physics
posted by Brette .
A positive charge q1 = 2.70 uC on a frictionless horizontal surface is attached to a
spring of force constant k as in the figure. When a charge of q2 = –8.60 uC is placed 9.50 cm away from the
positive charge, the spring stretches by 5.00 mm, reducing the distance between the charges to d = 9.00 cm.
Find the value of k.

k*x = Ke*q1*q2/r^2
where k is the spring constant, x is the distance stretched by the spring, Ke is Coulomb's constant, r is the distance between the charges.
r = 9.00 cm = 0.09 m
x= 5 mm = 0.005 m
q1 = 2.70 uC = 2.7 * 10^6 C
q2 = 8.6 uC = 8.6 * 10^6 C
Ke = 8.9 * 10^9 
4.62 x 10^3

k(.005m)=((8.99X10^9)*(2.70X10^6)*(8.60X10^6))/(.09m)^2
k=5.15X10^3