Bound surface charge on a linear dielectric half-cylinder

In summary, part d) is important to solving part b) of the homework. If we can use Gauss's law to find D then part b) should be easy. If we can't then I suppose I will have to use boundary values to solve for P somehow.
  • #1
Daniel Sellers
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Homework Statement


Problem statement in attached photo. This is an ungraded assigned problem which I am using to study for an exam, so I don't need the whole solution just help with a couple of points I am confused about.

One: Part d) is really important to how I will answer part b). If we can use Gauss's law to find D then part b) should be very easy. If we can't then I suppose I will have to use boundary values to solve for P somehow in order to find the bound charges.

I don't see any reason why we shouldn't be able to, but one (very flawed) online solution said we could not, and the fact that we are asked the question makes me nervous about proceeding this way, so I wanted to ask some of the critical experts that frequent this forum.

Two: Why is the fact Xe << 1 relevant? This also seems to point me towards using boundary value equations to solve the problem, but I'm simply not familiar enough with this material yet to confidently set up the problem.

Homework Equations



P = ε0χeE

Gauss's law for electric displacement?

D = ε0E + P = εE

The Attempt at a Solution



I am fairly confident that I have solved part a) correctly and that I can solve the rest if the above points are clarified
 

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  • #2
Daniel Sellers said:

Homework Statement


One: Part d) is really important to how I will answer part b). If we can use Gauss's law to find D then part b) should be very easy. If we can't then I suppose I will have to use boundary values to solve for P somehow in order to find the bound charges.

I don't see any reason why we shouldn't be able to...

In order to use Gauss' law, there needs to be enough symmetry so that there will exist a particular shape of the Gaussian surface that will allow you to simplify ##\oint \mathbf{D \cdot da}##. So, think about whether there is sufficient symmetry in this problem.

Two: Why is the fact Xe << 1 relevant?
This will allow you to use approximations in which you only need to be accurate to first order in ##\chi_e##. For example, it should be easy to find an expression for the polarization ##\mathbf{P}## that is accurate to first order in ##\chi_e## if you know ##\mathbf{E}## due to the line charge.
 
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  • #3
When considering the free charge only, we have cylindrical symmetry, but that symmetry does not apply to the entire problem. So for now I will abandon using Guass's law.

As for χe, assuming that it is very small let's us say that ε ≈ ε0. This would suggest that the electric field in the dielectric is approximately the field we would see from the charged wire alone.

Then P ≈ χeλ/(2πr) in the radial direction.

Is this close to what you meant?
 
  • #4
Daniel Sellers said:
When considering the free charge only, we have cylindrical symmetry, but that symmetry does not apply to the entire problem. So for now I will abandon using Guass's law.
Right.

As for χe, assuming that it is very small let's us say that ε ≈ ε0. This would suggest that the electric field in the dielectric is approximately the field we would see from the charged wire alone.

Then P ≈ χeλ/(2πr) in the radial direction.

Is this close to what you meant?
Yes, exactly what I meant. :oldsmile:
 
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  • #5
Terrific! Back to practicing electrostatics problems then.

Thanks for your help!
 

1. What is bound surface charge on a linear dielectric half-cylinder?

Bound surface charge on a linear dielectric half-cylinder refers to the electric charge that is accumulated on the surface of a half-cylinder made of a dielectric material. This charge is caused by the polarization of the dielectric material in response to an external electric field.

2. How is bound surface charge on a linear dielectric half-cylinder calculated?

The bound surface charge on a linear dielectric half-cylinder can be calculated using the formula: σ = χeε0E, where σ is the bound surface charge, χe is the electric susceptibility of the dielectric material, ε0 is the permittivity of free space, and E is the external electric field.

3. What factors affect the bound surface charge on a linear dielectric half-cylinder?

The bound surface charge on a linear dielectric half-cylinder can be affected by factors such as the electric field strength, the dielectric constant of the material, the shape and size of the half-cylinder, and the orientation of the half-cylinder with respect to the electric field.

4. How does the bound surface charge on a linear dielectric half-cylinder affect the electric field around it?

The bound surface charge on a linear dielectric half-cylinder can create an electric field that is opposite in direction to the external electric field. This can lead to a distortion of the electric field lines around the half-cylinder, causing a change in the overall electric field distribution in the surrounding area.

5. What applications does the study of bound surface charge on a linear dielectric half-cylinder have?

The study of bound surface charge on a linear dielectric half-cylinder has applications in various fields such as electrical engineering, materials science, and physics. It can help in understanding the behavior of dielectric materials in electric fields and designing devices such as capacitors, insulators, and electronic components.

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