Given: Relative permittivity of PE insulation ε = 2.2;
Cable outer diameter D = 10 cm;
HV conductor diameter d_{HV} = 2 cm (initial value);

Problem: Determine the optimal HV conductor diameter d_{HV} for getting minimal electric field stress in the cable insulation at the inner conductor surface.

Solution: Analytical solution states that the ratio D / d_{HV} should be equal to e (2.71828..).
Built-in LabelMover tool is used to perform the optimization. Electric field strength value is measured in the single reference point (labelled as "point") on the HV conductor surface.

Results: LabelMover indicates that the inner conductor diameter should be scaled up 1.834 times (which is 2 cm*1.834 = 3.668 cm)
The calculated ratio of outer to inner diameter is 10 cm / 3.668 cm = 2.726 (close enough
to the analytical result 2.718).
Electric field stress distribution in the insulation with optimal HV conductor dimensions is shown below