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Net charge

  1. Jan 18, 2014 #1
    Question:
    You’re 1.5 m from a charge distribution whose size is much less than 1 m. You measure an electric field strength of 282 N/C. You move to a distance of 2.0 m, and the field strength becomes 119 N/C. What’s 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.)

    I found another similar thread that was closed, and I still had questions.
    https://www.physicsforums.com/showthread.php?t=342378

    In terms of finding the net charge, I am slightly confused on how to proceed if it's unnecessary to calculate the charge itself. I have tried, though, and am not getting the correct solution. The answer is 0 from my textbook if anyone is interested, but I can't seem to come up with any mathematical explanation for the answer. From the linked thread, the last person comments on monopoles, dipoles, etc. I don't quite understand how that would allow us to determine the exact net charge given the data from the question. Any help would be greatly appreciated!
     
  2. jcsd
  3. Jan 18, 2014 #2
    Also, another question on the same topic:

    A thin rod lies on the x-axis between x = 0 and x = L and carries total charge Q distributed uniformly over its length. Show that the electric field strength for x > L is given by ## ε = \frac {kQ}{x(x - L)}##

    I've done:

    ## ε = \int_0^Q \frac {k}{(L-x)^2} dq ##
    ##ε = \frac {k}{(L-x)^2} \int_0^Q dq ##
    ##ε = \frac {kQ}{(L-x)^2} ##

    As you can see this is incorrect. It doesn't seem like an overly difficult problem but I am missing a step somewhere...any guidance on this question would be great too!
     
  4. Jan 18, 2014 #3
    Also, for the image attached, why would the apparatus not work? Although they all have q/m = a and this will be the only distinguishing factor, since m varies for isotopes, shouldn't it work? It would have to be very precise but I don't quite see why this machine would be dysfunctional.
     

    Attached Files:

  5. Jan 18, 2014 #4
    Sorry for filling this thread with questions. They are all on the same topic of electric fields and I didn't want to start multiple others.

    On this problem, I am only an order of magnitude away but can't seem to figure out why...

    Find the line charge density on a long wire if a 6.8-μg particle
    carrying 2.1 nC describes a circular orbit about the wire with speed 280 m/s.

    ## ε = \frac {2kλ}{r}## and ##F = qε = \frac {mv^2}{r}##
    ## \frac {2kλq}{r} = \frac {mv^2}{r}##
    ##2kλq = mv^2##
    ##λ = \frac {mv^2}{2kq}##

    I simplified this and calculated my answer as 1.41 x 10-3 C/m but the answer is -14 μC/m. I understand it must be negative since it is attractive and I did not include that in my vector for r, but why is the order go magnitude different?
     
  6. Jan 18, 2014 #5
    A dipole has a net charge of zero. A dipole field falls off according to the relationship r-3. If you can show from the data you're given that the field follows this relationship, you can conclude that it is a dipole, and therefore conclude that the net charge is zero.

    For the other one, I wouldn't integrate with respect to the charge. I would integrate with respect to x.
     
  7. Jan 18, 2014 #6
  8. Jan 18, 2014 #7
    As for your confusion on the centripetal force question, notice that 6.8 micrograms is 6.8 E-9 kilograms, not 6.8E-6 kilograms.
     
  9. Jan 18, 2014 #8
    Thank you for the help with the questions! I'm still doing the integration question but I understood the other 2 now.

    I keep getting more questions...

    We haven't really discussed structures with more than 2 constituents, so for water, does the dipole moment for each individual O-H bond equal qd still or is it assess differently. Either way, does this question still refer to the fact that ##ε ~ \frac {1}{r^3}##...I've been assessing the relationship but don't find anything meaningful...the answer is 0.4e, 0.03 e but i'm not sure why...
     

    Attached Files:

  10. Jan 18, 2014 #9
    The yahoo answer is incorrect.

    Notice how it gives the energy that it would take to "reverse the orientation". To go from the lowest potential energy to the highest potential energy, you have ΔU=pE-(-pE).
    Check out this wikipedia page - specifically, the section on torque on a dipole. Strictly speaking U is related to the dipole and the electric field by a dot product. The yahoo answer is oversimplistic, like usual. :wink:
     
  11. Jan 18, 2014 #10
    Aha thank you!

    For some reason, I decided to consider the initial as 0....:)

    Still having some trouble with the other qs...
     
  12. Jan 18, 2014 #11

    berkeman

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