Why separation of variables? (electrostatics)

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SUMMARY

The discussion focuses on the application of separation of variables to solve Laplace's equation within a sphere of homogeneous linear dielectric material placed in a uniform electric field E. The participants clarify that Laplace's equation can be applied in regions with zero free charge, as indicated by Griffiths' equation (4.22), which states that the divergence of the displacement field D is zero when free charge density ρ_f equals zero. They also explain that while Gauss's law can be applied, it may not yield useful results due to the absence of symmetries in the dielectric field, ultimately leading to the conclusion that the electric field inside the sphere is uniform and directed along the z-axis.

PREREQUISITES
  • Understanding of Laplace's equation and its applications in electrostatics
  • Familiarity with Griffiths' "Introduction to Electrodynamics" (3rd ed.) and its equations
  • Knowledge of electric displacement field (D) and its relationship with electric field (E) in dielectrics
  • Concept of bound and free charges in dielectric materials
NEXT STEPS
  • Study Griffiths' equation (4.22) in detail to understand the implications of zero free charge
  • Learn about the boundary conditions for electric fields in dielectric materials
  • Explore the application of Gauss's law in different dielectric scenarios
  • Investigate the method of separation of variables in solving partial differential equations
USEFUL FOR

Students and professionals in physics, particularly those specializing in electromagnetism, as well as engineers working with dielectric materials and electric fields.

syang9
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"A sphere of homogeneous linear dielectric material is palcced in an otherwise uniform electric field E. Find the electric field inside the sphere."

Griffiths uses separation of variables to solve laplace's equation in the interior of the sphere. I have two questions.

(1) How can you try to solve laplace's equation? isn't there bound charge in the dielectric (so the charge isn't 0?) laplace's equation can only be solved in regions where the charge is zero, right? or is that only free charge?

(2) Why can't we use gauss's law for dielectrics (integral of the displacement flux equals the enclosed free charge) to find the field? is it because there is no free charge?
 
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(1) it's true that there's bound charge in the dielectric. Would you agree however that there's no free charge in the dielectric. If so, then we can use equation (4.22) of Griffiths (3rd ed.) and obtain:

\nabla \cdot \textbf{D}=0
because \rho_f=0

Since the material is a linear dielectric, then equation (4.32) gives us

\textbf{D} = \varepsilon \textbf{E}

Substitution into our original equation gives us:

\varepsilon \nabla \cdot \textbf{E} = 0
so:
\nabla \cdot \textbf{E} = 0

Then using the conventional -\nabla \phi = \textbf{E}

gives us:
\nabla^2 \phi = 0

(However, the \varepsilon dependence comes into the boundary conditions)

(2) You can, but it won't be very useful because now you don't have any nice symmetries (for example, the field won't be radial, or something. It's true that in the end the field inside just points uniformly in the z direction, but you don't know that to begin with).
 

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