Re: Energy vanishing into air? (fwd)

[Steven Roys <sroys@xxxxxxxx>]

---------- Forwarded message ----------
Date: Sat, 17 Jan 2004 17:30:06 -0500
From: Alfred Erpel <alfred@xxxxxxxxx>
To: High Voltage list <hvlist@xxxxxxxxxx>
Subject: Re: Energy vanishing into air? (fwd)

Steven,

Could you please explain this better.  If I understand what you said, the
process of rotating the fully charged capacitor to the 0µF capacitance
position requires the expenditure of .05 joules to accomplish this.

equation 1       E =.5CV^2  and then solving for V

equation 2       V = squareroot[2E/C]

equation 3       Q = CV

When the air variable capacitor of .001µF charged to 10,000 volts has the
plates rotated towards 0µF you are lowering C and from equation 2 it appears
that V should increase toward infinity.  And how is equation 3 reconciled
with that?  At the starting values (.001µF charged to 10,000 volts) Q=CV=10;
as I am rotating the capacitor towards 0µF and reach .00001µF, does Q=CV
still = 10?

Regards,
Al



>  Steven Roys <sroys@xxxxxxxx>
> [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]


>> 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