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Ch 21: Electric Charge and Electric Field
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All textbooksYoung & Freedman Calc 14th EditionCh 21: Electric Charge and Electric FieldProblem 21

Chapter 21, Problem 21

Electric Field of the Earth. The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the center of the earth. (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 m? Is use of the earth's electric field a feasible means of flight? Why or why not?

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Hi, everyone. In this particular practice problem, we are asked to determine the force of repulsion between the two knights where we will have two knights actually wearing a metallic armor, 60 m apart on a charged road floating in the air using the road electric field pointing downward, which has a magnitude of 350 new term per Coolum. One of the knights is actually uniformly charged with a charge of minus two and the other one with a charge of minus 1.5 K. And we're asked to find a force of repulsion between the two. And also whether it is possible that one of the nights flights towards the other using the rote electric field. So in this particular example, we're actually justified in using formula for point charges because the sizes of the nights are assumed to be small compared to the distance that separates the two of them. So from Colo's law, the magnitude of the mutual force of repulsion of the two charges is going to follow this formula right here. So the force of repulsion of the two charges can be calculated with F equals K With absolute value of the first charge or Q one and the second charge of Q two over are squared or the distance. So Q is going to be the charge of the first night, which is minus two colum. Uh Q one, Q two is the charge of the second night, which is minus 1.55 colum. And the R is the distance that separates the two nights, which is 60 m. All of those are given in our problem statement right here. And the K is just going to be a constant. So we can actually find the repulsive force by substituting all these variables with the known value. So let's just start doing that. So for the first part of the question, the repulsive force is equals to K which is a constant, which is 8.99 times 10 to the power of nine Newton multiplied by meter squared over K squared, Multiplied by Q one and Q two, Which is uh -2 K Multiplied by -1.5 K absolute value. And then our squared is going to be 60 m squared and this will came out to be 7.5 times 10 to the power of six Newton. So the repulsive force between the two nights is going to be 7. times 10 to the power of six Newton. So we can kinda cross out option C and also option D. And now we have to determine whether um it is possible for one of the knights to fly, uh fly towards the other using the roads electric field. And the answer is no, it is not possible. So no, it is not possible that one of the knights actually flies towards the other because the repulsive force is actually really, really big. The repulsive force is really immense and this is not a feasible means of flight. F is big, not feasible means of flight. Anyway. So the answer of this question will actually came out to be option B right here with a repulsive force between the two nights of 7.5 times 10 to the power of six. And that the answer is no, the two uh none of the nights are possible to actually fly towards one or the other. And it is not possible to do that. So that will be all for this particular practice problem. And if you guys have any sort of confusion on this, please make sure to check out our other lesson videos on similar topics and that'll be all for this one. Thank you.
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Textbook Question
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