Re: Energy vanishing into air? (fwd)

[Steven Roys <sroys@xxxxxxxx>]

---------- Forwarded message ----------
Date: Sat, 17 Jan 2004 13:43:59 -0800
From: Jim Lux <jimlux@xxxxxxxxxxxxx>
To: High Voltage list <hvlist@xxxxxxxxxx>
Subject: Re: Energy vanishing into air? (fwd)

> Original poster: Steven Roys <sroys@xxxxxxxx>
> From: Alfred Erpel <alfred@xxxxxxxxx>
> Howdy All,
>
>     Imagine an air variable capacitor with plates fully engaged and the
> capacitor fully charged.
>
>     Let's use the example of .001µF charged to 10,000 volts. Stored energy
=
> joules = .5CV^2 = .05 watt=seconds.
>
>     What happens to the energy of the fully charged capacitor when the
> plates are rotated to the 0µF capacitance position?
>
> Regards,
>
> Al Erpel
>
> [Potential energy.  It would take that much energy to rotate the cap
> plates to the 0uF position (barring friction losses), and rotating them
> back would convert the energy back from potential to the energy stored in
> the electrical field.  SRR]
>
the voltage rises on the capacitor as the capacitance decreases, until,
eventually, the capacitor fails from breakdown of the air or dielectric.
This is how Wimshurst and other induction charging machines work.

The charge is constant, so CV has to remain the same.

As Steve adds, energy is CV^2/2 (or, if you like, QV/2), Since CV(Q) is
constant, and V is rising, the energy is increasing, so you have to put work
into the system