Energy lost in capacitor?

[symsane]

Connect a resistor across a capacitor(which has its own energy) to discharge it. After the capacitor is fully discharged, we know that there is not any energy left on the capacitor. So, where does the energy of the capacitor go???
I think, not all of the energy goes to heat, so what are the other possibilities?

[Naty1]

The energy stored in the capacitor IS basically utilized as heat in the resistor.

Although not included in common calculations, the connecting wire also carries resistance and therefore a tiny,tiny amount of heat would be disipated in the wire as well as the resistor. In fact a tiny amount of heat would also be created in the capacitor as well...as the plates would have some tiny,tiny resistance....and if the connecting wires sit in a loop in a lab on a table top one could argue a tiny,tiny amount of inductance would also be present. All these effects are normally negligible.

Another interesting situation: With a charged capacitor, no resistance in the circuit, a switch is closed.....now there is "no charge" , no energy, where did all that energy go??

[DaleSpam]

I think, not all of the energy goes to heatWhy not?

As far as other possibilities, electrical energy can be converted into any other kind of energy with the appropriate circuit element. Ideal resistors just convert energy to heat.

[Defennder]

Another interesting situation: With a charged capacitor, no resistance in the circuit, a switch is closed.....now there is "no charge" , no energy, where did all that energy go??Seems like you can't write an equation unless you take into account resistance of the wires. It's basically the same "problem" as what happens when you connect an ideal wire across a voltage source. Between any two points on the wires, is there or is there not a potential difference?

The energy is the capacitor in the case you described is best answered by considering the electrostatic case. On the one hand, there is clearly energy in the separation of charges (the very concept of a capacitor), the E-field does work to pull opposite charges closer together and the energy of the new charge configuration decreases (of course the change in potential energy of the setup is converted to kinetic energy of the movement of the charges).