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Electric Field calculation question

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data
    You're 1.5m from a charge distribution whose size is much less than 1m. You measure an electric field strength of 282 N/C. You move to a distance of 2.0m, and the field strength becomes 119 N/C. What is the net charge of the distribution?
    Hint: Don't try to calculate the charge. Determine instead how the field decreases with distance, and from that infer the charge.

    2. Relevant equations
    E = k*Q/r2


    3. The attempt at a solution
    E2-E1=k*Q(1/r22-1/r12)
    From this I got Q = 9.3E-8...however when I plug this into the relevant equation I don't get the right answers for the first or second charge distance pair. Also shouldn't the field strengths be different? I thought that the change of electric field was inversely proportional to the change in distance...so if you increase the distance by a factor of 1.33 shouldn't the field drop by a factor of 1.77? Shouldn't this give a value of 159 for the second electric field value?

    Any help or guidance is appreciated! :D
     
  2. jcsd
  3. Oct 4, 2009 #2

    kuruman

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    You completely ignored the hint "Determine instead how the field decreases with distance, and from that infer the charge." Take the ratio E1/E2. Is it equal to the ratio r22/r12?
     
  4. Oct 4, 2009 #3
    E1=282 N/C r1=1.5 m
    E2=119 N/C r2=2m

    E1/E2 = 282/119 = 2.4
    r22/r12 = 4/2.25 = 1.78
    Shouldn't these two values be equal? That's what I don't understand about the question...or maybe I'm calculating something wrong?
     
  5. Oct 4, 2009 #4

    kuruman

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    They would be equal if you have just one charge Q. Obviously, that's not what you have. This means that the charge distribution does not give rise to a monopole field. If that's not the case, what else could it be?
     
  6. Oct 4, 2009 #5

    gabbagabbahey

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    You might find my reply in this thread useful...

    P.S. hoycey wouldn't happen to be another account of yours would it?
     
  7. Oct 5, 2009 #6
    Sorry, I'm not quite sure what a multipole expansion is, but I'm assuming that since we're far away the term that's going to dominate is the distance, right?

    Also I found that the correct ratio here is E1/E2 = d23/d13...but I'm not sure how to incorporate that into the question...

    (And no, I only have one account on this website)
     
  8. Oct 5, 2009 #7

    gabbagabbahey

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    The multipole expansion, is basically a Taylor expansion of the field or potential in powers of [itex]\frac{1}{r}[/itex] (where [itex]r[/itex] is the distance from the origin)....

    The lowest order term in the expansion corresponds to the monopole moment, and it is proportional to [itex]\frac{1}{r^2}[/itex] (just like the field of a point charge!). The second lowest order term is the dipole term, and it is proportional to [itex]\frac{1}{r^3}[/itex] (just like the field of an ideal dipole!). The next term in the expansion is the quadrapole term (proportional to [itex]\frac{1}{r^4}[/itex])...and so on.

    At "large" distances (which means small 1/r), the lowest order non-zero terms will dominate (1/r^4 will be much smaller than 1/r^3 etc.)

    You really haven't learned about dipole moments yet?
     
  9. Oct 5, 2009 #8
    Hmm, so because this electric field decreases with the same proportionality as a dipole...can I assume that like a dipole the net charge over the distribution is zero?
     
  10. Oct 5, 2009 #9

    kuruman

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    You can. If there were a monopole term too, it will would dominate the dipole term and the spatial dependence of the electric field would not be what it is.
     
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