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

["Tesla list" <tesla@xxxxxxxxxx>]

---------- 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.