# 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.

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