   Chapter 15, Problem 6P

Chapter
Section
Textbook Problem

A molecule of DNA (deoxyribonucleic acid) is 2.17 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 1.00% upon becoming charged. Determine the effective spring constant of the molecule.

To determine
The spring constant.

Explanation

Given info: The charges are e. The length of DNA is 2.17μm .

The displacement is equal to 1% of the length.

x=(0.01)(2.17μm)=0.0217μm

Formula to calculate the spring force is,

F=kx       (I)

• k is the spring constant.
• x is the displacement.

From Coulomb’s Law,

F=keQ2r2      (II)

• ke is the Coulomb constant.
• Q is the charge.
• r is the distance of separation.

The spring force equals the Coulomb force at equilibrium.

From Equations (I) and (II)

kx=keQ2r2

On Re-arranging,

k=keQ2xr2

Substitute 8.99×109N.m2/C2 for ke , 0.0217μm for x, 1.6×1019C for Q and 2

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