Why Does Energy Decrease When Work Is Done?

[sachi]

We have a pair of capacitor plates that are being pulled apart against their attractive force at constant potential difference. We find that the energy stored between the plates decreases when this happens, and we are asked to explain why the energy decreases even though work is being done against the internal force.

I think this is because a battery is required to keep the plates at constant p.d. We find that when the energy between the plates is increased, half of the energy supplied by the battery goes to increasing the internal energy, and half goes to performing external work. I think therefore that if work is done on the system, this constitues the system doing negative work, and therefore we also get a negative increment in the internal energy (interestingly this would also mean that energy is being supplied to the battery.) I would appreciate it if someone would tell me if I'm going along the right lines, and also any pointers if I'm not. thanks very much.

Sachi

 

[Galileo]

For 'large' plate capacitors the E-field is constant and independent of the distance between the plates. So the potential difference across the plates is V=Ed. So if you increase the distance, E must decrease to keep V constant. So charge is flowing away from the the plates.

 

[Meir Achuz]

We have a pair of capacitor plates that are being pulled apart against their attractive force at constant potential difference. We find that the energy stored between the plates decreases when this happens, and we are asked to explain why the energy decreases even though work is being done against the internal force.

I think this is because a battery is required to keep the plates at constant p.d. We find that when the energy between the plates is increased, half of the energy supplied by the battery goes to increasing the internal energy, and half goes to performing external work. I think therefore that if work is done on the system, this constitues the system doing negative work, and therefore we also get a negative increment in the internal energy (interestingly this would also mean that energy is being supplied to the battery.) I would appreciate it if someone would tell me if I'm going along the right lines, and also any pointers if I'm not. thanks very much.

Sachi

You are basicly correct. At constant voltage, twice as much energy is put into the battery than the energy you expend. This is shown in advanced EM texts.