to move it over 0. m times the acceleration A is equal to 5. times 10 to the 13 m squared per second squared. Okay. That's at 7.2 times 10 to the six m per second all squared. Now, in order to get A, we divide by negative 0.72 and our acceleration A is going to be negative 7.2 times 10 to the exponents 14 m per second squared. Okay. We have meters squared per second squared divided by meter and so we get meter per second squared. Now this is our acceleration and this makes sense. The acceleration is negative because we're trying to slow down that electron. Okay. So the acceleration is going to act in the opposite direction of the motion to slow that electron down. All right. Now, our electric field e I recall is equal to the force divided by the charge Q. And we said that the force through Newton's law can be written as mass times acceleration. So we get em a divided by Q. This is going to be equal to the mass of an electron is 9.11 times 10 to the negative 31 kilograms. The acceleration we found was negative 7.2 times 10 to the 14 m/s squared. And we divide that by the charge of an electron negative 1. times 10 to the -19 columns. And this is going to give us an electric field of 4099.5. And our unit here is Newton K kilogram meter per second squared per cool. Um No, We know we have a magnitude of 4,099.5 newtons per Coolum. We've also included the direction of the acceleration and the sign of the charge. Okay. So because this is positive, that means that the electric field is going to be pointing in our positive direction which is to the right or eastward. However, we can also figure this out by thinking about it conceptually. So our charge is negative, we have an electron. So the charge is negative. So the electric force f and the electric field E are gonna point in the opposite directions. Okay. We know that our force points in the direction of the acceleration, which means that our force points left. Okay. So our electric field e points to the right. And again, we can see that because we end up with a positive value when we calculate the vector and not just the magnitude. Alright. So if we go back up to our answer choices, We found that we had approximately 40 100 newtons per column for the electric field strength and the direction was east. So we have answer choice. E thanks everyone for watching. I hope this video helped see you in the next one.