# Where does this energy go?

1. Feb 17, 2014

### VishalChauhan

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?

2. Feb 17, 2014

### DrStupid

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.

3. Feb 17, 2014

### VishalChauhan

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.

4. Feb 17, 2014

### DrStupid

5. Feb 17, 2014

### VishalChauhan

Yes.

6. Feb 17, 2014

### kittu1421

How are you calculating the energy of the system..

7. Feb 17, 2014

### Staff: Mentor

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.

8. Feb 17, 2014

### Staff: Mentor

Or it may radiate away, particularly if the time constant is very low.

9. Feb 17, 2014

### VishalChauhan

I am making a guess here, but ultimately these oscillations would damp out.Would then, the final energy loss be due to radiations?

10. Feb 17, 2014

### Staff: Mentor

Yes. If there is no radiation and no resistance then there would be no mechanism for damping.

11. Feb 17, 2014

### Staff: Mentor

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.