# Conservation of energy gone for a six?

## Main Question or Discussion Point

Let us charge a capacitor and disconnect it from the battery. Let the capacitance, charge and voltage of the capacitor be C, Q and V respectively. Now do some work and reduce the distance of the plates of the capacitor and make it half of the original distance. What are the energies stored in the capacitor before and after the reduction of distance? The energies can be calculated as square(Q)/2C and square(Q)/4C respectively. Where has the energy gone from the capacitor?

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Doc Al
Mentor
The energy was dissipated by whatever brought the plates together. Note that the plates are oppositely charged, thus one must do negative work to bring the plates together.

A similar situation can be had with gravity by lowering an object. The energy decreases. Where did it go?

You must be right. But I think I am missing a point.
Whien a stone is raised to a point, it has potential energy. If leave it there, it comes down.
If charge a capacitor, leave the plates as they are, they don't come closer on their own. Does the system do any work to bring the plates closer, or an external agency shoud do work?
Sorry, for the basic nature of my question. But I cant help asking, to convince myself.

Dale
Mentor
If charge a capacitor, leave the plates as they are, they don't come closer on their own.

D H
Staff Emeritus