Hooke's law and the Coulomb force

  • Thread starter Ampere
  • Start date
  • #1
44
0

Homework Statement



An unstrained horizontal spring has a length of 0.32 m and a spring constant of 220 N/m. Two small charged objects are attached to this spring, one at each end. the charges on the objects have equal magnitudes. Because of these charges, the spring stretches 0.020 m, relative to its unstrained length. Determine the magnitude of the charges.

Homework Equations



Hooke's law: F=-kx
Spring potential energy; E = 1/2kx^2
Coulomb's law; F=kq1q2/r^2
Voltage; V = kq/r

The Attempt at a Solution



I know energy is conserved so ΔPE + q*ΔV = 0

which gave me 0.5*220*0.02^2 = (8.99*10^9)*q^2*(1/0.34 - 1/0.32)

And q = 5.16 micro Coulombs.

But why do I get a different answer if I say that the forces are equal in equilibrium, i.e.

kx = (8.99*10^9)q^2/r^2
(220)(0.02)=(8.99*10^9)(q^2)/(0.34^2)?
 
Last edited:

Answers and Replies

  • #2
rude man
Homework Helper
Insights Author
Gold Member
7,904
807
What do you mean, 'energy is conserved'?

Your second method, equating spring force with Coulomb force, is correct.
 
  • #3
21,074
4,648
What do you mean, 'energy is conserved'?

Your second method, equating spring force with Coulomb force, is correct.
To elaborate on rude man's answer, if you are going to use energy considerations, you need to recognize that in the initial state, you need to have an externally applied force to hold the charges in place. As you ease up on this force, the spring stretches and the charges get farther apart. Your missing energy is the work done on the external agent as you ease up on the force.
 
  • #4
44
0
"Your missing energy is the work done on the external agent as you ease up on the force."

I don't understand this - any physics problem would require an "external agent" to set it up and/or hold it in place. Consider block-and-pulley problems. Before releasing the block, you would need to hold at least one of them in place, but there is no external agent in the energy equation.

It requires no work to hold the charges in place. If I simply release the charges, energy should be conserved from that point on. Aren't the spring and electric forces conservative? Yes, someone would have done work to push the charges together from infinity, but none of that work would be recovered if that person simply released the charges all at once.

The reason why I don't like using the forces is that the problem doesn't really say the spring is in equilibrium, just that the spring stretches. (Maybe it will stretch more before it's in equilibrium.) I would like to avoid making the equilibrium assumption if possible.
 
Last edited:
  • #5
21,074
4,648
"Your missing energy is the work done on the external agent as you ease up on the force."

I don't understand this - any physics problem would require an "external agent" to set it up and/or hold it in place. Consider block-and-pulley problems. Before releasing the block, you would need to hold at least one of them in place, but there is no external agent in the energy equation.

It requires no work to hold the charges in place. If I simply release the charges, energy should be conserved from that point on. Aren't the spring and electric forces conservative? Yes, someone would have done work to push the charges together from infinity, but none of that work would be recovered if that person simply released the charges all at once.

The reason why I don't like using the forces is that the problem doesn't really say the spring is in equilibrium, just that the spring stretches. (Maybe it will stretch more before it's in equilibrium.) I would like to avoid making the equilibrium assumption if possible.

In the examples you cited, there is also kinetic energy produced. The charged bodies have mass, so, if you release them suddenly, and no external force is subsequently being applied, the charged masses will oscillate about the equilibrium position. If you are unhappy with applying an external force, I have no problem with replacing the work provided by this force with kinetic energy of the charged masses. However, then the 0.02 meter displacement will only represent the equilibrium position, and you won't have enough information to ascertain the magnitudes of the charges. When the two charges are at the equilibrium location, the missing energy in your analysis will be equal to the maximum kinetic energy experienced by the charges over the oscillatory cycle.
 
  • #6
44
0
Okay that makes sense.

What you're saying is that, to prevent oscillation, someone must have "guided" the charges to the 2cm equilibrium position, and work was done as they did that.

Thanks for the help. I hadn't thought of it that way.
 

Related Threads on Hooke's law and the Coulomb force

  • Last Post
Replies
4
Views
11K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
14
Views
6K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
3
Views
6K
  • Last Post
Replies
0
Views
4K
Replies
2
Views
10K
  • Last Post
Replies
4
Views
5K
Top