Post a New Question

physics

posted by on .

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.

  • physics - ,

    -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

  • physics - ,

    4.62 x 10^3

  • physics - ,

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

    k=5.15X10^3

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question