Electric displacement of a uniformly polarized sphere

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SUMMARY

The discussion focuses on calculating the electric field of a uniformly polarized sphere with radius R and polarization vector \(\mathbf{P}\). The established solution is \(\mathbf{E} = \frac{-1}{3 \epsilon_0} \mathbf{P}\). The application of Gauss' Law reveals that \(\mathbf{D} = 0\) and leads to the incorrect conclusion that \(\mathbf{E} = \frac{\mathbf{P}}{\epsilon_0}\). The inconsistency arises from the lack of spherical symmetry in the polarization, necessitating a boundary value problem approach to accurately determine the electric potential and field.

PREREQUISITES
  • Understanding of electric displacement field (\(\mathbf{D}\)) and electric field (\(\mathbf{E}\)) concepts
  • Familiarity with Gauss' Law and its applications
  • Knowledge of boundary value problems in electrostatics
  • Basic principles of polarization in dielectric materials
NEXT STEPS
  • Study the derivation of electric fields in polarized materials
  • Learn about boundary value problems in electrostatics
  • Explore the implications of non-symmetrical polarization on electric fields
  • Review the mathematical formulation of Gauss' Law and its limitations
USEFUL FOR

Physicists, electrical engineers, and students studying electromagnetism, particularly those interested in the behavior of polarized materials and electric fields in non-symmetrical configurations.

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Homework Statement



Suppose we have a uniformly polarized sphere of radius R. The polarization vector is [tex]\mathbf{P}[/tex].

Goal: To find the electric field everywhere.

Homework Equations


I know the actual solution already. It should be [tex]\mathbf{E} = \frac{-1}{3 \epsilon_0} \mathbf{P}[/tex]

The Attempt at a Solution


However, if we use Gauss' Law [tex]\oint \mathbf{D} \cdot d\mathbf{A} = Q_{enc,free}[/tex].

We get [tex]Q_{enc,free}=0[/tex]

Hence, [tex]\mathbf{D}=0[/tex] and [tex]\epsilon_0 \mathbf{E}=\mathbf{P}[/tex]

Therefore, [tex]\mathbf{E} = \frac{\mathbf{P}}{\epsilon_0}[/tex].

But these two values are different.
 
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You can't use Gauss' law for the D-field if you don't have a symmetrical problem.In other words, since the polarisation is not spherically symmetric Gauss' law for the D-field with the free charge on the right hand side is not valid. Remember that curl of D is not zero, but equal to minus the curl of P if i remember correctly.

To find the E-field everywhere you can set up a boundary value problem for the electric potential with appropriate boundary conditions at the boundary of the sphere.
 

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