Hoping that this isn't homework, here goes:
Given a coaxial capacitor (I'm assuming that's what you are referring
to) with length much greater then diameter, outside diameter of inner
cylinder = a, inside diameter of outer cylinder = b and voltage across
the plates (inner and outer cylinder) = V
Ea = V/( a * ln ( b/a ) )
which is the voltage field strength at the surface of the inner
conductor. Assuming you know the system voltage V and the maximum field
strength (humidity and other factors will affect his for a practical
problem), I'll leave it up to you to solve for a.

--
Paul Hovnanian mailto: snipped-for-privacy@Hovnanian.com
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The "rule" is this class of homework problems was to understand what was
"conserved" (i.e.: constant).
In this case we have two "constants." The first, is the voltage drop
between the two electrodes and that is just the integral of the E-filed
between the two equi-potential surfaces. The second is the "flux" which is
the surface integral of E-field. The "surface" would be an imaginary
cylinder between the two electrodes.
When I "studied" this 45 years ago, I actually enjoyed solving these
problems.

I don't think the flux is constant for varying cylinder geometry. It
would depend on the capacitance between them and the resulting charge
stored for a given voltage. The capacitance varies with the geometry.

--
Paul Hovnanian mailto: snipped-for-privacy@Hovnanian.com
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The "flux" would be constant as E depends on rho/r where rho is the charge
per unit length and r is the radius and the surface area per unit length
would be proportional to the radius. It all works out (Shades of potential
coefficients for capacitance of power lines parallelling the inductance
"flux based" model).

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Don Kelly snipped-for-privacy@shawcross.ca
remove the X to answer

The capacitance of a coaxial line per unit length is
C = 2PIe / ln ( B/A )
where B is the outer cylinder radius and A is the inner cylinder radius.
For a fixed V across the plates and a fixed B, as A increases so does C.
As C increases, the charge per unit length increases (for the same V).
Since we assumed that B, and therefore the outer cylinder surface area
is constant, the charge per unit area and therefore flux increases.

--
Paul Hovnanian mailto: snipped-for-privacy@Hovnanian.com
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