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This is chapter 37 problem number 30.
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We're given an electron.
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And due to an electric field, there is the force acting on this electron and the magnitude of 5 times center per 0 .50 moutons.
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And we know the mass of the electron 9 .1 .1 times 10 per negative 31 kilograms.
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So we're asked to calculate the acceleration on this electron if a, speed of this electron is one kilometer per second, which is a thousand meters per second, which is not a relativistic speed.
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So you new 20 mechanics is going to be sufficient enough and times a equals the force.
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If you're after the acceleration, then if we divide both sides by the mass, ms are going to cancel.
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The acceleration is going to be equal to force.
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Divided by the mass forces 5 times 10 to per of negative 15 newtons.
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The mass is 9 .11 times 10 to per of negative 31 kilograms.
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In this case, our acceleration is going to be 5 .49 times 10 to per 15 meters per second squared.
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It's a very high acceleration.
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Part b, if again, find acceleration if now we're dealing with the relativistic speeds, 2 .5.
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Times 10 to curve 8 meters per second.
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And there is an important part to this problem where it says f is parallel with respect to the velocity vector because this is going to change everything...