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Review ConstantsAn electron travels with 5.00 x 10"; m/s Inrough point In space where B 0.110 j T The force on the electron at this point is (9.80 10-1i; 9.80 ...

Question

Review ConstantsAn electron travels with 5.00 x 10"; m/s Inrough point In space where B 0.110 j T The force on the electron at this point is (9.80 10-1i; 9.80 * 10-"k) NPart AWhat E the electric fiela?Express vector E in the form of Ez, Ey Ez, where the €, y,and components are separated by commas_Vicw Availablc Hint(s)AzdN/C

Review Constants An electron travels with 5.00 x 10"; m/s Inrough point In space where B 0.110 j T The force on the electron at this point is (9.80 10-1i; 9.80 * 10-"k) N Part A What E the electric fiela? Express vector E in the form of Ez, Ey Ez, where the €, y,and components are separated by commas_ Vicw Availablc Hint(s) Azd N/C



Answers

Review. An electron of mass $9.11 \times 10^{-31} \mathrm{kg}$ has an initial speed of $3.00 \times 10^{5} \mathrm{m} / \mathrm{s}$ . It travels in a straight line, and its speed increases to $7.00 \times 10^{5} \mathrm{m} / \mathrm{s}$ in a distance of $5.00 \mathrm{cm} .$ Assuming its acceleration is constant, (a) determine the magnitude of the force exerted on the electron and (b) compare this force with the weight of the electron, which we ignored.

Were given an electric field and we're told that an electron at rest at an initial point p moves along the path, ending at a point Q. Under the influence of this field, So the electric field in Newtons per Coolum is f X y Z equals 400 times X squared, plus z squared to the negative. First times the vector x zero z The electron begins at point P equals 537 and ends at point. Q equals 111 in part A. Where has to find a potential function for our field F well. One of the requirements is that the partial derivative of the potential V with respect to X should be equal to F one, which is 400 x over X squared plus z squared. Therefore, the potential function should have the form V of X y Z equals the integral of 400 x over X squared plus Z squared DX, which is 200 so 400 over two times the natural log of X squared plus z squared, plus some function G of wine z. Another requirement on the is that the partial derivative of fee with respect to why should be equal to F two, which is 400 x squared plus xi squared to the negative first time zero, which is just zero. And therefore, this implies that partial derivative of G with respect to why should be zero and so g itself should just be some function of h of Z. Therefore, potential function. The becomes 200 natural log of X squared plus Z squared plus hz in a final condition on V is that the partial directive V with respect to Z should be equal to the third component, F three, which is 400 z over X squared plus C Square's and therefore it should be hat taking the derivative. With respect to Z, we get to tens to hundreds 400 and then Z over X squared plus Z squared plus h Prime of Z is equal to 400 c over X squared plus Z squared. Therefore, H prime of Z is equal to zero and so h of C is just a constant C. And so our potential function v This is the function in general 200 natural log of X squared plus C squared plus C and of course, if we take see the constant to be zero that we obtain the particular potential function. V of X y Z equals 200 Natural log of X squared plus z squared, and this is our potential in Volts in Part B were asked to find the electrons speed at the point. Q. Well, the energy at point Q. I'll be charger than electron to me, times the change of potential. So we have potential at the Terminal Point Q, which is 111 man's potential. The initial point P, which is 537 and by conservation of energy. We have bet Mechanic Energy, which is the mass of the electron. In times the velocity squared over to this will be equal to Q E times, the 111 minus B 53 seven. In solving, we get that velocity V is equal to the square root of Let's see, we have two times. He charged Q E times V 111 minus fee of 537 over Mass M E. And we can rewrite this as see this is two times 200 which is 400 times the charge to mass ratio QE over Emmy which were given times. Then this is three natural log of one squared plus one squared minus the natural log of five squared plus seven squared. And if you plug this into a calculator, this simplifies to approximately 1.59 times 10 to the seventh and this is in meters per second.

