Hi kamenges
I must admit you make a compelling case for the lost energy in the charge transfer brainteaser. Good math back-up on it, too. I just don't know if I can buy it.
The reason for my skepicism is the two base equations you start with:
Q = C * V and E = 0.5 * C * V^2
have no means built into them to account for any of the losses discussed so far and yet they correctly predict the result.
Yes, conservation of energy using these equations can be used to correctly predict the energy dissipation without knowing the mechanism for energy dissipation. I see no problem with that. Regardless of where the energy goes… I know that if potential energy goes down by E0/2 than energy E0/2 has been transferred somewhere. That is the principle of conservation of energy.
None of the other energy conversion models correctly anticipate the final result of a test. If, for example, I roll a car down a hill, I should be able to predict is final speed using basic, Newtonian equations. But I can't because none of those equations account for friction or other losses. So the car will not go as far as I predict it would. How can we assume that the idealised capacitance equations, which do not account for losses, correctly predict the final outcome and also attribute 1/2 the total energy distribution in the system to losses?
Let’s say the car starts on top of a hill and then is allowed to roll into a valley between two hills where it will oscillate and finally come to rest. I don’t have to know the exact energy dissipation mechanism (air friciton, rolling friction, etc) to know that the total energy dissipated is equal to the gravitational potential energy (m*g*h).
Knowing the initial capacitance C and voltage V is equivalent to knowing the initial height of the car… it allows me to calculate initial potential energy.
Looking at it a different way, if I simply reduced the voltage applied from an infinite energy source (which would replace any lost energy) to 1/2 it's original value, the single original cap would store 1/4 of the original energy, just like it would if we discharged it into another equivalent cap. This in spite of the fact that it is connected to an infinite energy source that will replenish any losses.
I agree with that. No matter how you change the voltage, the potential energy formula PE=0.5*C*V^2 will still hold. I’m not sure what your the dilemma or contradiction is.
Maybe here is a different way to look at it. PE=0.5*C*V^2. Substituing V=Q/C we have PE = 0.5*C*(Q/C)^2 = 0.5*Q^2/C. The basic behavior which gave rise to our dlima is that potential energy does NOT vary linearly with Q (it actually varies with Q^2). So there is NOT a fixed amount of PE per charge regardless of voltage. So I can NOT expect in general to be able to transfer charges from one cap to another without changing the total energy.
Why shouldn’t each unit of charge carry an equal unit of PE? deltaPE = V*deltaq….where V itself is a function of total stored charge Q (V=Q/C). As a charge up the cap my V is increasing so each incremental unit of charge requires more and more energy to place it on the cap. Total PE is integral V*dq from q = 0 to Q = integral q/Cdq from 0 to Q = [0.5q^2/c] evaluated at Q minus [0.5q^2/c] evaluated at zero = 0.5*Q^2/C. Possibly drawing that integral graphically as a right triangle area shows the intuition that each unit of charge added (or removed) from a cap does not require the same amount of energy.
Once again... modeling resisitve energy dissipation was my first choice based on math ease in modeling and relative simplicity to understand. I don't rule out other possiblities like electromagnetic which will may also dissipate significant energy depending on the experimental setups. No matter what, we can rest assured that energy conservation will hold.
