Non-linear isotropic dielectric capacitor

AI Thread Summary
The discussion revolves around the complexities of calculating fields in a non-linear isotropic dielectric capacitor. Participants clarify that the relationship between polarization (P) and electric field (E) can be expressed without assuming linearity, using the equation P = ε₀(χ₁ + χ₃E²)E. The mentor emphasizes the need to start with the electric field (E) when calculating fields between infinite plates and questions whether the voltage-to-distance relationship still holds with a dielectric present. There is also a clarification regarding the direction of polarization and electric fields, with one participant acknowledging a mix-up in their understanding. Overall, the thread highlights the challenges and nuances in working with non-linear dielectrics in capacitor scenarios.
milkism
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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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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 \mathbf{P} relates to \mathbf{E}; the question states <br /> \mathbf{P} = \epsilon_0(\chi_1 + \chi_3E^2)\mathbf{E}. The definition <br /> \mathbf{D} = \epsilon_0\mathbf{E} + \mathbf{P} does not assume a linear relationship between \mathbf{P} and \mathbf{E}.
 
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pasmith said:
You're told how \mathbf{P} relates to \mathbf{E}; the question states <br /> \mathbf{P} = \epsilon_0(\chi_1 + \chi_3E^2)\mathbf{E}. The definition <br /> \mathbf{D} = \epsilon_0\mathbf{E} + \mathbf{P} does not assume a linear relationship between \mathbf{P} and \mathbf{E}.
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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You are told to calculate the fields between the plates. Start with \mathbf{E}. 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?
 
pasmith said:
You are told to calculate the fields between the plates. Start with \mathbf{E}. 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?
 
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 🤣🤣.
 
This is my new solution:
425d9dc3a7bcbe0f6e1a9de8188cf8ca.png

Are these correct?
 
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?
 
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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