Re: Evidence for energy density

On Feb 27, 12:27 am, David Rutherford <drutherf...@xxxxxxxxxxx> wrote:
Richard D. Saam wrote:
Thanks, guys. After a capacitor is charged from a source, I would agree
that the energy density is (1/2)e_0 E^2. But this is only _half_ of the
original energy density of the charge distribution. The other half was
dissipated as heat by the resistor through which the capacitor was
charged.

Modern practice is to design "switching" power supplies, which you can
look up. These are able to convert current from one voltage to another
with negligible losses. The problem of designing a switcher that
would give negligible losses for variable voltage conversion (i.e. so
you could charge a capacitor without resistive losses) is easy
provided
you allow variable capacitors and or variable inductors. So there's no
electrical engineering reason to assume any loss of energy in the
charging.

Switchers got much cheaper in the 1970s. Before then, designs
used transformers and voltage conversion was done with transistors,
which essentially are no better than a resistor. After that time,
higher levels of integration allowed the more complicated switching
circuits to be cheaply manufactured (I think the idea for how to make
a switcher, like most good ideas in electronics, probably dates to the
1930s). Hmmm.

I've always thought the interesting part about the relation between
voltage and energy was in the square. One can think of the similar
relation for the stored energy in a spring, k x^2 / 2 where k is the
spring
constant and x^2 is the square of the displacement.

What's going on in general here is that the force from a potential is
given by the gradient. For the spring, the force is F = kx, so the
potential must be the integral of this, k x^2 / 2. For the capacitor,
the potential is the voltage V = Q/C where Q is the charge and C
is the capacitance. In this case, the displacement is the amount
of charge, and therefore the energy relation looks like the integral
of charge, i.e. Energy = Q^2 / ( 2 C ). You can then convert Q into V
by the previous relation and get the V^2 / 2 relationship.

Please forgive me if this is gibberish; it is quite late, I am quite
tired, and I never did quite get around to taking an undergraduate
E&M class so I've left off units etc.
.

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