Non-linear isotropic dielectric capacitor

In summary, the conversation discusses how to calculate the fields between two infinite flat plates with a prescribed potential difference, taking into account a non-linear dielectric material. It is noted that the formulas for linear dielectrics do not apply and the definition of \mathbf{D} = \epsilon_0\mathbf{E} + \mathbf{P} is used instead. The direction of \mathbf{P} is also discussed, with clarification on the direction of polarisation and dipole moment.
  • #1
milkism
117
15
Homework Statement
a. Identify the location of all the free and bound charges present in the region between the plates of the parallel-plate capacitor, and determine the surface charge densities associated with them.

b. Determine the values of the polarization (P), electric displacement (D), and electric field (E) in the region between the plates.
Relevant Equations
See solution.
Question:
390be9a42062505da4c826f1c6296336.png

Solution first part:
bf91efd69f7327b257515688a07ad370.png

Have I done it right?

I don't know how to begin with second part since the dielectric is non-lineair, and most formulas like $$
D=\epsilon E$$ and $$P= \epsilon_0 \xhi_e E$$, only apply for lineair dielectrics. What to do?
 
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  • #2
milkism said:
I don't know how to begin with second part since the dielectric is non-lineair, and most formulas like $$
D=\epsilon E$$ and $$P= \epsilon_0 \chi_e E$$, only apply for lineair dielectrics. What to do?

You're told how [itex]\mathbf{P}[/itex] relates to [itex]\mathbf{E}[/itex]; the question states [tex]
\mathbf{P} = \epsilon_0(\chi_1 + \chi_3E^2)\mathbf{E}.[/tex] The definition [tex]
\mathbf{D} = \epsilon_0\mathbf{E} + \mathbf{P}[/tex] does not assume a linear relationship between [itex]\mathbf{P}[/itex] and [itex]\mathbf{E}[/itex].
 
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  • #3
pasmith said:
You're told how [itex]\mathbf{P}[/itex] relates to [itex]\mathbf{E}[/itex]; the question states [tex]
\mathbf{P} = \epsilon_0(\chi_1 + \chi_3E^2)\mathbf{E}.[/tex] The definition [tex]
\mathbf{D} = \epsilon_0\mathbf{E} + \mathbf{P}[/tex] does not assume a linear relationship between [itex]\mathbf{P}[/itex] and [itex]\mathbf{E}[/itex].
So basically $$\mathbf E = \frac{\mathbf P}{\epsilon_0 \chi_1 + \chi_3 E^2}$$
and
$$\mathbf D= \mathbf P \left( \frac{1}{ \chi_1 + \chi_3 E^2} + 1 \right)$$
The question doesn't really clarify what the fields should be on terms of what. Would you do what I did also? (Have no idea why it won't latex.)

Mentor (@Mark44) note: I fixed the LaTeX. Please let me know if it's what you intended.
 
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  • #4
You are told to calculate the fields between the plates. Start with [itex]\mathbf{E}[/itex]. You are asked to assume that the plates are infinite. Do you know how to find the field between a pair of infinite flat plates a fixed distance apart with a prescribed potential difference between them?
 
  • #5
pasmith said:
You are told to calculate the fields between the plates. Start with [itex]\mathbf{E}[/itex]. You are asked to assume that the plates are infinite. Do you know how to find the field between a pair of infinite flat plates a fixed distance apart with a prescribed potential difference between them?
Does V/d still apply when there's a dielectric between?
 
  • #6
milkism said:
So basically $$\mathbf E = \frac{\mathbf P}{\epsilon_0 \chi_1 + \chi_3 \mathbf E^2}$$
and
$$\mathbf D= \mathbf P \left( \frac{1}{ \chi_1 + \chi_3 \mathbf E^2} + 1 \right)$$
The question doesn't really clarify what the fields should be on terms of what. Would you do what I did also? (Have no idea why it won't latex.)

Mentor (@Mark44) note: I fixed the LaTeX. Please let me know if it's what you intended.
Yes, thank you, but E shouldn't be vectored, we don't want to divide vectors by vectors 🤣🤣.
 
  • #7
This is my new solution:
425d9dc3a7bcbe0f6e1a9de8188cf8ca.png

Are these correct?
 
  • #8
milkism said:
1682795495735.png

From the relation ##\mathbf{P} = \epsilon_0\left( \chi_1 + \chi_3 E^2 \right) \mathbf E##, shouldn't ##\mathbf P## have the same direction as ##\mathbf E##?

You say that "##\mathbf P## goes from negative to positive". Can you elaborate on this? Which positive and negative charges are you referring to here?
 
  • #9
TSny said:
From the relation ##\mathbf{P} = \epsilon_0\left( \chi_1 + \chi_3 E^2 \right) \mathbf E##, shouldn't ##\mathbf P## have the same direction as ##\mathbf E##?

You say that "##\mathbf P## goes from negative to positive". Can you elaborate on this? Which positive and negative charges are you referring to here?
Doesn't polarisation go from negative charge to positive charge, whereas electric field goes from positive to negative?
 
  • #10
milkism said:
Doesn't polarisation go from negative charge to positive charge, whereas electric field goes from positive to negative?
Omg I mixed up direction of dipole moment with direction of polarisation, sorry.
 
  • #11
milkism said:
Yes, thank you, but E shouldn't be vectored, we don't want to divide vectors by vectors 🤣🤣.
Fixed. :smile:
 
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1. What is a non-linear isotropic dielectric capacitor?

A non-linear isotropic dielectric capacitor is an electronic component that stores electrical energy by utilizing a non-linear dielectric material, which has varying electrical properties depending on the strength of the electric field. This type of capacitor is different from a linear capacitor, which has a constant dielectric constant regardless of the electric field strength.

2. How does a non-linear isotropic dielectric capacitor work?

A non-linear isotropic dielectric capacitor works by storing electrical energy in the form of an electric field between two conductive plates separated by a non-linear dielectric material. When a voltage is applied, the dielectric material becomes polarized, creating an electric field that stores energy. The amount of energy stored is determined by the dielectric constant of the material and the distance between the plates.

3. What are the advantages of using a non-linear isotropic dielectric capacitor?

One advantage of using a non-linear isotropic dielectric capacitor is that it can store more energy compared to a linear capacitor of the same size. This is because the non-linear dielectric material can withstand higher electric fields, allowing for a higher energy density. Additionally, non-linear capacitors can also have a wider range of operating temperatures and frequencies compared to linear capacitors.

4. What are the applications of a non-linear isotropic dielectric capacitor?

Non-linear isotropic dielectric capacitors are commonly used in high voltage and high frequency applications, such as in power electronics, telecommunications, and medical equipment. They are also used in energy storage systems, such as in electric vehicles and renewable energy systems, due to their high energy density.

5. Are there any limitations or drawbacks of using a non-linear isotropic dielectric capacitor?

One limitation of using a non-linear isotropic dielectric capacitor is that the non-linear dielectric material can introduce non-linearities in the circuit, which can affect the performance of the overall system. Additionally, these capacitors can be more expensive compared to linear capacitors and may have a shorter lifespan due to the stress placed on the dielectric material.

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