Capacitors, charge and energy: a question

[Kalee20]

I have a feeling there is something wrong with the V squared part of the formula

Ni, there's nothing wrong with that.

The energy in a capacitor is ½CV²

The charge in a capacitor is CV.

Charge is conserved, but some of the energy is converted to heat.

Kalee20 - again, thank you for your contribution, but to try to bring this matter to a conclusion, I must ask you this: you state that half of the energy is lost: where has it gone?

It's converted to heat. You start with 50µJ, you connect the capacitors via a 1MΩ resistor, you end up with 12.5µJ in each capacitor (25µJ total), and the other 25µJ is lost as heat in the resistor.

Connecting with a 1kΩ resistor, exactly the same applies. You lose half as heat.

As energy lost = 25µJ, independent of R, take the limit as R tends to zero - you get 25µJ lost even with zero resistance.

Where has it gone? Well with non-zero resistance, it's trivial to see it's gone as heat. But try another thought-experiment scenario - a very long superconductor to zap the capacitors in parallel.

In this case, the thing will oscillate (as the superconductor has inductance). But there will be some radiation losses, so eventually the oscillations will decay - approaching the 25µJ retained, 25µJ lost situation. Now shorten the superconductor. Shorter means less inductance, so higher frequency oscillation, but the end result is the same, with things happening over a shorter time frame. In the limit, as the length approaches zero, the frequency of oscillation approaches infinity, the time for it to decay approches zero - but there is still a finite, non-zero loss of energy from the capacitor pair, which gets radiated as a pulse of radio waves, as infra red, as light, as UV, as Xrays, as gamma rays etc, as the length is reduced from very long to very short..

There will always be a mechanism for some of the energy to be converted from electrical energy, to something else.