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4) A Proton is released from rest from the upper plate of a capacitor The thickness of the capacitor is 0.10 m. The electric potential difference between the two pl...

Question

4) A Proton is released from rest from the upper plate of a capacitor The thickness of the capacitor is 0.10 m. The electric potential difference between the two plates is V= 5.0 V(k = 8.99 109 N m2/C2, e = 1.60 10-19 C , m proton = 1.67x 10-27 kg,a) What is the initial electric potential energy of proton?b) What is the initial kinetic energy of the proton?c) What is the final electric potential energy of proton?What is the final kinetic energy of the proton?e) What is the speed of the proton wh

4) A Proton is released from rest from the upper plate of a capacitor The thickness of the capacitor is 0.10 m. The electric potential difference between the two plates is V= 5.0 V (k = 8.99 109 N m2/C2, e = 1.60 10-19 C , m proton = 1.67x 10-27 kg, a) What is the initial electric potential energy of proton? b) What is the initial kinetic energy of the proton? c) What is the final electric potential energy of proton? What is the final kinetic energy of the proton? e) What is the speed of the proton when it reaches to the second plate?



Answers

The potential difference between the plates of a capacitor is
175 V. Midway between the plates, a proton and an electron are released.
The electron is released from rest. The proton is projected perpendicularly
toward the negative plate with an initial speed. The proton strikes the
negative plate at the same instant that the electron strikes the positive
plate. Ignore the attraction between the two particles, and find the initial
speed of the proton.

Question 31 has this shown with this figure here, Right? Proton is fired at the velocity given through a parallel plate capacitor. And we're told the electric field is pointing downward with this magnitude. So we're gonna find out a couple of things. For part a gonna find what magnetic field for both strength and direction must be applied to allow the proton to pass through without changing its speed or direction. So what many feel must exist such that this proton disco street Sorry, different color here. Just go straight through and doesn't seemingly is unaffected by the electric field. So because the, um, forced the election field will be pulling it, Pushing it downward means we know that the force on the magnetic field must be acting upward. This we know that be in order for that decree because in the cross product of if you think the force due to be and getting my q v cross be in order for the forced to be acting any upward direction and you know the particles traveling to the right, that means the magnetic field be must be into the page. First balls will determine the direction automatically B is into the page. Great. So in order for this to occur, you know that the force due to the electric field must equal in magnitude the force in the magnetic field. Those two balance. We have the force due to electric eels. Just charge times E and for magnetic field already kind of spoiled it. It's Q B Cross be are charges cancel out so even to solve for the magnetic field to being electric field divided by the velocity of our charge. And so we go in this plug those values in, and you can find it. Can't, um, that the harming I field turns out just by looking at night is now It's going to be 0.1 Tesla. And again, that's into the page based on the right hand roll of needing the magnetic force to be pushing the charge upward. Great perp even find the electric and magnetic field in the protons reference frames. So the protons reference frames it's the one we think of. It's the prime. It's the one moving along with the changes in the scenario. It's the electric field from an outside point of view, plus V across B and So if we have our electric field from our um, point of view for the protons point of view, the luxury field is, of course, putting in the negative direction are greater what downward, which we can call to get a J but didn't label it that way. So it's negative, plus the across B at the positive term. So, um, these 1.0 times 10 to the six we know they're particularly each other, so need aware about the cross product multiplied by be 0.1. Well, if I have a negative plus this positive of the same magnitude than in the protons from the reference there's no electric field, which makes sense because it feels an electric field, so it does not acted on it by the peril plate capacitor and similar, it was his b still not see and for the magnetic field, Big prime that's one over C squared, be across E and so they can evaluate this. Oh, is minus Sorrento looking to use their My apologies. This is be prime minus when something happens to be, say, a couple of mistakes. My apologies for that. This is being a standard cream just into the page and then remind us we have, um, 0.10 testily here, minus one over C squared and be cross. He's a one time center, six crack movie that cross mark because me very taking that to account two stories multiply times the ultra killed of 1.0, I was 10 to the five Oprah meter. So you know this is this term is times 10 to 11 but divided by term times tend the big of 16 cause it's C squared, so it's a very, very minimal effect. That term is essentially zero, and it will have a very, very minimal effect on the negative field. So this in terms of just about zero, So you can just say that this the person feel Proton experience is a 0.10 Tesla magnetic field. Rid New York. I believe that if it be so, we can look at part C now and you want to know. How does an experimenter in the protons reference frame explain that it does experience no force? Well, it's not. Accelerating is traveling at a constant speed. As we stated in the question, and be not Harvey's constant. I shouldn't care that. So, since it's not changing direction and not changing its velocity or orientation, all we know that the force acting on this part of all is, of course, sometimes a but because it experiences experiencing no acceleration and traveling at the same velocity, therefore must experience no forces. Well, that's that's what I come from. And that's why it seems kind of odd to say there's near no electric field and experience is no one should feel a little. Of course it does. Just in its reference, training is not so. Yeah, so there you go.

