I don't see how lifting the bucket fits.
Lifting it is more like offsetting with a bias.
If you aren't going to add energy to the bucket then your only option to placing the bucket on a higher ladder step is to lower the ladder.
What happens now is the bucket and the water are stretched taller & more narrow OR shorter & wider but not both.
Lifting is analogous to adding voltage to the charge. However, if we are going to look at this like we are talking about a capacitor, there are some tweaks that need to be done. I was a little bit bothered by "W=VQ" and in Grumpy's link the formula is actually W = V*deltaQ and it applies to a case where the electric field constant. In reality when you charge a capacitor by putting current into it the electric field is changing, and the higher the voltage, the harder it is (i.e.; the more work you have to do) to charge it.
For a capacitor we know W = 1/2 C V^2, and C = Q/V.
So that means W = 1/2 (Q/V) V^2 = 1/2 VQ. So in the capacitor case it's "W=1/2VQ" and not "W=VQ."
The following link shows this for a capacitor where you integrate on the charge you put into the capacitor (i.e.; current flowing into the capacitor irrespective of time) and you get the same formula.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c1Another comment with respect to lifting the bucket up to a higher height and the energy that adds to the system. We know that the energy stored in a capacitor is proportional to the square of the voltage (W = 1/2CV^2). However, we also know that the energy in the bucket of water increases linearly with the height (W=MGh). Therefore to keep the "capacitor analogy" holding true for the bucket of water, you have to lift it proportionally higher to reflect the voltage. Suppose one volt equals one foot of height. Then two volts has to equal four feet of additional height so that the amount of energy in the bucket of water resembles the energy associated with the voltage in a capacitor.
In other words, the height of the bucket of water is actually represents the square root of the amount of energy in the equivalent capacitor. i.e.; lifting the bucket up from one foot high to two feet high only represents 1.414 volts for the equivalent capacitor. You have to lift the bucket from one foot high to four feet high to represent two volts in the equivalent capacitor.
Grumpy:
That's it. I am proposing that time-dependent electrostatic induction can accomplish this feat.
Good luck but those are just words. If you actually did experiments on the bench you would find that no matter how you approached the problem, you still end up paying the real energy price to charge the capacitor. There is just no way around it. I am assuming that for both of us this is an academic discussion so let's not fret about it.
This story is analogous to how I showed that an inductor acts exactly like a physical spring, we are just using different "through" and "across" variables to describe the behaviour. Otherwise they are identical. Any person with common sense knows that a spring is not a potential source of energy.
In this case I showed how charging a capacitor is equivalent to lifting up a bucket of water. So the question is can you lift up the bucket of water with less energy than MGh dictates that you have to expend to move it up by "h." Unless you have a source of Upsidasium, the answer is no.
Not that I am stopping anyone from dreaming here... After all Santa is getting ready...
MileHigh
Author
Topic: The Patent Of William Barbat (Read 178793 times)