We begin this question by calculating what is the Net force that acts on the electron and how can you do that? Well, remember that from Newton's second blow, the net force is given by the massed finds declaration. We know the mass of the electron, the mask off letter. That is the question 9.7. I'm stand to minus 30 world kilograms. Now we have to get her mind. What is the exploration and how can we do that? We'll take a look at the data, so the record goes from 5.4 times 10 to 15 meters per second, 22.1 time step into six meters per second. Then the variation in the velocity is 2.1 time stamped realistic minus 5.4. I'm Stan, this is and this happens while the electron was traveling a distance off 0.0 38 meters. So all we knew is no Cassie Anelka. The stents. How can you know what is exploration? We can use your Charlie's equation, which tells us that the final velocity squared is because the initial they lost its were plus two kinds declaration times. It's the displacement. Then we have the following the final velocity Q 0.1 time standing six square is equal to the initial velocity 5.4 I understand square plus true times acceleration times 0.0 38 Then we have to solve this equation for acceleration. In order to do that, we have first to do some things. So we sent this term to the other side to get the following. So we get true 0.1 time stamped with six. We're minors. 5.4 time Stand for the fifth square. Is it close to two times acceleration times 0.0 38. Damn, we had this. We can do the following send both be true on the 0.0 30 exit on the other side. So we got true 0.1 time stamped into six squared minus 5.4. I'm standing Feud queer divided by true kind 0.0 38 Is it close to the acceleration men? They have given this an acceleration off approximately 5.42 time Stand to the 13 meters per second square. So these these declaration now we can go back in complex in that force when that force is. Then it costs 29.11 kind. Stand to minus 31 times five. 40 to understand the 13. Then that force is approximately 4.94. Understand? To minus 17 mu Tums. And this is the answer for the first item. Now, on second. Like them, we have to calculate what is the strength off the force at two. And how can we do that? Well, note that the force that one swinging to the right and the force at two is pointing to the left. Sure, in that force is equal. Stoop at one minus two. The Net Force. We know it's 4.94 Understand? To my 17 the force s one we know a 7.5. Thanks then to the miners. 17. Now, I only have to do to complete the value off, too. We can do that. Like sending his current to the side. His term with the site. So we got to is equal to seven twice. Understand? To my 17 minus 4.9. Before time Stan to mind. 70. And these gives this force too equals true. 2.6. I'm Stan to minus 17 new tops on this. The answer off the second night

In this question. We are given uh this situation. Okay. Yeah, electron that's moving uh a long exhale direction. Yeah. And then he is moving through a region with uniform E. And electric fuel and many other few key. And then he is given to be along y axis. A long white heads. Then we have a graph of the net force along the white component, a function of the speed of the electron. So we want to find a many to of E. And direction B in unit vector notation. Okay, so to do a okay, discussion is about Lawrence force. Okay. To be using the Lorentz force, he M. Is equal to two times G. As Crosby. Okay, so to find a community of E. Okay, we will set ah B zero. So that and just be Q. E. Right. And then from the graph um Okay, F Y is negative two times 10 to the negative 19 in the negative direction. So we have negative two times tended to how negative 19 request to negative. 1.6 times negative 19. Hey, thanks e to reserve the jihad factor. Okay, So and then two is the charge of an electron. So negative. So E Okay. 1.25 jihad, was that? Okay? So the magnitude of E. Okay. Is 1.25 What? Uh So this is the unsellable party and maybe we want to find out they're actually not B. Okay. So using the orange spots on the again. Okay. I I would use the the situation when when A. Is zero. Okay. So we have E. Class B. Crosby because to you know, okay. And when you go to zero think he is 50 m per second. And last year I had Uh huh. So and E is 1.25 J heads. That would be ahead cross B is equal to zero. Uh huh. So um there if we draw out coordinate system. Okay, so this is X. This is why 15. Right? So we want uh huh. Our 50 I had Crosby is going to be negative 1.25 jihad. So I cross some direction needs to get negative jihad. Okay, so this means that mhm From bagram he be will be alone. Kids direction king. So the baby is I'm going to be 1.25 by 15. Um He had and the answer is 0.0 to finish. He had Hessler. So this is the answer for a bee and that's all this question.

Using the farm with we find a sled leading spread. That's acceleration. It's this man. We find that accident for these we finest 12 months in the field. Scored over. Thanks. You'll find Force. He plays in a week list over to its why we financed by spreading bath is nine for one month, 31 divided by you. See, the five may done. We find where my feeling is. It will be seven led by. Yes, please. Blair, I stand by you. It's glad we don't split four seconds. Glad on that. Pounds out of the people on 64 way is the 4 p.m. D. It was fine for 11 you know, grand line for a dirty place. That sled. It was a for 9 30 brother. So we can find that absolutely being force divided, by the way, once out to be for 11 Yeah, that


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