[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

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



---------- 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=  C Vsquared  and W = Fd  then: F =  C Vsquared / d



The E field is said to be uniform then: V = E d  and  Vsquared = Esquared
dsquared



So:   F =  C  Esquared d  =  e A Esquared



We can find an expression for pressure by dividing by Area:



Prx =   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



 CVsquared / Csquared = Mass



 e A Esquared dsquared/ d (Csquared) = Mass



 e  E squared Ad / (Csquared) = Mass



Thus:   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.