Conductors separated by interface

[captain.joco]

Homework Statement

The Plane z=0 forms an interface between two conductors of conductivity σ1 and σ2 respectively. In the 1st conductor a steady current J1 flows and it makes an angle θ1 with the normal of the interface. Find the current in the 2nd conductor and the charge density on the interface.

Homework Equations

The Attempt at a Solution

I don't know the shapes of the conductors and for some reason this poses a problem for me. I know that from Ohms lawJ = σE, so the current and the electric field are in the same direction. I see no reason why the second current shouldn't be just
J2 = σ2E where the electric field is the one from the 1st current ( unless there should be a contribution from the field from the interface...?! )

Any help will be much appreciated. Thank you!

 

[Jhero]

Thank you for your question. First, it is important to clarify that the electric field in a conductor is actually zero, as any electric field will cause the electrons to move and redistribute themselves until the electric field is canceled out. Therefore, in a steady state, the electric field inside a conductor is zero.

Now, to answer your question, the current in the second conductor can be found using the continuity equation, which states that the current flowing into a point must be equal to the current flowing out of that point. In this case, the current flowing into the interface from the first conductor must be equal to the current flowing out of the interface into the second conductor. Therefore, we can say that J1 = J2.

Next, to find the charge density on the interface, we can use the fact that the electric field is perpendicular to the interface and the electric field is related to the charge density by the equation E = σ/ε0, where σ is the charge density and ε0 is the permittivity of free space. Therefore, we can say that the charge density on the interface is σ = ε0E.

I hope this helps. If you have any further questions, please let me know.