An infinitely long line of charge has linear charge den­sity 5.00x10^-12 C/m. A proton (mass 1.67x10^-27 kg, charge +1.60x10^-19 C) is 18.0 cm from the line and moving directly toward the line at 3.50x10^3 m/s. (b) How close does the proton get to the line of cha rge?

Video transcript

Welcome back, everyone. We are making observations about a very tall rod that is lying vertically. We're told that it has a linear charge of micro couls per meter. And we are told that an alpha particle who has a mass of 6.64 times 10 to the negative kg and a charge of positive two E which is equivalent to two times one .6 times 10 to the negative 19th coons is emitted from a decaying nucleus and it moves towards the rod. Now at a distance which I will just call X A of centimeters away from the rod, it has a speed of 1.8 times to the seventh meters per second. And we are tasked with finding at what distance will the alpha particle reach before it is repelled backwards? A K, it turns around well, at that point, once it turns around, if it's traveling in one direction, it will have to pause for just a second before traveling back the other direction. So really what we're trying to find is at what distance from the rod Is the velocity of the Alpha Particle zero. And let's go through our answer choices here. We have a 0.1 oh six centimeters. B 9.39 centimeters. C 0.235 m D 2.7 centimeters or e it hits the rod and goes all the way. All right. Well, here's what we're going to do here. We are going to use the conservation of energy principle that states that the total energy before is equal to the total energy after A K initial kinetic and potential energy is equal to final kinetic plus potential energy. Let's go ahead and sub in the formulas that we know for each of these terms here have that our K A or our initial kinetic energy is one half M va squared plus our charge times our initial uh potential here or our sorry, our initial va here, then we have that this is equal to one half M V B squared plus Q V B. Now, here's what I am going to do here. I am going to get similar terms on the same side. So I'm gonna subtract U V B from both sides and I'm going to subtract one half M va squared from both sides. And here is what we get from that we have that Q times va minus V B is equal to one half M V B squared minus va squared. And what I'm gonna do is I'm going to divide both sides by Q and you'll see that on the left hand side that cancels out. Now, we are ready to find a change in our VA and V B here. So let's go ahead and plug in all of the values we know we have that this is one half times 6.64 times 10 to the negative 27th times our final velocity of zero squared minus our initial velocity of 1.8 times 10 to the seventh squared all divided by our charge of two times 1.6 times 10 to the negative 19th. And what we get from this is a difference of negative 3.36 times 10 to the six volts. That is going to be our potential difference there. All right. So now that we have our potential difference, we can actually use one other relationship here. We know that our difference in potentials is equal to our linear charge density divided by two pi times epsilon knot times the natural log of the final position over our initial position. Remember what we're trying to find is this final position right here. So let's set all of this equal to our negative 3.36, which we know to be the potential difference times 10 to the sixth. And then I'm going to let me actually move this part of the equation over and you'll see why in just a second, I'm gonna multiply both sides by two pi epsilon knot over lambda two pi oops, sorry, two pi epsilon knot over lambda, these terms cancel out and we're able to solve for the natural log of X B over X A. So let's go ahead and plug in some values here to solve for that. We have that this is equal to negative, 3.36 times 10 to the 6V times two pi times epsilon knot, which is 8.85 times 10 to the negative 12 all divided by our linear charged density of 84 micro Kums. So we're gonna multiply this by 10 to the negative six just to get cools. And from this, we get negative 2.22. Now, in order to solve for X B, we can use a relation of natural logs that says that E raised to our constant power is gonna be equal to what's inside the natural log. So we have E to the power of negative 2.22 is equal to X B over X A. And all I have to do is just multiply both sides by X A. And we finally have a formula for our final position. Let's go ahead and plug in our X A and solve from there. We have that X B is equal to E to the negative 2. times 25 centimeters. And when you plug this into your calculator, you get a final position of 2.7 centimeters corresponding to our answer. Choice of d Thank you all so much for watching. I hope this video helped. We will see you all in the next one.