from the center each sphere moves 8 - 4 = 4 cm = .04 m
that is how far each spring is compressed.
F = k x = 0.04 k
now what is the electrostatic force when the spheres are 16 cm = 0.16 meters apart?
F = [ 9*10^9 ] Q1Q2/(0.16)^2
so
0.04 k = [ 9*10^9] [ 2.5*10^6] [2.5*10^6] / [ 1.6^2*10^-2 ]
Those charges are huge. Typo?
Two spheres are attached to two identical springs and
separated by 8.0 cm, as in Figure 14. When a charge of
2.5 106 C is placed on each sphere, the distance
between the spheres doubles. Calculate the force constant
k of the springs.
2 answers
The forces being used in this scenario are the:
-electric force (Fe)
-spring force (Fs)
In each sphere, since the other is pushing away from it (due to the like charge), and since the spring is trying to restore equilibrium by pushing the sphere back, the equation will be Fnet = Fe - Fs.
Fnet = Fe - Fs
Fnet = 0 (the sphere's are stationary)
Fs=Fe
-kx = K(q1)(q2) / r^2
-kx = K(q)^2 / r^2
x is negative because it is being compressed (as stated by Hooke's law)
x= 0.04m (because this is how many meters each spring retracts back)
q1=q2= 2.5 * 10^-6 C
K= 9 * 10^9
r= 0.16 (the new distance between the two sphere)
-k(0.04) = 9 * 10^9 (2.5 * 10^-6 C)^2 / (0.16)^2
k = 54.9316 N/m
k= 55 N/m
-electric force (Fe)
-spring force (Fs)
In each sphere, since the other is pushing away from it (due to the like charge), and since the spring is trying to restore equilibrium by pushing the sphere back, the equation will be Fnet = Fe - Fs.
Fnet = Fe - Fs
Fnet = 0 (the sphere's are stationary)
Fs=Fe
-kx = K(q1)(q2) / r^2
-kx = K(q)^2 / r^2
x is negative because it is being compressed (as stated by Hooke's law)
x= 0.04m (because this is how many meters each spring retracts back)
q1=q2= 2.5 * 10^-6 C
K= 9 * 10^9
r= 0.16 (the new distance between the two sphere)
-k(0.04) = 9 * 10^9 (2.5 * 10^-6 C)^2 / (0.16)^2
k = 54.9316 N/m
k= 55 N/m
3 answers