Rotational Kinetic Energy and Conservation of Momentum

[cmb]

The quote below comes from the textbook:

Nelkon & Parker Advanced Level Physics Fourth Edition

"The energy comes from the battery.This supplies an amount of energy equal to QV during the charging process.Half of this energy goes to the capacitor.The other half is transferred to heat in the circuit resistance.If it is a high resistance the transfer is made quickly;if it is a low resistance the transfer is made slowly.In both cases,however,the total amount of heat produced is the same,0.5QV".

Whatever the source, IT IS STILL WRONG in terms of how the capacitor stores charge. I've explained how the energy in a capacitor builds up, by the masses-on-steps analogy. Think for yourself and recognise some things in print are not right. This is one of them.

 

[Ken G]

Actually, that paragraph from Nelkin and Parker is not wrong, it simply makes a key assumption that we must recognize: it assumes the battery maintains the voltage V the whole time. That's what creates the 50% loss of heat, the very same thing would happen if we stretch a spring into equilibrium by maintaining a fixed force F on the end of the spring the whole time, using kinetic friction to keep the spring stretching at a slow speed.

 

[cmb]

But if the voltage is at V the whole time, then no matter how small the initial resistance, an energy QV must appear somewhere, as that is the work done by a battery that is always at V.

Quite so, but bear in mind that circuits very rarely contain a capacitor exposed to a source so stiff that the voltage stays 100% whilst it is low on the cap. There are always inductive and internal resistance effects such that the circuit never sees 'V' until the capacitor is charged. Whatever the resistance in the circuit, as soon as the capacitor starts charging, the voltage across the load, thus the current, diminishes yet the argument saying '0.5QV always goes into heating' is, I think, must be predicated on a constant current load because the losses across the resistor do not drop off linearly with an increase in the capacitors voltage, but by the square of the differential voltage across the resistance load.

{edit; Ken G beat me to the same point whilst I was typing!}

 

[Ken G]

Quite so, but bear in mind that circuits very rarely contain a capacitor exposed to a source so stiff that the voltage stays 100% whilst it is low on the cap. There are always inductive and internal resistance effects such that the circuit never sees 'V' until the capacitor is charged.

I think the standard situation is what Nelkin and Parker have in mind-- there is some slight resistance in the circuit, such that the current is low enough that the battery can support V the whole time. If one makes that assumption, their statement is correct (and even cute-- it doesn't matter how fast the capacitor is charged, 50% of the energy is lost-- given the above assumptions). Of course, the voltage across the capacitor is not V the whole time, that's what we have to integrate self-consistently-- it is the voltage across the battery that stays V, and that's what controls the work done by the battery, not the work done on the capacitor. The same could be said for a spring attached to a mass on sandpaper being stretched into equilibrium with a constant force F. So we agree that Dadface was incorrectly interpreting the significance of the Nelkin and Parker statement in terms of this thread, but the Nelkin and Parker statement is correct when properly interpreted.