Electric Field in dielectric

In summary, The electric field inside dielectric and surface charge density on plates can be found using Gauss's law.
  • #1
unscientific
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Homework Statement



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Find the electric field inside dielectric and surface charge density on plates.

Homework Equations


The Attempt at a Solution



[tex]\rho_b = -\vec {\nabla} . \vec{P} = -\vec {\nabla} . \epsilon_0 \chi \vec E = -\vec {\nabla} . (\epsilon_r - 1)\epsilon_0 \vec E[/tex]

[tex]\int \rho_b dV = (1-\epsilon_r)\epsilon_0 \int \vec {\nabla} . \vec E dV[/tex]

[tex] \int \rho_b dV = (1-\epsilon_r)\epsilon_0 \int \vec E . d\vec S [/tex]

[tex]\rho_0 \int (d^2 - x^2) A dx = (1-\epsilon_r)\epsilon_0 EA[/tex]

[tex]E = \frac{\rho_0}{(1-\epsilon_r)\epsilon_0} \int_{x'}^{d} d^2-x'^2 dx'[/tex]

[tex]E = \frac{\rho_0}{(1-\epsilon_r)\epsilon_0}\left[ \frac{2}{3}d^3 - d^2x + \frac{1}{3}x^3\right][/tex]

To find surface charge density,

[tex]\epsilon_0 \nabla . E = \rho_f + \rho_b[/tex]
[tex]\epsilon_0 \int \nabla . E dV = \int \rho_f dV + \int \rho_b dV[/tex]
[tex]\epsilon_0 \int E.dS = \sigma A + \int \rho_b A dx[/tex]
[tex]\epsilon_0 EA = \sigma + \int \rho_b dx[/tex]

Rearranging,

[tex]\sigma = \epsilon_0E - \epsilon_0(1-\epsilon_r)E = \epsilon_0 \epsilon_r E[/tex]
[tex]\sigma = \frac{\rho_0 \epsilon_r}{1-\epsilon_r} \left[ \frac{2}{3}d^3 - d^2x + \frac{1}{3}x^3\right][/tex]
 
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  • #2
The embedded charge density given in the problems is "free charge" density ##\rho_f## of the system, not bound charge density ##\rho_b##.

In your integrals, you are integrating from ##x'## to ##d##. But the upper limit can be tricky to deal with since there is bound surface charge density on the surface of the dielectric at ##x = d## as well as free surface charge density on the surface of the conductor at ##x = d##.

There is quite a bit of symmetry in the problem. Can you use the symmetry to tell you anything about D and E at x = 0? If so, you might want to set up your integrals from ##x = 0## to ##x = x'## and avoid ##x = d##.
 
  • #3
TSny said:
The embedded charge density given in the problems is "free charge" density ##\rho_f## of the system, not bound charge density ##\rho_b##.

In your integrals, you are integrating from ##x'## to ##d##. But the upper limit can be tricky to deal with since there is bound surface charge density on the surface of the dielectric at ##x = d## as well as free surface charge density on the surface of the conductor at ##x = d##.

There is quite a bit of symmetry in the problem. Can you use the symmetry to tell you anything about D and E at x = 0? If so, you might want to set up your integrals from ##x = 0## to ##x = x'## and avoid ##x = d##.

Ah, I see the problem now.

Gauss's law in Dielectric reads:
[tex]\nabla . D = \epsilon_0 \epsilon_r \nabla . E = \rho_f[/tex]

[tex]\epsilon_0 \epsilon_r EA = \rho_0 \int_0^{x'} (d^2 - x'^2) A dx' [/tex]

[tex]E = \frac{\rho_0}{\epsilon_0 \epsilon_r}\left[d^2x - \frac{1}{3}x^3\right][/tex]

For the surface charge, we must compute the bound surface charges on the dielectric first.

2nsumte.png


Gauss's Law reads:

[tex]\epsilon_0 EA = Q - Q_b[/tex]
[tex]\epsilon_0 EA = \sigma A - A\vec P . \hat n[/tex]
[tex]\epsilon_0 E = \sigma - \epsilon_0 (\epsilon_r -1)E[/tex]
[tex]\sigma = \epsilon_0 \epsilon_r E[/tex]

Or in other words, ##E = \frac{\sigma}{\epsilon_0\epsilon_r}##, the electric field is reduced by a factor of ##\epsilon_r## due to the presence of the dielectric's polarized field.

Electric field at d is ##\frac{2}{3}\frac{\rho_0}{\epsilon_0\epsilon_r} d^3##.
Thus, ##\sigma = \frac{2}{3}\rho_0 d^3##.
 
Last edited:
  • #4
I get essentially the same result except for some signs. I get a positive amount of bound surface charge at the surface of the dielectric (due to the polarization of the medium). I get a negative surface charge density on the surface of the metal plate.

If I pick a Gaussian pill box as you did, I would write Gauss' law as

##-\epsilon_0 E_x A = Q_{tot} = Q + Q_b = \sigma A + P_x A = \sigma A +\epsilon_0(\epsilon_r - 1)E_x##
 
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  • #5


I would like to clarify the context of this problem. It appears to be a calculation for the electric field and surface charge density inside a dielectric material, specifically between two parallel plates. The equations used are based on the relationship between the electric displacement field (D), the electric field (E), and the polarization of the dielectric material (P). The first equation shows the relationship between the bound charge density (ρb) and the polarization vector (P), which is then related to the electric field through the dielectric constant (εr). The second equation is a general expression for calculating the electric field inside a dielectric, and the third equation is used to find the surface charge density on the plates. The final equation calculates the surface charge density using the values from the previous equations. Overall, this is a valid approach to solving for the electric field and surface charge density in a dielectric material.
 

1. What is an electric field in a dielectric?

An electric field in a dielectric is a region in space where electrically charged particles, such as electrons and protons, experience a force due to the presence of an electric charge. It is a vector quantity that describes the direction and magnitude of the force on a charged particle.

2. How is the electric field affected by the presence of a dielectric material?

The presence of a dielectric material, such as an insulator, can alter the strength and direction of the electric field. Dielectrics have a property called permittivity, which determines how much the electric field is reduced when it passes through the material. This can either enhance or weaken the overall electric field in the surrounding space.

3. What is the difference between an electric field in a vacuum and in a dielectric?

In a vacuum, the electric field is not affected by any surrounding material and is solely dependent on the electric charge. In a dielectric, the permittivity of the material can alter the strength and direction of the electric field, making it different from the field in a vacuum.

4. How does the concept of polarization relate to the electric field in a dielectric?

Polarization is the process by which the electric dipoles in a dielectric material align themselves in response to an external electric field. This alignment can either enhance or weaken the overall electric field in the surrounding space. The degree of polarization depends on the permittivity of the material and the strength of the electric field.

5. What are some real-world applications of electric fields in dielectric materials?

Electric fields in dielectric materials have a wide range of applications, including capacitors, transformers, and electronic devices. They are also used in medical imaging, such as MRI machines, and in particle accelerators for scientific research. Additionally, the concept of dielectric breakdown is important in understanding the behavior of electrical insulation in high-voltage systems.

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