What is the effective spring constant of a charged DNA molecule?

[jybe]

Homework Statement

A molecule of DNA (deoxyribonucleic acid) is 2.10 μm long. The ends of the molecule become singly ionized: negative on one end, positive on the other. The helical molecule acts like a spring and compresses 1.09% upon becoming charged. Determine the effective spring constant of the molecule.

Homework Equations

Q = 1.9 E-19

F = KQ^2/r^2

F= kx

The Attempt at a Solution

F = KQ^2/r^2

F = ((8.99E9)*(1.6E-19)^2)/(2.1E-6)^2

F = 5.224489796E-17 N

F = kx

k = (5.224489796E-17)/(0.0109*2.1E-6)

k = 2.282433288E-9 N/m

This is the answer I get but the software keeps saying my answer is wrong BUT within 10% of the correct answer. I've done it about 10 times now and don't understand. Can anybody see anything wrong?

 

[SammyS]

Homework Statement

A molecule of DNA (deoxyribonucleic acid) is 2.10 μm long. The ends of the molecule become singly ionized: negative on one end, positive on the other. The helical molecule acts like a spring and compresses 1.09% upon becoming charged. Determine the effective spring constant of the molecule.

Homework Equations

Q = 1.9 E-19

F = KQ^2/r^2

F= kx

The Attempt at a Solution

F = KQ^2/r^2

F = ((8.99E9)*(1.6E-19)^2)/(2.1E-6)^2

F = 5.224489796E-17 N

F = kx

k = (5.224489796E-17)/(0.0109*2.1E-6)

k = 2.282433288E-9 N/m

This is the answer I get but the software keeps saying my answer is wrong BUT within 10% of the correct answer. I've done it about 10 times now and don't understand. Can anybody see anything wrong?

Almost good.

When you use Coulomb's law to get the force, the distance between charges is reduced to slightly less than 2.10 μm. Right?