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

["Tesla list" <tesla@xxxxxxxxxx>]

---------- Forwarded message ----------
Date: Tue, 31 Jul 2007 06:40:06 -0400
From: Scott Bogard <teslas-intern@xxxxxxxxxxx>
To: tesla@xxxxxxxxxx
Subject: RE: Parallel Capacitor Plates: Force, Pressure and Mass (fwd)

Jared,
     That is cool.  Perhaps you can also tell me why grass blades are 
attracted to and stick to wires charged with 60 Hz high voltage.
Scott Bogard.


>From: "Tesla list" <tesla@xxxxxxxxxx>
>To: tesla@xxxxxxxxxx
>Subject: Parallel Capacitor Plates: Force, Pressure and Mass (fwd)
>Date: Mon, 30 Jul 2007 21:45:03 -0600 (MDT)
>
>
>---------- Forwarded message ----------
>Date: Mon, 30 Jul 2007 23:21:40 -0400
>From: Jared Dwarshuis <jdwarshuis@xxxxxxxxx>
>To: Pupman <tesla@xxxxxxxxxx>
>Subject: Parallel Capacitor Plates: Force, Pressure and Mass
>
>Parallel  Capacitor Plates: Force, Pressure and Mass
>
>
>
>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
>
>
>
>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.
>
>

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