a molecule of DNA (deoxyribonucleic acid) is 2.17 µm long. The ends of the molec

Question

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 1.8% upon becoming charged. Determine the effective spring constant of the molecule. coulombs constant = 8.98755 x10^9 amd the acceleration due to gravity is 9.8 ms2.






Ive found several questions/answers on here but none of them are helping me get to the correct answer. Also, none of them use the acceleration and Im sure it needs to be used.

Explanation / Answer

But electrical force is given by, F = Kq^2/r^2 The q is the charge of electron and is given by 1.6 * 10^-19 C The distance, r = 2.17 m = 2.17 * 10^-6 m So, F = kq^2/r^2 = (9*10^9)(1.6*10^-19)^2 / (2.17*10^-6)^2         F = 4.9 * 10^-17 N Spring force is given by, F = kx -------- (1) Where k is the spring constant and the x is compression Spring force is given by, F = kx -------- (1) Where k is the spring constant and the x is compression x = 1.8% of ( 2.17 * 10^-6 m) = 0.018 * 2.17 * 10^-6 m = 0.04* 10^-6 m F = kx K = F/x = (4.9 * 10^-17 N) / ( 0.04* 10^-6 m) = 1.22 * 10^-9 Nm Electrical force
Fe = (K * q1 * q2 ) / (d^2)

Spring Force
F_s = k * x ,


set the forces equal to each other and solve for k

k=Kq^2/(d^2)(x)

k=((8.99*10^9)((1.6*10^-19)^2))/((((.99*(2.17*10^-6))^2)(.01*(2.17*10^-6)))
k=2.29800022 ×10-9
k=2.30x10^-19

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