It's very simply, this is an energy conservation problem, so change in potential energy is equal to change in kinetic energy. Therefore, electric potential in James's, given by Q Delta V is equal to kinetic energy, which is 1/2 m times V squared displaying the velocity were interested in. So Velocity V will recall to the square root of two times cute had still to be over. And, um, we further note that Delta View is, ah, electric potential Is the electric field strength I should say Times distance d Mom played separation off play capacitors on So this is equal to now playing in values. Cue the elementary charge charge of a proton 1.6 10 Some of the negative 19 cool ums A mass of a proton is 1.67 times 10 to the negative. 24 um, that 27 kilograms. The times he which is far 50,000 Newtons per cool times, um, distant play capacitor a separation which is 2/10 ahead of the three meters bait. And so all of this is in a square root. So this gives you velocity of foam about 1.4 times 10 to the five eaters per second

This question. You are given that proton with an initial speed off it times 10 to the 5 m per second spot to rest in an electric field. Mhm. So, uh, actually, parts in the question part A. Did a proton move into a region or higher potential or lower potential? The the answer is higher potential. Okay, maybe we need to calculate the potential difference. Okay, So to find a potential difference using conservation of energy. Okay, the change in kinetic energy is you go to minus Q times Delta bi. So the change in kinetic energy is minus have and the i square because the final kinetic energy zero Okay, so final kinetic energy minus the initial kinetic energy you get minus have FBI square is equal to minus Q times Delta v so that obviously go to and b I square by two Q So the Mets off Proton is 1.67 times 10 to the my next 27 k g B. I is eight times 10 to the five damage. I am the charge this 1.6 times 10 to the minus 19. Can you calculate this? You get 3340 words. Okay, So which makes sense? Because we say that we must tow the final points off. Is that higher potential? So we finally Is that a higher potential than the initial? So the w b positive. Yeah. And then proxy, um, what was the initial kinetic energy? An electron votes. Okay, so from this equation, we can immediately right now that the initial kinetic energy is 2340 E v. Yeah, and that's all for this question.

All right. So by energy conservation, the change in potential energy will recall to the change in kinetic energy. Therefore, a que turns out of you, you will be equal to 1/2 mess has velocity squared. And so this means sad velocity is he called a squared of two Q Delta beauty over em. Um, and since we know that tell to be they called it your teens was two Q turns et over em. Uh, so if the charges increased by two I mean, since we know the beauty is proportional to the square root of Q, this means that if Q goes from, you know, one times cute two times q than, uh, that would mean that velocity would go from one times view to square root of two times. So therefore, being new will be squared of two times. Be old, so sweat of two times five times 10 to go for 50,000 meters per second. So this would be about, um to be about 70,000 about 70,700 meters per second for the new velocity X squared of two times sealed


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