Review of the current field junction at the crossover of two linear surface , such as
Figure 6. The ratio of the relative permitivity environment is εr2 / εr1 = 3. The current density vector J1 in the he first environment , closes with the normal to the split surface angle
α1 = π / 6, and the intensity of this vector is J1 =1μA/mm^2 .
On crossover surface, there is no surface free of charge. Determine the intensity of vector J2 in the 2nd environment
J = σE
tan(α1)/ tan(α2) = ε1 / ε2
The Attempt at a Solution
I don't have relation between σ1 and σ2 , so I don't know how to solve this.
Correct solution is d) 1.732 μA/mm^2