Re: electric strength for "x" cm of "y" material

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Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>

Hi Steve,

I think you got it right.  Another way to think of it is the 
displacement current is the same on both sides of the boundary 
between the two diaelectrics (at some arbitrary frequency).  So the 
displacement field must also be the same across the boundary.  Since 
E = D/permittivity, the E field must be lower in the poly than in the 
air.  Since the voltage has to all add up to the same (integral of 
E.dl), the E field must be larger in the air with the poly present 
than without.

Gerry R.

> Original poster: Steve Ward <steve.ward@xxxxxxxxx>
>
> Well, say there is (just throwing out random numbers) 50kV between 2 
> conductors and they are seperated by 10cm.  The electric field (with 
> just air) would then be 500kV/m.  But say we introduce a 5cm thick 
> sheet of some poly material where its dielectric constant = 2.  That 
> would make the electric field in the air become twice that of inside 
> the plastic, right? But we must still arrive at our 500kv/m 
> figure.  So there should be 16.6kV "dropped" across the plastic, and 
> 33kV "dropped" across the air... except now we only have 5cm of air, 
> so now the E-field in the surrounding air would be 666kV/m.  I 
> believe this can cause problems in some cases with corona 
> forming.  Please correct me if my theory/understanding is incorrect here.