The electric potential in a region of space is given by V=V₀[(x²+...

Question

The electric potential in a region of space is given by V=V₀[(x²+2y²)/(0.10 m)²], where V₀ is a constant. A proton released from rest at (x, y)=(20 cm, 0 cm) reaches the origin with a speed of 7.5×10^5 m/s .b. At what value of y on the y-axis should a He+ ion (charge +e, mass 4 u) be released from rest to reach the origin with the same speed?

Accepted Answer

Hey, everyone in this problem, we're told that a two D experimentation setup has electric potential given by V is equal to V, not multiplied by three X squared plus two Y squared where V knot is a constant, A beryllium ion, A be two plus with a mass of nine new initially resting at X and Y position of 10 centimeters and two centimeters has a speed of 6.3 multiplied by 10 to the exponent 5 m per second when it arrives at the origin. And we're asked to determine the positive Y position from which a lithium ion lithium as a mass of seven U must be released such that it arrives at the origin with the same speed as the beryllium ion. We're told to assume the initial exposition of the lithium ion is zero. The charge on the beryllium ion is plus two E, non lithium is plus one E. We're given four answer choices all in centimeters. Option A 22 option B 0.027 option C 0.019 and option D 15. So let's start what we're looking for is something to do with this lithium ion and we're looking for the initial y position it needs to be released from so that it gets to the origin with the same speed. So let's start there. And if we're thinking about this problem, we have some electric potential. OK. So there's gonna be some electric potential energy there, we have things moving. So there's gonna be some kinetic energy. So we're gonna think about our mechanical energy and what we have, we can assume there are no net external forces is a conservation of mechanical energy. So we can say that the initial mechanical energy emac I is going to be equal to the final mechanical energy emac f and again, this is dealing with the lithium ion. So we've already hinted at this, but the mechanical energy is made up of two components here, we have the kinetic energy, we have the potential energy and in this case, it's electric potential energy. So we're gonna have Kiku I is equal to KF plus UF where K is kinetic energy U is a potential energy. And then we have those subscripts denoting the initial and final time points. Now, initially, this lithium ion is going to be at rest and we're told that it's released. And so initially it's at rest, which means it has no initial speed. So the kinetic energy K I is going to be zero. Now at the end, right, the final electric potential energy. OK. Well, let's write this out and we'll see what happens there. OK. But recall that the electric potential energy can be given by the charge Q multiplied by the electric potential. The kinetic energy can be given by one half MV squared. And so what we get is that the charge QL of lithium and multiplied by the initial electric potential of lithium Lvil is gonna be equal to the final kinetic energy. So one half ML V LF squared, who SQL VF now because our electric potential V we're multiplying by three X squared plus two Y squared. And our final position is at the origin where X and Y are both zero, the final electric potential will be zero, which means the electric potential energy finally will be zero here. So that entire term is gonna go to zero. So what's happening here is we're converting all of this electric potential energy into kinetic energy. Let's go ahead and fill in what we know about the electric potential. So we have QL multiplied by V not multiplied by three X squared plus two Y squared. That's going to be equal to one half Ml VLF squared. OK. So if we wanna solve for the Y position, OK. The first step is to divide by our QL and B not. So we get three X squared plus two Y squared is equal to ML VLF squared divided by two QLVN. Next step will be to subtract three X squared from both sides. We can write that to Y squared is equal to ML VLF squared divided by two QLV, not minus three X squared. And as a last step, we are going to take the square root after dividing by two. So Y is gonna be plus or minus the square root of MLVLF squared divided by four QLV, not minus three halves X squared. I know this is how we're gonna find why we can start to substitute in our values. But what you're gonna notice is that we don't know this value of the, not just yet. OK. So what we need to do is switch gears and we're gonna use what we know about beryllium to find this VNO value. And then we can come back to this equation and solve for Y. And what we're gonna do is call this equation star so that we can come back to it in just a minute. So we're gonna switch over to beryllium and we're gonna do the exact same thing dealing with bellum. OK. So if we look at Merii, so we had be two plus, hey, again, we're gonna have a conservation of mechanical energy. So emac I will be equal to emac f again, our mechanical energy is made up of kinetic K I plus potential U I that's equal to KF plus UF just like before our initial kinetic energy is zero. And our final electric potential energy will be zero because again, that potential will be zero because the X and Y positions are both zero. OK. So because of the equation we were given for the potential, OK. So again, converting that potential energy into kinetic energy. So we get QB viv is equal to one half MB VBF squared. No, we have to be careful. The first V this is our potential. The second V on the right hand side, that's a speed. So we just have to be careful that we know which ones are potentials and which ones are speed. But we can go ahead and substitute in what we know. Now, the charge we're told is two E. So we have two multiplied by E which is 1.6 multiplied by 10 to the exponent negative 19 cos multiplied by our potential. That's gonna be that V not value multiplied by three X squared. So three multiplied by our initial exposition which is 10 centimeters. We can write this in terms of meters. So 0.1 m squared plus two multiplied by our initial Y position which is two centimeters or 0.02 m squared. Hey, this is equal to one half multiplied by the mass we're told is nine new converting to our standard unit, we wanna convert to kilograms. So we have nine multiplied by 1.66 multiplied by 10 to the exponent negative 27 kg. And we're multiplying by that final speed squared, which we're told is 6.3 multiplied by 10 to the exponent 5 m per second squared. OK. So what we can do is work out everything on the right hand side and we can divide everything on the left hand side that's surrounding that V knot. And when we do that, we're gonna find that V knot is going to be equal to about 30,000, sorry 300,816 in our unit here is kilograms per coulomb second squared and the meter squared will divide it. So we have our V not value. Now, we have everything we need to solve for Y that posi initial position, that initial y position of our lithium ion. So we can go back to our third equation substitute in these values Y is going to be equal to colos or minus the square root of the mass. Now, lithium has a mass of seven U. So seven multiplied by 1.66 multiplied by 10 to the exponent negative 27 kg multiplied by the final speed squared. That's that same speed. We want to have 6.3 multiplied by 10 to the exponent 5 m per second, all squared divided by four multiplied by the charge of lithium. And that's just equal to E 1.6 multiplied by 10 to the exponent negative 19 coulombs multiplied by this V not value. We just found 300,816 kg per Coulomb second squared. Right now our initial X value we're told is zero, right? So if the initial X value of lithium is zero, then this entire three half X squared term is zero. OK. All right. So that's everything we need. We can substitute this into our calculator. And we're gonna find that Y is going to be equal to plus or minus 0.15 4 7 7 6 m. And that's that Y position. And we have both the positive and negative root here in the answer choice or in the question we're asked to find the positive value. So we're gonna take that positive route to correspond with the correct answer. In order to convert from meters into centimeters. Like the answer choices, we're gonna multiply by 100 we will get that, we have an initial Y position of about 15 centimeters so that this lithium ion will get to the origin with that same speed as the brilliant ion. This corresponds with answer choice. D. Thanks everyone for watching. I hope this video helped see you in the next one.

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