--- title: "Electric vectors E, P and D" subject: "Engineering Physics" chapter: 7 key: "semester_1/engineering_physics:dielectric-vectors-d-e-p" source: https://www.bctengineer.com/q/semester_1/engineering_physics/dielectric-vectors-d-e-p site: BCT Engineering --- # Electric vectors E, P and D Engineering Physics · Chapter 7: Electrostatics - Asked in: 2074 Chaitra, 5 marks; 2072 Kartik, 5 marks; 2068 Shrawan, 5 marks (asked 3 times) - Video: [Relation between dielectric polarisation, displacement vector and electric field intensity](https://www.youtube.com/watch?v=NaYUFVZ0DKs) — Physics by Kapoor Sir - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/dielectric-vectors-d-e-p - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7 ## Question Define the three electric vectors E, P and D and develop a relation between them (D = ε0E + P). (Discuss the behaviour of dielectrics in a parallel plate capacitor using Gauss' law.) ## Short answer E is the total electric field intensity (V/m), P the polarization (dipole moment per unit volume, C/m²), and D = ε₀E + P is the electric displacement (C/m²) whose flux depends only on free charges. In a parallel-plate capacitor with dielectric, Gauss’s law gives σf = ε₀E + P, leading to D = ε₀ε_r E. ## Answer #### The three vectors - **Electric field intensity E:** force per unit positive test charge; it is the total field in the dielectric, produced by all (free and bound) charges. Unit: V/m. - **Polarization P:** electric dipole moment per unit volume of the dielectric. Unit: C/m². For a linear isotropic dielectric P = ε₀ χₑ E, where χₑ is the electric susceptibility. - **Electric displacement D:** the vector whose flux depends only on free charge; defined as D = ε₀E + P. Unit: C/m². #### Relation D = ε₀E + P using a parallel-plate capacitor ```text + + + + + + + + + plate, free charge +q (σf = q/A) - - - - - - - - - bound charge -σp (dielectric surface) [ dielectric ] + + + + + + + + + bound charge +σp - - - - - - - - - plate, free charge -q ``` 1. Plates carry free surface charge density ±σf. A dielectric slab fills the gap. It becomes polarized, and bound charges of surface density −σp (near the positive plate) and +σp (near the negative plate) appear on its faces. 2. The bound-charge density equals the polarization: σp = P. 3. Apply Gauss' law with a pillbox enclosing the surface of the positive plate and part of the dielectric. The net charge per unit area enclosed is σf − σp: ε₀ E = σf − σp = σf − P 4. Rearrange: **σf = ε₀E + P** 5. Define D by the Gauss' law for free charge, ∮ D·dA = q_free, so D = σf. Therefore **D = ε₀E + P** #### Dielectric behaviour in the capacitor - With P = ε₀χₑE: D = ε₀(1 + χₑ)E = ε₀ε_r E = εE, where ε_r = 1 + χₑ is the dielectric constant. - The field falls to E = E₀/ε_r (E₀ is the field without the dielectric) because the bound charges oppose the free charges. - At the same charge, the potential difference falls by ε_r, so the capacitance increases to C = ε_r C₀. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the polarization P? A. Force per unit test charge B. Total free charge per unit area on a plate C. Electric dipole moment per unit volume of the dielectric D. Electric dipole moment per unit charge Answer: C. P is dipole moment per unit volume (C/m^2); for linear dielectrics P = ε0 χe E. (Key point: Polarization) 2. What is special about the displacement vector D? A. It depends only on bound charge B. Its flux depends only on the free charge: ∮D·dA = q_free C. It equals E in all media D. It is the same as P Answer: B. D is defined so that Gauss's law for D involves free charge only. (Key point: Meaning of D) 3. In the capacitor derivation, which step leads to D = ε0E + P? A. Gauss's law gives ε0E = σf - σp with σp = P, so σf = ε0E + P = D B. ε0E = σf + σp, so D = ε0E - P C. σp = ε0E, so D = P D. The bound charge is ignored in the pillbox Answer: A. The enclosed charge per unit area is the free charge minus the bound charge on the slab face, and σp = P. (Key point: Deriving D = ε0E + P) 4. At the same plate charge, what does inserting the dielectric do? A. E rises to εr E0 and C falls B. E stays the same and C rises C. E falls and C falls by εr D. E falls to E0/εr and C increases to εr C0 Answer: D. Bound charges oppose the free charges, lowering E and V while C = q/V rises by εr. 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