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Tricky Charge/Net Force Problem

  1. Sep 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Estimate the net force between the CO group and the HN group shown in the figure. The C and O have charges ±0.40e and the H and N have charges ±0.20e where e = 1.6 × 10-19 C. [Hint: Do not include the "internal" forces between C and O, or between H and N.]

    http://www.webassign.net/gianpse4/21-70.gif


    2. Relevant equations

    Felec=kQ1Q2/r^2

    k=9*10^9

    3. The attempt at a solution

    First I tried going between O- and H+ as in: F=(k(-.4*.2)(1.6*10^-19)^2)/(.18*10^-9)^2
    which didn't work.

    I could try the charges on each by the opposite two poles, but if I'm doing that why not try include the internal forces while I'm at it?

    Basically, without including the internal forces I'm not sure where to start. Also, I have only one attempt left so I wanted to check on here to see if I can get a push in the right direction.

    Thank you in advance!
     
  2. jcsd
  3. Sep 5, 2012 #2
    What is the force acting H? Force acting on N?
     
  4. Sep 5, 2012 #3
    H: F=(k(.40*-.20)e^2)/.18^2

    N: F=(k(-.40*.20)e^2)/.3^2
     
  5. Sep 5, 2012 #4
    H has two forces acting on it: from C and from O. Ditto for N.
     
  6. Sep 5, 2012 #5
    True. My mistake.

    H: F=k(.40)(-.2)e^2/(.18^2) +k(-.4)(-.2)e^2/.3^2
    N: F=(k(-.40*.20)e^2)/.28^2 + k(.4)(.2)e^2/.4^2

    Does that seem right? The mistake was not accounting for both charges?
     
  7. Sep 5, 2012 #6
    I think the signs of the charges are not accounted for properly.
     
  8. Sep 5, 2012 #7
    My mistake, H and N's signs are reversed.

    Would the net force between the two groups just be the sum of the forces on H and the sum of forces on N?
     
  9. Sep 5, 2012 #8
    The forces are applied along the same line so, yes, the resultant force is simply the sum of them.
     
  10. Sep 5, 2012 #9
    Oh, ok. Thank you very much for the help
     
  11. Sep 5, 2012 #10
    Keep in mind in problems of this kind signs are paramount. This applies to every stage of computation, so always keep an eye on them.
     
  12. Sep 5, 2012 #11
    I'll keep that in mind, it's just been one of those days...
     
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