How do I get this interface condition?

[svletana]

Homework Statement

I have an infinitely long cylinder of a dielectric material, surrounded by another dielectric material and coated with graphene which has surface conductivity $\sigma$, implying it has a superficial current. The sheet of graphene is very thin, and the dielectrics are asumed to be isotropic. There's an incident p-polarised monochromatic plane wave in the x direction, and I have to show the interface conditions are:
$$\frac{1}{\epsilon_1} \frac{dF_1}{dr} = \frac{1}{\epsilon_2} \frac{dF_2}{dr}$$

$$F_2 - F_1 = \frac{4i \pi}{\omega \epsilon_1} \sigma \frac{dF_1}{dr}$$

where F is the H field in the z direction, the subindex 1 indicates the material inside the cylinder and 2 outside the cylinder. I'm having trouble with the second one.

Homework Equations

The boundary conditions given is that the tangential component of E (tangential to the normal vector of the surface of the cylinder) is continuous, and that the tangential component of H is proportional to the surface conductivity density.

I have tried using Maxwell equations: $$\nabla \times H = \frac{4 \pi}{c}J - \frac{i \omega}{c} \epsilon E$$

The Attempt at a Solution

The $\phi$ component is: $$-\frac{dH_z}{dr}= \frac{4 \pi}{c}J_{\phi} - \frac{i \omega \epsilon}{c} E_{\phi}$$

If I do the integral between R- and R+ (R is the radius of the cylinder) then the left side would look like $F_2 - F_1$, which is what I'm looking for, but what about the right side? I feel like I have to use the boundary conditions given, but I'm not sure how. Please help! Thank you!

 

[mark726]

Thank you for your question. I understand that you are having trouble understanding the second interface condition for the infinitely long cylinder of dielectric material surrounded by another dielectric material and coated with graphene.

Firstly, I would like to clarify that the boundary conditions you have mentioned are correct. The tangential component of E and H must be continuous across the interface, and the tangential component of H is proportional to the surface conductivity density.

To solve for the second interface condition, we can start by using the tangential component of H. We know that the tangential component of H is proportional to the surface conductivity density, which in this case is represented by \sigma. Therefore, we can write:

H_{t2} = \sigma E_{t2}

Where H_{t2} is the tangential component of H outside the cylinder and E_{t2} is the tangential component of E outside the cylinder.

Next, we can use Maxwell's equations to relate the tangential components of H and E. From \nabla \times H = \frac{4 \pi}{c}J - \frac{i \omega}{c} \epsilon E, we can obtain:

H_{t2} = \frac{4 \pi}{c} \sigma E_{t2} - \frac{i \omega}{c} \epsilon E_{t2}

Substituting this into the first equation, we get:

\frac{4 \pi}{c} \sigma E_{t2} - \frac{i \omega}{c} \epsilon E_{t2} = \sigma E_{t2}

Solving for E_{t2}, we get:

E_{t2} = \frac{4 \pi}{i \omega \epsilon} \sigma E_{t2}

Now, we can use this expression to solve for the second interface condition. The tangential component of E is continuous across the interface, therefore:

E_{t1} = E_{t2}

Substituting the expression for E_{t2} into this equation, we get:

E_{t1} = \frac{4 \pi}{i \omega \epsilon} \sigma E_{t1}

Finally, we can use this expression to obtain the second interface condition:

F_2 - F_1 = \frac{4 i \pi}{\omega \epsilon_1} \sigma \frac{dF