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Determining Work from E-field Graph

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img413.imageshack.us/img413/8307/prob01a.gif [Broken]

    Determine the work done on a 0.5 µC charge moved from A to B.

    2. Relevant equations

    delta V= -integral(E*dl), aka the area under the curve on an E-field graph from A to B equals the potential difference. (Sorry, don't know how to display math properly on these forums yet.)

    (delta V)*(q)=Work

    3. The attempt at a solution

    I've attempted to aproximate the the area under that curve 9 times now (from A to B) using a series of triangles/rectangles, and have gotten several results for the area (delta V) that ranged from 12V(certainly to small) to 18V (certainly to big).

    I've been largely ignoring the sign's of things throughout, as the field is clearly from a positive charge, and so moving a 0.5*10-6 C charge closer to it will surely result in a positive amount of work done in the end. Is there any error in this reasoning, such that I should only be trying negative results?

    Is there a better way to find the work done beyond attempting to aproximate the area? It makes theoretical sense to me so it 'should' work, but the absence of particularly precise y-values on the graph make it difficult, and it seems unreasonable of them to want a very accurate answer (which I presume they do, based on my first 9 tries being rejected) considering they didn't give us the equation of the graph.

    Final answers I've tried ranged from 6.00*10-6 J to 8.00*10-6 J, but I've foolishly neglected to write down exactly what all I've tried because I was confident that my method was sound and it'd have a large range of acceptable answers to account for the aproximation of the area, but apparently not.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 21, 2009 #2
    As you said the area under the graph is potential difference. Is there some way to present the electrical field so that it would help you to intergrate?
     
  4. Feb 21, 2009 #3
    I believe the equation of that line would be something for the form KQ/r2, where K is coulombs constant (8.85*109), Q is the unknown charge producing that field, and r is of course the radial distance in meters.

    That would integrate to -KQ[1/rA-1/rB], (assuming I'm not doing something wrong), so the answer I'd want is KQq[1/rB-1/rA]. (could someone check that integration, it 'feels' wrong for some reason?)

    However, there doesn't appear to be any clear point from which I could determine the value of Q from in the first place, so I'm hesitant to switch my aproximations from area to aproximations of charge, as I only have 6 more tries left.
     
  5. Feb 21, 2009 #4
    According to Newton's 3rd law every force has its counter-force. So couldn't you just reverse your thinking, and bring the electrical field closer to the charge? This is just my reasoning, so don't take my ideas for granted. I'm brainstorming with you. =)
     
  6. Feb 21, 2009 #5
    I'm pretty sure that while correct, that would still have the same issue of the unknown charge that's producing the field, and would perhaps switch the sign of the answer.

    Speaking of sign, are we in agreement that the final answer should be positive?
     
  7. Feb 21, 2009 #6
    Yep, that should hold true.
     
  8. Feb 21, 2009 #7
    Alright, I managed to get 'close enough' by taking (2,25) as a point on the graph to find Q, and then using that knowledge to integrate. The actual answer was slightly less (7.49E-6 C versus my input 7.50E-6 C) suggesting that even that point wasn't perfectly accurate, but at least it was close enough. :smile:
     
  9. Feb 21, 2009 #8
    Was that the method they were looking for or was it solved some other way?
     
  10. Feb 21, 2009 #9
    I don't know of any other relationship between E and work (and I certainly looked!), so I presume they must have expected us to use the method of finding Q and integrating. I imagine it'd be possible to get it through finding an aproximate area, but one would have to get fairly lucky and/or be fairly accurate to get it this way.

    I'll inevitably be helping a few of my classmates with this problem in a week or two (that is to say, the day it's due), and I expect I'll be encouraging the method of finding Q from the closest 'good' point and integrating with that. If that doesn't work though, I don't know what I'd recommend. I suspect it'll work though.
     
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