Question

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.

Answer 1

-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

Answer 2

4.62 x 10^3

Answer 3

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

k=5.15X10^3

Answer 4

A positive charge q1 = 3.10 µC on a frictionless horizontal surface is attached to a spring of force constant k as in the figure shown below. When a charge of
q2 = −8.900 µC
is placed 9.50 cm away from the positive charge, the spring stretches by 5.00 mm, reducing the distance between charges to d = 9.00 cm. Find the value of k.

A spring with spring constant k is attached to a wall on its left end and positive charge q1 on its right end, where q1 rests on a horizontal surface. A negative charge, q2, rests on the surface a distance d to the right of q1.