Calculating streamer breakout of top-loads

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Original poster: "Gavin Dingley" <gdingley-at-ukf-dot-net> 

Hi all,
I was playing around with some math to figure out the voltage at which 
corona will form around a sphere. First, am I right in thinking that in 
general air breaks down at an electric field strength of 3MV/m?

Here is that math I was playing around with:

Electric field strength (E) in free-space is given by    E = D / e0    equ 1

where D is the electric flux density in C/m^2, and e0 is the permitivity of 
free space equal too 8.85e-12 F/m

But electric flux density is given by    D = Q / A    equ 2

where Q is the charge on the top-load, and A is the surface area.

For a sphere the surface area is given by    A = 4 pi r^2

so equ 2 becomes    D = Q / ( 4 pi r^2 )    equ 3

The charge on a top-load (Q) is given by    Q = C V    equ 4

where C is the isotropic capacity of the top-load, and V is the voltage 
applied to it (the potential between the top-load and ground)

For a sphere the isotropic capacity is given by    C = 4 pi r e0

So equ 4 becomes    Q = 4 pi r e0 V

and in turn equ 3 becomes    D = ( 4 pi r e0 V ) / (4 pi r^2)

which reduces to    D = ( eo V ) / r

Substituting this into equ 1 becomes

     E = D / e0    equ 1

     but

     D = ( eo V ) / r

     therefor

     E = ( e0 V ) / r e0

     E = V / r

     and so

     V = E r


For air this becomes    V = 3e6 r

if V is in volts and r is in metres

or    V = 75 r

if V is in kV and r is in inches.

That is a 2 inch radius sphere will break out at V = 75 * 2 = 150 kV

Is this correct?


Thanks in advance,

Gavin