Speed of Electron in Electric Field

by Patrick Ford
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Hey, guys. So in this problem, if take a look at it, we have a new electron that is initially at rest and that it starts moving inside of a uniform electric fields. And then after some displacement, we're supposed to figure out what is the electrons Speed. Alright, so first things first and these kinds of problems where you're not really told the direction where the electric field points, just feel free to draw it yourself. So what I'm gonna do is I'm just gonna assume that the electric field points off to the right for this problem. So I mean, that is R E Field. And the other thing is that we know that this actually a constant e field. So that's really important, because that determines which work equation we're going to use. So let's figure out what's going on here. You have an electron, so we have sort of a charge like this, and we know that that charge little Q is equal to the negative elementary charge. So we know that the Q we're working with is negative 1.6 times 10 to the minus 19 cool homes, and we have a uniformed electric field and we know that the field strength there is 500 Newtons per Coolum. Now the displacement here or this travel distance is going to be D, and that's equal to 10 centimeters. So that's equal to 0.1 m. And we're supposed to figure out what is the electrons speed. So it means the variable that we're looking for is V. But how do we relate that Back to the work equation? Remember what happens is the work is equal to the change in the kinetic energy, and the change in the kinetic energy is just final kinetic energy minus initial. So that's gonna be one half mass of the electron V final squared, minus one half massive electron, the initial squared. Right, So it's just final minus initial. We're working with the kinetic energy equation. Now what happens is this electron is initially at rest inside of this electric field, so that means that the initial Velocity V not just equal to zero. So what that does is that basically just cancels out this term over here. So that means that the electrons speed after some displacement is actually this V final that we're looking for So that means that V Final is our target variable right here. And the way we solve that is relating it back to the work equation for a moving charge inside of a constant electric field. All right, so let's take a look at our equation. Right. So we know that the work equation we're gonna use is gonna be Q e d times the co sign of data. Remember, we have those two work equations. This is not point charges. This is actually constantly fieldwork. So we have the charge. That is the electron. We have the strength of the electric field. We have the displacement. Now we just need to figure out what the coastline of the angle is. And to do that, we need to look at our diagram and figure out which the displacement. What's that? Is that What direction is the displacement? Right. So we have this electron that is in a constant electric field. Now, whenever you have positive charges, positive charges want to move along the field lines. But negative charges always want to do the opposite of those field lines. So that means in this case, what happens is that the electron is gonna have a force that pulls it to the left. So that means that this distance over here is actually my displacement. Now, remember that the angle this CO sign of data over here is always the angle between the displacement and the electric field. Now, in this case, what happens is that my electric field points to the right and my displacement actually points to the left. So that means that the angle between these two things is exactly 180 degrees because they're totally opposite. So that means that this fatal here is 180 degrees. And what happens is that the co sign of this 180 degrees silly me, right that the co sign of 180 degrees is just equal to negative one. So what happens is that this work equation just picks up a negative sign. So it's negative. Q e d. Right. So we have that this, uh, this angle here is equal to 180 degrees. So now what happens is we're ready. Just plug everything and figure out the work. So this is just gonna be equal to negative. And now our charge is negative. 1.6 times 10 to the minus 19. Now we have the electric field of 500 the displacement is 0.1. So if you plug all of this stuff in, what happens is that the negative signs will cancel out in this equation and you should just get a work of eight times 10 to the minus 18 jewels. So now that we have the work here, we can relate that to the final velocity using this equation. So we're not quite done yet. So we have one extra step to Dio, and that is that the work that's done is equal to one half times the mass of the electron V final squared. So we're actually looking for this V final, so we have to move everything over to the other side. So the one half goes over and the mass of the electron goes underneath and gets divided. So we end up with is two times the work divided by the mass of the electron is equal to the final squared. And now the last thing we need to dio is just take the square root. So that means that the final is equal to the square roots of two times eight times 10 to the minus 18 jewels divided by the mass of the electron. Now, just in case you're not given this in a worksheet, most likely you will. The mass of the electron is 9.11 times 10 to the minus 31. So if you go ahead and work all that stuff out in your calculator, you get a final velocity of 4.19 times 10 to the sixth meters per second. And that is our final answer. That is the electron speed after it gets accelerated through this electric field. Our guys, that's it for this one. Let me know if you have any questions.