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

Original poster: Finn Hammer <f-h@xxxx>

Antonio, Stephen,All,

This opened my eyes to the source of some problems i have been facing lately.
I put the concept into a spreadsheet and found out that with an air/poly sandwich in an electric field, the field strength in the remaining slot of air increases toward a maximum, which is the total field multiplied by the poly`s dielectric constant.

Electric field in air - polymer sandwich

Poly dielect. constant= 4
Total field kV = 100 Distance mm= 100
Field MV/m 1 airthick polythick airdrop kV polydrop kV airfield
polyfield
1 99 3,88
96,12 3,88 0,97
10 90 30,77 69,23
3,08 0,77
20 80 50,00 50,00
2,50 0,63
30 70 63,16 36,84
2,11 0,53
40 60 72,73 27,27
1,82 0,45
50 50 80,00 20,00
1,60 0,40
60 40 85,71 14,29
1,43 0,36
70 30 90,32 9,68
1,29 0,32
80 20 94,12 5,88
1,18 0,29
90 10 97,30 2,70
1,08 0,27
99 1 99,75
0,25 1,01 0,25


I don`t understand how to add the information in Antonio`s  post into this.

Cheers, Finn Hammer


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.