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This is college Physics Answers with Shaun Dychko. We have an electron with some initial rate of 3 times ten to the six meters per 2nd and the initial rate is completely in the x direction. And it’s going to pass v this electric field the is in between these fee plates, and also it will leave the plates going at some velocity v1 and also it’ll have a vertical component come its velocity as well as a horizontal component. For this reason in part a, we figure out what is this upright displacement and also in component b, we figure out what is the upright component the its velocity after the passes v the plates and then component c is what is this angle. So part a, an initial of all we need to number out how long does that spend in between the plates and also we use that to eventually calculate the upright displacement since it’s walk to it is in one fifty percent times the vertical acceleration times time squared. So let’s very first figure out the time. We know that the horizontal ingredient of that velocity will be constant because this electric field will impart a vertical velocity and it’ll have no effect on that horizontal velocity that it had initially therefore the horizontal component of velocity stays constant. And also so we understand the x displacement which we’re provided is 4 centimeters, is walking to same the horizontal ingredient of velocity main point by time, and also so we’ll deal with for t by splitting both sides by vx here. And so t is dx split by vx, and the horizontal displacement is four centimeters written as time ten come the minus two meters, and also divided by the x ingredient of the velocity i m sorry is three times ten to the six meters per second. And we gain 1.3333 times ten to the minus eight seconds, maintaining lots of digits in there because this is an intermediate calculation. Now the upright component of its displacement will certainly be the early stage y component of the velocity time time, to add one half times vertical acceleration times time squared, yet there is no initial component to that is velocity vertically, and so this term is zero. So this functions out come one fifty percent ay t squared. Now the y component of the network force equals mass time the y ingredient of acceleration and also this needs to equal the pressure exerted by the electric field on the charge. And also so we’ll solve for the upright acceleration by separating both political parties by m here, and also so upright acceleration is the electric field strength times the charge split by that mass, and then we substitute that into the upright acceleration, we carry out that here. So the upright component that its displacement after passing v the bowl is one fifty percent Eq over m time time squared. Therefore it’s one fifty percent times 100 newtons every coulomb, electric field strength, times 1.6 time ten to the minus 19 coulombs, primary school charge, split by the mass of an electron 9.11 time ten to the minus 31 kilograms, times the time squared, and we gain 1.56 millimeters. Climate in part b, we desire to know what is the vertical component the its velocity after that passes in between the plates. Well that’s walking to same the initial component of the velocity the is vertical which is zero, to add the vertical acceleration time time. And the acceleration we already found is Eq over m from increase here, therefore we deserve to substitute the again down below so we’re acquisition that multiplying it by time. And also so electric field stamin times primary school charge split by massive of electron main point by time, offering us 2.34 times ten to the 5 meters per 2nd will be the vertical component that its velocity. And also the angle of its velocity will be the inverse tangent that the y component separated by the x component, so that’s inverse tangent of this answer for component b, divided by the x component the we’re given, and this is 4.46 degrees above the hopeful horizontal.
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