Re: Parallel Capacitor Plates: Force, Pressure and Mass (fwd)

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---------- Forwarded message ----------
Date: Tue, 31 Jul 2007 13:39:19 +0000
From: stork3264@xxxxxxxxxxx
To: Tesla list <tesla@xxxxxxxxxx>
Subject: Re: Parallel Capacitor Plates: Force, Pressure and Mass (fwd)

Please define your terms.  What are Einstein's mass energy equations for masses at rest and at relativistic velocities?  Lastly, your analysis measures volume between the capacitor plates, but somehow you interpret this volume to be related to an increase mass of a capacitor plate.

Can you give a quantitative experimental example?

RWW

> Larry Morris,  Jared Dwarshuis   (August  07)
> 
> 
> 
> We will examine the force between two parallel capacitor plates where C = e
> A/d.
> 
> 
> 
> Since W= ½ C Vsquared  and W = Fd  then: F = ½ C Vsquared / d
> 
> 
> 
> The E field is said to be uniform then: V = E d  and  Vsquared = Esquared
> dsquared
> 
> 
> 
> So:   F = ½ C  Esquared d  = ½ e A Esquared
> 
> 
> 
> We can find an expression for pressure by dividing by Area:
> 
> 
> 
> Prx = ½  e Esquared
> 
> 
> 
> Now we find an expression for Mass from energy arguments
> 
> 
> 
> Since Energy = M  Csquared    ( this applies when there is no rest mass!)
> 
> 
> 
> Energy / Csquared = Mass
> 
> 
> 
> ½ CVsquared / Csquared = Mass
> 
> 
> 
> ½ e A Esquared dsquared/ d (Csquared) = Mass
> 
> 
> 
> ½ e  E squared Ad / (Csquared) = Mass
> 
> 
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> Thus:  ½ e Esquared Volume/ Speed of light squared = Mass
> 
> 
> 
> Analysis:
> 
> A charged capacitor plate experiences Force and Pressure and has an increase
> in Mass. The Force, Pressure and Mass  decrease as the capacitor discharges.
> 
>