Current density in two dielectric environments

[Ivan Antunovic]

Homework Statement

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

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Homework Equations

J = σE
tan(α1)/ tan(α2) = ε1 / ε2

The Attempt at a Solution

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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

 

[anathema]

Hello,

Thank you for your post. After reviewing the information provided, it appears that the problem is asking for the intensity of vector J2 in the second environment. To solve this, we can use the equations J = σE and tan(α1)/ tan(α2) = ε1 / ε2.

First, let's find the electric field in the first environment using J = σE, where J1 = 1μA/mm^2 and σ1 is unknown. This gives us E1 = J1 / σ1.

Next, using the given ratio of relative permitivity (εr2 / εr1 = 3) and the equation tan(α1)/ tan(α2) = ε1 / ε2, we can find the value of tan(α2). This will give us the angle α2 for the second environment.

Now, using the known angle α2 and the known electric field E1, we can solve for the intensity of vector J2 in the second environment using J = σE.

I hope this helps and please let me know if you have any further questions.