# Spring constant of molecule

1. Mar 14, 2008

### bastige

1. The problem statement, all variables and given/known data
A molecule of DNA lies along a straight line. It is 1.12 μm long. The ends of the molecule become singly ionized; negative on one end, positive on the other. The helical molecule acts as a spring and compresses .4% upon becoming charged.
The Coulomb constant is 8.99 X 109 N*m2/C2.
Determine the effective spring constant of the molecule. Take into account the compressed length when calculating the distance between the ends of the molecule. Answer in units N/m.

2. Relevant equations
Fe= ke * ( q1q2 / d^2 )

3. The attempt at a solution
d = 1.2*10-6 - (1.2*10-6 * .004) = 1.1952 μm
Fe = ke * ( q1q2 / d2 ) = (8.99*109) * (( 1.6*10-19 )2) / (1.1952*10-6)2 = 2014 N
Fs = kx
x = d
Fe = Fs ---> 2014 = (1.1952*10-6) k ---> k = 1.685*109 N/m

2. Mar 16, 2008

### Staff: Mentor

Is the distance 1.2 or 1.12 μm?
In Hooke's law (F = kx), x is the amount of compression from the uncompressed/unstretched position. So x does not equal d!

3. Mar 16, 2008

### bastige

distance is 1.12μm , so
d=(1.12E-9) - (1.2E-9 * .004) =1.11552E-9

& i totally butchered the rest. I'm not sure what its supposed to look like
Fe=8.99E+9 * ????? / 1.11552E-9

Then
Fs=kx
Fs*x =K

4. Mar 16, 2008

### Staff: Mentor

Good.

You had the right idea before:
$$F_e = k_e q^2/d^2$$

F = kx
so, k = F/x.

What's x? (By what distance is the "spring" compressed?)

5. Feb 25, 2009

This guy is right. x does not equal d. x equals (1.12 μm*.004)