Charged wire inside a linear dielectric

In summary, the boundary conditions for the E field in this scenario are that the tangential components of the electric field must be continuous across the boundary surface and the surface charge density must be equal on both sides of the boundary.
  • #1
jennyjones
35
0

Homework Statement



see attachment 2

Homework Equations



a.


The Attempt at a Solution



see atachment 1


i am stuck at problem a. the question is to write down the boundary conditions for that the E field has to satisfy infinetly far away from the wire and at the boundary of the dielectric z = 0.

I'm not sure how to do this, any help is welcome

thanks

Jenny
 

Attachments

  • Foto op 10-05-14 om 12.28.jpg
    Foto op 10-05-14 om 12.28.jpg
    31.6 KB · Views: 532
  • Schermafbeelding 2014-05-10 om 12.34.04.jpg
    Schermafbeelding 2014-05-10 om 12.34.04.jpg
    49.5 KB · Views: 570
Physics news on Phys.org
  • #2
Dat mag niet. Groetjes B. N.
 
  • #3
You have the following equations, valid everywhere
[tex]\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{D}=\rho, \quad \vec{D}=\epsilon \vec{E}.[/tex]
In your case you have
[tex]\epsilon=\Theta(z)+\epsilon_{r} \Theta(-z).[/tex]
To get the boundary conditions you just apply Stokes's theorem to the first equation with an arbitrary rectangular surface perpendicular to the surface. This leads to the condition that the tangential components of the electric field must be continuous across the surface, i.e., in your case
[tex]\vec{e}_z \times [\vec{E}(x,y,0^+)-\vec{E}(x,y,0^-)]=0,[/tex]
because [itex]\vec{e}_z[/itex] is the normal vector of the dielectric's boundary surface.

Further you use an infinitesimal cube parallel to the boundary on the 2nd equation. This yields the boundary condition
[tex]\vec{e}_z [\vec{D}(x,y,0^+)-\vec{D}(x,y,0^-)] = \sigma.[/tex]
Here [itex]\sigma[/itex] is the surface charge density on the surface of the dielectric.
 

FAQ: Charged wire inside a linear dielectric

What is a charged wire inside a linear dielectric?

A charged wire inside a linear dielectric refers to a conducting wire that has an electric charge and is surrounded by a material that can be polarized by an electric field, known as a dielectric. This creates an electric field around the wire, causing the dielectric to become polarized.

How does a charged wire inside a linear dielectric affect the electric field?

A charged wire inside a linear dielectric can significantly alter the electric field. The presence of the wire causes the electric field lines to bend and concentrate near the wire, resulting in a higher electric field strength compared to the same wire in a vacuum.

What is the significance of a linear dielectric in this scenario?

The linear dielectric plays a crucial role in this scenario as it allows for the polarization of the material around the charged wire. This polarization results in a net electric field that can either enhance or reduce the overall electric field around the wire, depending on the properties of the dielectric.

How is the strength of the electric field affected by the distance from the charged wire?

The strength of the electric field decreases as the distance from the charged wire increases. This is because the electric field lines spread out as they move away from the wire, resulting in a weaker electric field at greater distances.

Can a charged wire inside a linear dielectric produce a magnetic field?

Yes, a charged wire inside a linear dielectric can produce a magnetic field. This is due to the movement of the electric charges in the wire, which creates a magnetic field that is perpendicular to the direction of the electric field. However, the strength of this magnetic field is typically much weaker than the electric field produced by the wire.

Back
Top