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
118
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?
 
Physics news on Phys.org
  • #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].
 
  • Like
Likes vanhees71 and berkeman
  • #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.
 
Last edited by a moderator:
  • #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:
 
  • Like
Likes milkism

FAQ: Non-linear isotropic dielectric capacitor

What is a non-linear isotropic dielectric capacitor?

A non-linear isotropic dielectric capacitor is a type of capacitor that uses a dielectric material with non-linear properties, meaning its permittivity changes with the applied electric field. The term "isotropic" indicates that the dielectric material has the same properties in all directions.

How does the non-linearity of the dielectric material affect the performance of the capacitor?

The non-linearity of the dielectric material causes the capacitance to change with the applied voltage. This can lead to a more complex relationship between voltage and charge compared to linear dielectrics, potentially affecting the energy storage and discharge characteristics of the capacitor.

What are the applications of non-linear isotropic dielectric capacitors?

Non-linear isotropic dielectric capacitors are used in applications where variable capacitance is desirable, such as in tunable circuits, voltage-controlled oscillators, and dynamic energy storage systems. They are also used in power electronics and signal processing where non-linear behavior can be exploited for specific functionalities.

What materials are typically used for the dielectric in non-linear isotropic dielectric capacitors?

Materials such as ferroelectric ceramics (e.g., barium titanate), certain polymers, and composite materials are commonly used as dielectrics in non-linear isotropic dielectric capacitors. These materials exhibit significant changes in permittivity with applied electric fields.

What are the challenges in designing non-linear isotropic dielectric capacitors?

Designing non-linear isotropic dielectric capacitors involves challenges such as accurately modeling the non-linear behavior of the dielectric material, ensuring stability and reliability under varying operating conditions, and managing thermal effects due to non-linear dielectric heating. Additionally, manufacturing processes must be carefully controlled to maintain the isotropic properties of the dielectric material.

Back
Top