A small sphere of charge0.784 iC hangs from the end of a spring as in figure a.

Question

A small sphere of charge0.784 iC hangs from the end of a spring as in figure a. When another small sphere of charge q2- -0.594 C is held beneath the first sphere as in figure b, the spring stretches by d 3.30 cm from its original length and reaches a new equilibrium position with a separation between the charges of r = 4.55 cm, what is the force constant of the spring? 2.02268x Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N/m 1 92 Need Help?Readit

Explanation / Answer

Using force balance:

Fe = Fs

Fe = electrostatic force = k*q1q2/r^2

Fs = spring force = K*x

k = 9*10^9

K = force constant = ?

r = 4.55 cm = 0.0455 m

d = 3.30 cm = 0.033 m

So,

kq1q2/r^2 = K*d

K = k*q1*q2/(r^2*d)

K = 9*10^9*0.784*10^-6*0.594*10^-6/(0.0455^2*0.033)

K = 61.35 N/m

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