Question

An uncharged molecule of DNA (deoxyribonucleic acid) is 2.14 µm long. The ends of the molecule become singly ionized so that there is a charge of −1.6 × 10−19 C on one end and +1.6 × 10−19 C on the other. The helical molecule acts like a spring and compresses 1.4% upon becoming charged. Find the effective spring constant of the molecule. The value of Coulomb’s constant is 8.98755 × 109 N · m2 /C 2 and the acceleration due to gravity is 9.8 m/s 2 . Answer in units of N/m.

Answer 1

K = 1.72 *10^{-11} N/m

Step-by-step explanation:

given data:

L= 2.14*10^{-6} m

`\Delta = 1.4%* of L`

`= (1.4)/(100) *2.14*10^(-6)`

= 2.99*10^{-8} m

`L' = L - \Delta L`

`L ' = 2.14*10^(-6) -2.99*10^(-8)`

`L = 2.11*10^(-6) m`

electrostatic force = spring force

`( kq^2)/(L'^2) = K \Delta L`

putting all equation to get spring constant K value

`(8.98*10^9 *1.6*10^(-19))/(2.11*10^(-6)) = K 2.99*10^(-8)`

K = 1.72 *10^{-11} N/m