Answer to Question #68846 in Electricity and Magnetism for thabang sewagodimo

Question

A molecule of DNA (deoxyribonucleic acid) is 2.19 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 1.20% upon becoming charged. Determine the effective spring constant of the molecule.

answer in N/M

Answer 1

Data: L=2.19*〖10〗^(-6) m,ε=1.2%
As its said that molecule is ionized slightly: we can assume that both charges equals to elementary charge: e=1.6*〖10〗^(-19)

The electrical force is compensated by force of elastisity:

k*(q_1 q_2)/L^2 =æ∆L
æ=k*(q_1 q_2)/〖∆L*L〗^2 =k*(q_1 q_2)/〖ε*L〗^3 =9*〖10〗^9*1.6*〖10〗^(-19)*1.6*〖10〗^(-19)/(〖2.19〗^3*〖10〗^(-18)*0.012)=1.82*〖10〗^(-9) N/m

Answer: æ=1.82*〖10〗^(-9) N/m