Boundary Conditions - Cylinder in dielectric

In summary, the boundary conditions for this problem are that the tangential components of the electric field must be equal inside and outside the surface, and the difference in perpendicular components of the electric displacement is equal to the surface charge density divided by the permittivity of free space. To find the potential everywhere, we first find the potentials inside and outside the surface. For the potential inside, we have a finite value at r=0 and 0 at r=0, leading to the coefficients a_0 and b_0 being equal to 0. For the potential outside, we have a term that goes to 0 as r approaches infinity, leading to only the n=1 term existing. The coefficients c_n is also equal to
  • #1
unscientific
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Homework Statement



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Part (a): List the boundary conditions
Part (b): Show the relation for potential is:
Part (c): Find Potential everywhere.
Part (d): With a surface charge, where does the Electric field disappear?

Homework Equations


The Attempt at a Solution



Part (a)

Boundary conditions are:
[tex]E_{in}^{||} = E_{out}^{||}[/tex]

[tex]D_{out}^{\perp} - D_{in}^{\perp} = \frac{\sigma}{\epsilon_0}[/tex]

2jd3yae.png


Part (c)

First we find potential inside and outside.

1. For potential inside ##V_{in}##, it is finite when r = 0, so ##b_n = 0##. When r = 0, ##V_{in} = 0##, so ##a_0 = b_0 = 0##.

2. For potential outside ##V_{out}##, as r → ∞, ##V \propto \frac{1}{r}## (From part (b)). So only term n = 1 exists.

3. For n=1, and ##\phi = \frac{\pi}{2}##, ##V_{in} = V_{out} = 0##, so sine term goes away: ##c_n = 0##.

Now we have the two potentials:

[tex]V_{in} = a_1 r cos(\phi)[/tex]
[tex]V_{out} = \frac{d_1}{r} cos(\phi)[/tex]

Applying first boundary condition at r = a,

[tex]\frac{\partial V_{in}}{\partial \theta} = \frac{\partial V_{out}}{\partial \theta}[/tex]

[tex] a_1 a^2 = d_1[/tex]

Applying second boundary condition at r = a,

[tex]D_{out}^{\perp} = D_{in}^{\perp} [/tex]
[tex]\epsilon_0 \frac{\partial V_{in}}{\partial r} = \frac{\partial V_{out}}{\partial r}[/tex]
[tex] \epsilon_0 a_1 = -\frac{\epsilon d_1}{a^2}[/tex]

Solving them, this implies ##\epsilon_0 + \epsilon = 0##, which doesn't make sense.
 
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  • #2
I've got a feeling that my potential inside is missing a term like the one in part (b): ##V = \frac{\lambda}{2\pi \epsilon_0}\frac{d cos \phi}{r}##

Because it shouldn't strictly depend only on ##cos \phi##?
 
  • #3
I think the potential inside should be something like:

[tex]V_{in} = Ar cos \phi + \frac{B}{r^2}cos \phi[/tex]

That should solve it.
 

1. What are boundary conditions in relation to a cylinder in a dielectric?

Boundary conditions refer to the set of rules that govern the behavior of electromagnetic fields at the interface between two different media, such as a cylinder in a dielectric material. These rules determine how the electric and magnetic fields behave at the boundary between the two media.

2. How do boundary conditions affect the behavior of a cylinder in a dielectric?

The boundary conditions at the interface between a cylinder and a dielectric material determine the distribution of electric and magnetic fields around the cylinder. They also affect the capacitance and inductance of the cylinder, as well as the propagation of electromagnetic waves through the material.

3. What are the boundary conditions for a cylinder in a dielectric?

The boundary conditions for a cylinder in a dielectric are the continuity of tangential electric and magnetic fields, as well as the discontinuity of normal electric and magnetic fields at the interface between the two media. This means that the tangential components of the fields must be continuous across the boundary, while the normal components will have a jump in value.

4. How do boundary conditions differ for a cylinder in a dielectric compared to other shapes?

The boundary conditions for a cylinder in a dielectric are similar to those for other shapes, such as a sphere or a plane. However, the specific values of the fields and the equations used to calculate them will differ based on the geometry of the object. For example, the boundary conditions for a cylinder will involve the Bessel functions, while those for a sphere will involve spherical harmonics.

5. What are some practical applications of understanding the boundary conditions for a cylinder in a dielectric?

Understanding the boundary conditions for a cylinder in a dielectric is essential for designing and analyzing devices such as transmission lines, antennas, and waveguides. It also has applications in areas such as radar, telecommunications, and medical imaging. Additionally, understanding boundary conditions can help predict and control the behavior of electromagnetic waves in complex systems.

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