Electric displacement in plate capacitor

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SUMMARY

The electric displacement field D in a parallel plate capacitor with dielectric material and plate charge densities ±σ is defined as D=σ, not D=2σ. The reasoning that considers contributions from both plates leading to D=2σ is flawed, as it results in inconsistencies when deriving the electric field from displacement and bound charges. The correct approach involves recognizing that the displacement field accounts for the free charge only, thus confirming D=σ as the accurate representation.

PREREQUISITES
  • Understanding of electric displacement field (D)
  • Familiarity with Gauss's Law in electrostatics
  • Knowledge of bound and free charges in dielectrics
  • Concepts of parallel plate capacitors
NEXT STEPS
  • Study the derivation of electric displacement field in dielectrics
  • Learn about Gauss's Law applications in electrostatics
  • Explore the relationship between bound charges and electric fields
  • Investigate the behavior of parallel plate capacitors with varying dielectric materials
USEFUL FOR

Students of electromagnetism, electrical engineers, and physicists interested in understanding electric fields and displacements in capacitive systems.

musemonkey
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Suppose there's dielectric material in inside a parallel plate capacitor with plate charge densities ±σ. What's the electric displacement D?

I've seen solutions that draw a Gaussian surface around one of the surfaces and argue that the enclosed charge is Aσ and the flux of D is Aσ therefore D=σ.

But what about the contribution of the other plate? If one takes it into account, it makes an equal contribution yielding a total D=2σ. How is this reasoning flawed? I know it to be wrong because one can derive the field from the displacement and the bound charges and then get the field again by considering the bound and free charges. The two field calculations are inconsistent if D=2σ.

Thanks.
 
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