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Registered Member #1792
Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527
IamSmooth wrote ...
In the water column example, potential energy is M*g*h.
What is the integral of the potential energy?
I didn't explain myself well enough. For each particle of mass m in the column, it will have a potential equal to m*g*h, where h is the height. For any given height, every particle on that plane will have the same PE, so multiply by the cross-sectional area of the column for the total PE of water at that height. Let M = m*A, so PE as a function of h is M*g*h. Now each of these thin planes of water is at a different PE, because they are at different heights. Integrate the function M*g*h with respect to h, between h=0 and h=h1, where h1 is the height of the column. You then get a total PE of 1/2*M*g*h^2.
Radiotech: in this two capacitor problem, energy is indeed conserved. However, some of initial electrostatic potential energy will be converted into heat through parastic resistance, or flow out of the circuit from radiation. A Marx Generator is a different system because you are not moving energy between the capacitors. If you do the calculations for two caps of size C with voltage V, E=1/2*2*C*V^2 = C*V^2. When you put them in series, voltage double, and there is now only one capacitor of half the size, so E=1/2*(C/2)*(2V)^2 = CV^2 as before, so no energy moved anywhere.
Registered Member #2463
Joined: Wed Nov 11 2009, 03:49AM
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Posts: 1546
I see your point on the Marx. How about an experiment with too circular plates made like a folded up fan,ie an insulated shaft could open up the fan plates to increase the area facing when the device was charged at say A1 to A2, A being area, yet maintaining constant separation, the voltage between the plates will drop, the energy in the condenser maintained and the work in expanding the area, torque * distance per unit time, should equal the temperature rise in the metal plates due to charge redistribution. But if the process was adiabatic and the plates did not increase in temperature, would the voltage between the plates change disproportionately indicating the mechanical energy was being stored?
As an aside I have a wind-up charger for a dosimeter, which is an electrostatic generator, and like a Van de Graf column needs work to turn the belt pulley.
That old 2 capacitor problem is akain to the arguments in explaining how the electrophoresis works, considering the work needed to lift the handle.
Fun stuff that gets serious in these days of nanotechnology where at that size the electric forces are a hugh factor.
Registered Member #1792
Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527
In the case of time-varying capacitance what matters is conservation of charge. If there is no conduction path (no current in or out), then the net charge on the + and - plates will remain constant. So if capacitance halves, Q=CV, so voltage must have doubled for charge to remain constant. But since E=0.5CV^2, that operation will double your energy. So you had to do work equivalent to what was in the capacitor. Equivalently, if capacitance doubles, the voltage halves, and the capacitor is charging half the energy it used to be.
You can actually make a generator using this principle if you want to. It takes a fair amount of effort though to get any useful kind of power density.
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where did the energy go?
