Lost energy when capacitor discharges

[Tekk]

Homework Statement

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I have two capacitors: 1-Farad capacitor A, and 9-Farad capacitor B.
In state (a), I charged A with 1 Coulomb of charges, as depicted in figure (a).
In state (b), I paralleled A with B, by doing this I discharged A, as depicted in figure (b).

I try to find the total energy of state (a) and state (b). From what I have calculated, the system lost 90% of its energy when it goes from (a) to (b). Where does the energy go?

Homework Equations

Q=C\phi
U=(1/2)C\phi^2

The Attempt at a Solution

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Please refer to the image below.

 

[mfb]

Where does the energy go?

You heat the wires and the capacitors from the resistances in the setup. A tiny bit is also lost to electromagnetic radiation.

 

[Tekk]

You heat the wires and the capacitors from the resistances in the setup. A tiny bit is also lost to electromagnetic radiation.

I think my approach to find energy assumes there is no resistance on wires. So I suspect that energy would not turn to heat on these wires.

What I see during the discharge process is a flow of electrons. Can you explain how electromagnetic radiation is produced?

 

[mfb]

Your approach is independent of the resistance.
Connecting capacitors like that without any resistors is a very special case, it will give you an extremely good oscillator (as you always have a small inductance) and all the lost energy will be emitted as radiation over time.

Can you explain how electromagnetic radiation is produced?

You have a variable current flow = accelerated charges. With small resistances, current will oscillate back and forth for a while.