Physics problem - Spring Constant

In summary, two spheres with a charge of +1.60 uC each are mounted on identical horizontal springs and face each other on a frictionless table. When the spheres are uncharged, the spacing between them is .05m and the springs are unstrained. When charged, the spacing doubles and the springs compress. To determine the spring constant, the force equation (F = (kq1q2)/r^2) and Hooke's law (F = -k_spring x) can be used by finding the amount of compression.
  • #1
Two spheres are mounted on two different identical horizontal springs and rest on a frictionless table (one spring is connected to the left side of a wall with a sphere attached to the other end & the other sphere is connected to the other spring which is connected to the other wall -- both spheres are facing each other) When the spheres are uncharged, the spacing between them is .05m, and the springs are unstrained. When each sphere has a charge of +1.60 uC, the spacing doubles (springs compress). Assuming that the spheres have a negligible diameter, determine the spring constant of the springs.

Okay so I know F= -kx, F=k [((q1)(q2)) / r^2] but I am not quite sure how to connect the dots...and what would q1 and q2 be? I am confused...

Please HELP!
 
Physics news on Phys.org
  • #2
allyson6018 said:
Okay so I know F= -kx, F=k [((q1)(q2)) / r^2] but I am not quite sure how to connect the dots...and what would q1 and q2 be? I am confused...
The charge on each sphere is given. How much does each spring compress?
 
  • #3
q1 and q2 are the charges of the spheres, which is +1.60 micro C. r is the distance between the charges, which is double the original spacing between the spheres. Keep in mind that the two k's you have written in the equations above are not the same!

Anyways, you have the force F exerted onto the spheres (from F = (kq1q2)/r^2). Using F = -k_spring x, you can find the spring constant. All you need is x, which is the amount of compression.
 
Last edited:

Suggested for: Physics problem - Spring Constant

Replies
12
Views
259
Replies
18
Views
393
Replies
10
Views
461
Replies
14
Views
2K
Replies
3
Views
46
Replies
1
Views
717
Replies
5
Views
463
Back
Top