1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Hydrogen bonding force problem

  1. Jan 19, 2015 #1
    1. The problem statement, all variables and given/known data

    In a DNA molecule, the base pair adenine and thymine is held together by two hydrogen bonds (see figure below).


    Let's model one of these hydrogen bonds as four point charges arranged along a straight line. Using the information in the figure below, calculate the magnitude of the net electric force along this hydrogen bond.


    2. Relevant equations


    3. The attempt at a solution

    Since we are looking at 4 point charges, we need 4 equations. They are:

    k = 8.99x10^9 (N*m^2)/C^2
    N = -3x10^-10 C
    O = -4x10^-10 C
    H = 3x10^-10 C
    C = 4x10^-10 C

    F_no = (k*|N|*|O|) / ((3x10^-10 m)^2) = -1.20x10^10 N -> negative as repelling force
    F_nc = (k*|N|*|C|) / ((4.2x10^-10 m)^2) = 6.12x10^9 N
    F_ho = (k*|H|*|O|) / ((1.8x10^-10 m)^2) = 3.33x10^10 N
    F_hc = (k*|H|*|C|) / ((3x10^-10 m)^2) = -1.2x10^10 N -> negative as repelling force

    Since we are asked for find the magnitude of the net electric force, we need to sum all 4 forces together.

    ΣF = F_no + F_nc + F_ho + F_hc = 1.54x10^10 N

    .... Opps. So I must have done something wrong.

    The hint given was "You need 4 eqns: pair N&O, N&C, H&O, H&C and use the charges & distances appropriate for each pair to get a net magnitude between the bases." ... which to my knowledge I have here.

    So not sure what I did wrong. Any help would be much appreciated.
  2. jcsd
  3. Jan 19, 2015 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Check these numbers. What does "e" stand for in the picture?
  4. Jan 19, 2015 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I'm not following your arithmetic.
    It would be clearer to leave out k and the unit charge of an electron as common factors to bring in later. E.g. for N::O write -0.3*0.4/(0.12+0.18)2.
  5. Jan 19, 2015 #4
    Ah! Small but critical oversight on my part. "e" meant electron ... so that each base pair is a factor .3 or .4 charge of an electron.

    This would make:

    N = (-.3)(1.602x10^-19 C) = -4.806x10^-20 C
    H = (.3)(1.602x10^-19 C) = 4.806x10^-20 C
    O = (-.4)(1.602x10^-19 C) = -6.408x10^-20 C
    C = (.4)(1.602x10^-19 C) = 6.408x10^-20 C

    Using these new values for N, H, O, C I get

    F_no = -3.08x10^-10 N -> repelling force
    F_nc = 1.57x10^-10 N
    F_ho = 8.55x10^-10 N
    F_hc = -3.08x10^-10 N -> repelling force

    Such that,

    ΣF = 2*(-3.08x10^-10 N) + 1.57x10^-10 N + 8.55x10^-10 N = 3.99x10^-10 N which ended up be the correct answer.

    Thanks a lot.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted