How Is Energy Lost in a Resistanceless System?

  • Context: Graduate 
  • Thread starter Thread starter VishalChauhan
  • Start date Start date
  • Tags Tags
    Energy
Click For Summary

Discussion Overview

The discussion revolves around the behavior of charges in a system of two concentric shells connected by an ideal, resistance-less wire. Participants explore the implications of energy conservation, potential differences, and the role of inductance in this context, questioning where energy is lost in such a system.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that charges will flow until the potentials of the two shells are equal, questioning where the energy goes since the wire is resistanceless.
  • Another participant argues that charges will continue to flow until the initial potential energy is reached, leading to oscillations where total energy remains constant.
  • Some participants assert that a potential difference is necessary for current flow, challenging the idea of continuous flow without it.
  • Inductance is introduced as a factor, with one participant explaining that the magnetic field generated during current flow can induce voltage, sustaining current even after the potential difference is reduced.
  • There is a suggestion that energy may radiate away, especially if the time constant is low, contributing to energy loss in the system.
  • One participant speculates that oscillations would ultimately damp out, leading to energy loss due to radiation.
  • Another participant notes that in a real system, resistance would lead to energy being lost as heat, but in a hypothetical zero-resistance scenario, energy loss would be attributed to radiation.

Areas of Agreement / Disagreement

Participants express differing views on the behavior of charges and energy conservation in the system. There is no consensus on the mechanisms of energy loss, with multiple competing perspectives on the role of inductance, potential differences, and radiation.

Contextual Notes

Participants discuss the implications of ideal versus real systems, highlighting the assumptions involved in considering a resistanceless wire and the potential for energy to be lost through radiation or resistive heating in practical scenarios.

VishalChauhan
Messages
30
Reaction score
1
Say, i have two concentric shells,each carrying a different charge.If i connect an ideal, resistance less wire between the two, charges will flow till potential of the two shells is equal.For an arbitrary selection of charges, the final and initial energy is not same(try it with two shells of radius r and 2r, with the inner one charged initially). Since the connecting wire is resistanceless, it does not cause any loss of energy.

Where does did my energy go?
 
Physics news on Phys.org
VishalChauhan said:
If i connect an ideal, resistance less wire between the two, charges will flow till potential of the two shells is equal.

No, it will continue to flow until the initial potential energy is reached. Than the charges will flow backward and the process starts again. During this oscillation the total energy remains constant.
 
That is quite contrary to what i have read.The only "motivation" for charges to flow will be a potential difference, and once that does not exist,there should be no flow of current.
 
VishalChauhan said:
That is quite contrary to what i have read.The only "motivation" for charges to flow will be a potential difference

Have you read about inductance?
 
Yes.
 
How are you calculating the energy of the system..
 
DrStupid said:
Have you read about inductance?

VishalChauhan said:
Yes.

Then you have your answer. The setup you described has non-zero inductance, so as the current flows a magnetic field forms; as the potential difference is reduced the magnetic field collapses generating an induced voltage which continues to drive the current even after the potential difference has fallen to zero. Then when the magnetic field is completely collapsed, the two previously low-potential shell will be at a higher potential and the cycle will repeat in the other direction.

You've just built an capacitor/inductor oscillating circuit - google will find you plenty more explanations.

And the answer to your question "where does the energy go" is that it goes into the magnetic field, and then back into potential energy to drive the next cycle of the oscillation.
 
  • Like
Likes   Reactions: 1 person
Or it may radiate away, particularly if the time constant is very low.
 
I am making a guess here, but ultimately these oscillations would damp out.Would then, the final energy loss be due to radiations?
 
  • #10
Yes. If there is no radiation and no resistance then there would be no mechanism for damping.
 
  • #11
VishalChauhan said:
I am making a guess here, but ultimately these oscillations would damp out.Would then, the final energy loss be due to radiations?

In any real system, the resistance of the wire won't be quite exactly zero, so most of the energy will end up as resistive heating of the wire. But if you really could build something with zero resistance... yes, the energy loss would be due to radiation.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 12 ·
Replies
12
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 64 ·
3
Replies
64
Views
10K
  • · Replies 6 ·
Replies
6
Views
13K