Electric field inside a polarized sphere

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SUMMARY

The discussion focuses on determining the electric field inside and outside a polarized sphere with a polarization defined as \(\vec{P} = k\vec{r}\). It is established that the electric field outside the sphere is zero due to the cancellation of polarization charges. Inside the sphere, the electric field cannot be assumed to be \(-\frac{1}{3\epsilon_0} \vec{P}\) because the polarization is not uniform. Instead, the bound charge density \(\rho_b = -\vec{\nabla} \cdot \vec{P}\) must be calculated, and Gauss' law should be applied to find the electric field as a function of the radial distance \(r\).

PREREQUISITES
  • Understanding of electric polarization and bound charge density
  • Familiarity with Gauss' law in electrostatics
  • Knowledge of vector calculus, specifically divergence
  • Basic concepts of electric fields in dielectric materials
NEXT STEPS
  • Calculate the bound charge density \(\rho_b\) for the given polarization \(\vec{P} = k\vec{r}\)
  • Apply Gauss' law to derive the electric field \(E(r)\) inside the polarized sphere
  • Explore the implications of non-uniform polarization on electric fields
  • Study the behavior of electric fields in uniformly polarized spheres for comparison
USEFUL FOR

Students and professionals in physics, particularly those studying electromagnetism, as well as educators seeking to clarify concepts related to electric fields in polarized materials.

Fabio010
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Homework Statement



A sphere of radius R carries a polarization
\vec{P}= k\vec{r},

where k is a constant and \vec{r} is the vector from the center.


Find the field inside and outside the sphere.


In solution, the field outside sphere is 0.

I interpreted that as the field produced by the polarization charges.

Their sum is 0. So, the outside field is zero.

My problem is inside sphere.

Can i consider always the electric field inside a sphere as -\frac{1}{3εo}*\vec{P}
??


Because the formula -\frac{1}{3εo}*\vec{P} is obtained by a uniformly polarized sphere.

In the exercise \vec{P} is not uniform.
 
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Since there is no free charge anywhere inside the sphere, I would think the E field, by Gauss, would be zero everywhere inside the sphere.
 
Fabio010 said:
Can i consider always the electric field inside a sphere as -\frac{1}{3εo}*\vec{P}
??

Not sure where you're getting that expression. The polarization produces a bound charge density inside the sphere: ##\rho_b = -\vec{\nabla}\cdot \vec{P}##

From the bound charge you can get the field using Gauss' law.
 
So you are telling me that \vec{E}(A) = \frac{∫-∇.\vec{P}dv}{εo}
 
Last edited:
Fabio010 said:
So you are telling me that \vec{E}(A) = \frac{∫-∇.\vec{P}dv}{εo}

Yes (although I'm not real clear on your notation in this equation). See what you get for the bound charge density as a function of ##r## and then use it in Gauss' law to find the magnitude of the field inside, ##E(r)##, as a function of ##r##.
 

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