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!

Calculate the Effective Spring Constant using Coulomb's Law

  1. Apr 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider a segment of a molecule of DNA that is a coil with an overall length of 2.1*10^-6 m. If the ends of the molecule become singly ionized (one end loosing a single electron and the other end gaining a single electron) the helical molecule acts like a spring and compresses by 1.08% of its original length. Determine the effective spring constant of the molecule.

    2. Relevant equations

    F(elec)= F(spring)
    ke* [(Q1*Q2)/r]= k*(change in r)

    3. The attempt at a solution
    The DNA will have a charge of +e and one end and -e at the other end which we will plug in for our Q1 and Q2 values.

    I'm particularly concerned if I calculated the radius and change in radius correctly:
    Calculate change in r:
    .0108*(2.1*10^-6 m)= 2.3*10^-8 m

    Calculate r:
    2.1*10^.6 m- 2.3*10^-8 m= 2.1 *10^-6 m
    It seems fishy that my r is the same as my r initial.

    Plug in to equation: ke* [(Q1*Q2)/r]= k*(change in r)
    8.99*10^9 Nm^2/C^2 *[(-1.60*10^-19C)*(+1.60*10^-19C)/ (2.1*10^-6 m)]= k*(2.1*10^-6 m)

    k= -4.76*10^-15
  2. jcsd
  3. Apr 23, 2009 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    For the electrostatic force, it should be r2 rather than r.

    Note that r is known to about 5% accuracy, while the change in r is only about 1%. Or put another way, the 2.3*10-8 would affect the 2nd decimal place of 2.1*10-6, but we are only accurate to 1 decimal place there.

    Couple of errors being made here. The r2 I mentioned earlier. Also, shouldn't the change in r, 2.3*10-8 m enter in here somewhere?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook