# Capacitor Discharge Energy

## Main Question or Discussion Point

One of the discussion questions (ie. questions to think about) on a lab that I just did in physics asked where the energy in a charged capacitor goes when the capacitor is discharged. I know the energy is calculated using 1/2*C*(dV)^2, but I cannot figure out where this energy is transferred as we bring dV to zero. Qualitatively (by feeling the discharge wire) I do not think that all the energy lost could be thermal. Any ideas of where else the energy could have gone will be helpful.

Thanks

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What about the kinetic energy of the electrons when the capacitor is discharged? For there to no longer be a potential difference across the plates, charge must flow in the circuit.

I think it is akin to holding a ball in the air. It has potential energy. When you release it (discharging the capacitor), that potential energy is being converted into kinetic energy as it drops.

Kinetic energy for the electrons is right, the energy is used for balancing out the potential difference, i.e. moving the electrons to the other side of the capacitor through the wire, which means release of thermal energy unless you're using a superconductor.

Presumably, I would answer the potential energy (due to the electric field set up between the plate) is used to move the electrons. The case of using a battery to "push" electrons by setting up a potential difference is similar to the discharge of capacitor,while the potential difference drops with time expotentially.
However, How about superconductor? I am not very familiar to superconductor because it is quite rare for me to touch this field. I am only an advanced level student.
What is the kinetic energy of the electrons when current is detected? in the circuit connected to different potential difference?
Is the kinetic energy of the electrons is higher when the e.m.f of the cell is higher?
Is there any relationship between the k.e. of the electrons and the e.m.f of the cell connected and otherwise, the (overall impedance or)resistance of the wire?

If you take a large capacitance from a power supply it may have a 0.01 F capacitance.
If you put 10 volts on it, the energy contained in it is only 0.5 J. If the heat gets produced in a 1 g copper wire, the temperature of it will only go up by 0.5/(0.3846) = 1.3 K (heat capacity of copper 0.3846 J g^-1 K^-1). Try it with a very short stretch of really thin wire and make the rest of the circuit of very thick wire.
If you use a superconductor for the entire circuit you will get a RL circuit that will oscillate for a long time. The circuit will eventually lose energy because of radiation of radio waves, the induction of current in nearby materials, and leakage current.