Grumpy:

I am just going to comment on your "off topic" discussion here for the fun of it. You said:

So, based on the simple equation V=W/Q, means that W=VQ.

W = work in joules

V = volts (between two points and considered a force here)

Q = charge

Using arbitrary numbers, if it normally takes 10 joules of work to move 1 Coulomb of charge with a force of 10 volts then:

W=VQ and 10=10*1

So, if I use a different method of moving this charge that requires less work, then I can move more charge with the same amount of work. This is the same result if I increase the "force" (voltage) and the work remains the same.

W=V(10*Q) or W=(10*V)Q

Either way, you get more for less.

Let's try to look at this using good old analogies. The equation states that work is equal to voltage times charge.

Now we know that power is the "through variable" times the "across variable." In the past I have stated that the "through variable" is current (which flows through something) and voltage is the "across variable" (you measure voltage across two points).

Also, to get to energy, we know that it is equal to power times time, or, the "through variable" times the "across variable" times time.

It all makes sense, because (current x time) = charge.

Therefore, work = power x time = (voltage x current) x time = voltage x (current x time) = voltage x charge.

So, you can look at that as saying it takes work to take a certain amount of charge and "lift it up" to a higher voltage. (Side note, when you think about it that's exactly what a battery does)

So let's use a simple analogy where electric current is equivalent to the flow of water. Charge is then simply an amount of water, say for example a bucket of water. Let's say that voltage is equivalent to the height. (Note that you always measure height between two points.)

So here is an analogy for W=VQ plugging in your values where 10 = 10*1:

You have a big bucket of water sitting on the floor. You lift the bucket up and put it on a 10-foot high shelf. That's it.

Note that the act of moving the bucket of water up 10 feet could have been done by pumping the water through a hose also. Hence you can envision "current flow" when you lift the bucket up by hand and move it.

So leading into the question: You often speculate that somehow you can do a work-around to "trick" the system and as a result gain energy.

So, if I use a different method of moving this charge that requires less work

So the question is how do you raise the bucket of water 10 feet up in the air with a work-around that uses less energy?

MileHigh