Two stationary point charges +3.00 nC and +2.00 nC are separated ...

Question

Two stationary point charges +3.00 nC and +2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 cm from the +3.00-nC charge?

Accepted Answer

Hey, everyone today, we're doing the problem about electric potential, electric potential energy and kinetic energy. So we're being told that we have two charged particles with charges of positive 3. micro columns and positive 5.5 micro columns that are fixed in place 0.4 m apart. A proton has dropped at a rest. But at the midpoint between the two charges, If the proton moves purely along the line joining the two charged particles were being asked to determine its speed when it's at a distance of 18 cm from the .38 micro column particle. So let's before doing anything else, let's sort of draw this out, right? So I'll draw two states will draw state one when the proton has just moved in state two after the proton has moved a certain distance to end up at 18 centimeters from the 180.38 micro colon particle Or 3.8 Mexico and particle. Sorry. So let's just say that this is our first oops. No, let me get a better circle. That's a square. Let me just get a nice circle. There we go. So we'll say that this is Q Would say that position X is equal to zero. Let us say that this is still not a circle. That'll do, let's say this is cute too. Keep in mind that or we'll say this is 0.4 m away. And finally, we'll have our proton that was dropped at the midpoint that's still at the midpoint. And the midpoint between them would be 0.2 m. And we'll say this is positive E and it's a proton, it has a positive charge of an electron. So let's also write down what these are so far. Q one, He is 3.8 Micro columns. And we can recall that one micro column Micro Coolum is nothing but 10 to the negative 6th columns. Oops. And just for future reference, I'm also gonna note that one centimeter is equal to 10 to the 92nd meters. But if one micro column is equal to 10 to the negative six columns, then Q one will therefore become 3.8 times 10 to the negative 6th columns. Similarly, you too will become Plus 5. times 10 to the native 6th columns. And E will be 1.6 times 10 to the negative 19th columns. It's a negative 19th columns. No, from here, once the particle is moved, the system will look a little different, We'll still have our Q1 particle and X is equal to zero. We'll also have our Q2 particle Q two at X is equal to 0.4. However, our proton has moved right because now we're being told that it is at a position of positive E is now the position of 18 centimeters or 0.18 m from Q one, the 3.8 micro colon particle. So with this, now that we've sort of conceptualized it, we know that it started at rest, the proton had originally started at a rest. It had. And let's let's think about this because the particle isn't moved, which means we have to deal with kinetic energy. But we're also dealing with electric potential energy because it started at a stationary spot. So let's think about this. We have an initial kinetic energy of the entire system of the proton of zero. It was not at rest or it was it was address, which means it had some electric potential energy. What it is what that is we have yet to figure out, we know that there is some potential energy um at the beginning and there is some potential electric potential energy at the end of the process. So with this, we can say that we're dealing with inter conversion between potential and kinetic energy, which means that we can say that the final Connecticut or sorry, the initial kinetic energy plus the initial potential energy should also be equal to the sum of the final kinetic energy plus the final potential energy. Because the energy within the system can neither be created nor destroyed. It has to be converted from one form to another or transferred between two or more objects. So with this in mind, our target, what we're looking for the speed can be found in K F. Well, why is that? Well, that's because kinetic energy can be described as one half mass times velocity squared. While electric potential energy can be described as Q multiplied by an electric potential, which can also be rewritten as K which is cool um constant nine times 10 to the ninth needn't meter squared of cool M squared multiplied by two charges over R the distance between set charges. So with all this in mind, we can go ahead and start working towards the final kinetic energy because that will eventually help us find our final speed. So we know that our initial kinetic energy is zero jewels because we're not in motion, which means to find or isolate the final kinetic energy will simply be the initial potential energy minus the final potential energy. Now, the whole potential energy will include both the effect of Q one and Q two on the electric chart. So we're going to need to factor that in. So this will be K E E being the charge of the proton is Q one over are excuse me, we'll just say our one, our one initial will say because this will be this distance here. This is our one initial let's just label these. We'll say this is our two initial, this distance will be our one final after its change and this will be our one or our two final. Does that make sense? So it's K E now we have Q two over R two initial Subtracted by the final electric potential energies which would be very similar KEQ one over R one final plus que E que two Over our two final. So with this, if we factor out our cake us, we get okay. E Q one, Q one multiplied by one over are not minus or R one. Excuse me are one initial minus one over R one final plus que E que tu Into one over R two initial - over R to final. So with all of this in mind, let's go ahead and substitute in our values we'll have and I'll write this starting over here. We'll have nine times 10 to the ninth Newton meter squared over cool and squared Multiplied by 1.6 Times 10 to the -19 Coolum multiplied by 3.8 times 10 to the night of six school um Multiplied by one over, excuse me, 0.2 m minus 0.18 m plus and we'll start in the next line. It'll be nine times 10 to the ninth. Again, Newton meters squared over cool them squared into the charge of the proton is 1.6 times 10 to the NATO 19th. Cool um Oops 19th cool. Um This will be 5.5 times to the negative sixth. Cool. Um That is cute too. I have one over R2 initial which is also 0.2 m and R two F initial or R R. Sorry R two final will be the initial plus 0.2 because the proton moved 0.2 or 0.2 m to the left. So that will become 0.22 m. So with all this in mind, plug this into our calculator. And solving gives us an answer of 5.5938 times 10 to the negative 16th jewels, which means now we can take our kinetic energy final because that's what this is equal to, this is equal to final kinetic energy. Well, that means this is equal to one half M, the final squared oops MV, final square, Which means the final, the final speed when it's at this distance, the final distance that we're calculating will be the square root of 2K, the potential energy over em. So again, we're determining the speed of the proton. So we need the mass of the proton mass of a proton or an electron. In this case, either both have very similar masses but the mass of a proton mass or let's say mass of a proton will be 1.67 times 10 to the negative 27th kilograms. So utilizing that we get the square root of two times 5.59, 3, 8 times 10 to the negative 16th jewels right over A mass of 1.6, 1.67 times 10 to the negative 27th kilograms, Which gives us a final answer of 8. times 10 to the 5th m/s. So what this means is the final velocity, the final speed of the proton when it's at a distance of 18 centimeters from the 3.8 micro colon particle will be answer choice C 8.18 times 10 to the fifth meters per second. I hope this helps and I look forward to seeing you all in the next one.

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Two stationary point charges +3.00 nC and +2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 cm from the +3.00-nC charge?

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