Question

1. A 2 µC charge is located 0.1 mm from a - 5 µC charge. (a) What is the magnitude of electric force between the charges? (b) Is the force repulsive, or attractive?

2. The deoxyribonucleic acid (DNA) is 2.17 µm long helical molecule. The ends of the molecule become ionized, with one end being negatively charged and the other end being positively charged. Upon becoming charged, the molecule acts like a spring, and compresses by about 1% of its length. What is the effective spring constant of the molecule?

Answer 1

1.

a.)

electrostatic force is given by

F = kQ1*Q2/d^2

Given, Q1 = +2 µC, Q2 = -5 µC

d = distance between both charges = 0.1 mm = 1*10^-4 m

k = 9*10^9

then, F = (9*10^9)*(2*10^-6)*(5*10^-6)/(1*10^-4)^2

F = 9.0*10^6 N

b.)

Force will be attractive if both charge have different signs, and force will be repulsive if both charge have same signs.

Since, Q1 and Q2 have oppsite signs.

So, Force will be attractive.

2.

By force balance on ions,

Electrostatic force(Fe) = Spring force(Fs)

k_e*q1*q2/r^2 = k_s*x

here, k_e = 9*10^9

q1 = q2 = charge on ions = e = 1.60*10^-19 C (taking, both ions are singly ionized)

r = final distance between charges = 99% of length of molecule = 0.99*2.17 µm

k_s = spring constant = ??

x = change in length of molecule = 1% of it's length = 0.01*2.17 µm

then,

k_s = (9*10^9)*(1.60*10^-19)^2/((0.01*2.17*10^-6)*(0.99*2.17*10^-6)^2)

k_s = 2.30*10^-9 N/m

"Let me know if you have any query."