Questions: Dielectric Susceptibility and Permittivity

Question 1

A parallel-plate capacitor holds fixed free charge Q on its plates. The gap is filled with a dielectric of relative permittivity εᵣ = 4. Compared to vacuum, the electric field E inside the dielectric is:

• A. Four times larger — the dielectric enhances the field
• B. Four times smaller — the bound surface charges partially cancel the free charges
• C. Unchanged — the field only depends on free charge
• D. Zero — the dielectric screens the field completely

Answer: (B)

When the free charge is fixed, Gauss's law in terms of D gives ∇·D = ρ_free, and D is determined by free charge alone. In the dielectric, D = ε₀εᵣE, so E = D/(ε₀εᵣ) = E_vacuum/εᵣ. With εᵣ = 4, the field is one-quarter of its vacuum value. The polarization of the dielectric creates bound charges that oppose the applied field — this is the physical mechanism. A common misconception is that D and E scale together; they don't, because D = ε₀εᵣE, not D = ε₀E.

Question 2

What is the primary purpose of introducing the displacement field D = ε₀E + P in electrostatics?

• A. D is easier to measure experimentally than E
• B. D satisfies ∇·D = ρ_free, separating the contribution of free charges from bound (polarization) charges
• C. D is always continuous across dielectric interfaces, unlike E
• D. D replaces E entirely inside dielectric materials

Answer: (B)

The D field is constructed specifically so that its divergence equals only the free charge density: ∇·D = ρ_free. This isolates the sources you control (the free charges you place on conductors or in space) from the bound charges created by polarization, which are harder to track directly. This clean separation is what makes boundary value problems tractable — you apply boundary conditions on D (normal component continuous across uncharged interfaces) and E (tangential component continuous) without needing to account for bound charge explicitly. D does not replace E; both are needed for a complete description.

Question 3

In vacuum, the electric susceptibility χ = 0 and therefore D = ε₀E.

• T. True
• F. False

Answer: True

Vacuum has no material to polarize, so P = 0. Since P = ε₀χE, we have χ = 0. Then D = ε₀E + P = ε₀E + 0 = ε₀E, and εᵣ = 1 + χ = 1 + 0 = 1. This is consistent: in vacuum, D and E carry exactly the same information, differing only by the constant ε₀. The D field only 'adds value' over E when there is a polarizable medium present.

Question 4

A material with larger relative permittivity εᵣ has a stronger electric field E inside it for the same distribution of free charges, because larger permittivity means a stronger material response.

• T. True
• F. False

Answer: False

This is exactly backwards. A larger εᵣ means a stronger polarization response, which generates bound charges that partially cancel the applied field. For fixed free charges, E = D/(ε₀εᵣ); since D is fixed by the free charges (∇·D = ρ_free), a larger εᵣ gives a smaller E. Water (εᵣ ≈ 80) reduces the electric field inside it to 1/80 of its vacuum value. High permittivity means strong polarization, which means strong field reduction — the opposite of what the misconception claims.

Question 5

Explain why the displacement field D is useful for solving electrostatic problems involving dielectrics, rather than working with E alone.

Model answer: When a dielectric is present, the total electric field E is generated by both free charges (the ones you control) and bound charges (induced by polarization, distributed throughout the material). Tracking bound charges separately is cumbersome. The displacement field D = ε₀E + P is defined so that ∇·D = ρ_free — its sources are only the free charges. This means you can apply Gauss's law using only the known free charge distribution to find D, then use D = ε₀εᵣE to find E. Boundary conditions also separate cleanly: the normal component of D and tangential component of E are each continuous across an uncharged interface.

The D field is a computational convenience that captures the effect of polarization implicitly through the permittivity. It does not represent a new physical field independent of E and P — it is a linear combination of them. But that combination happens to satisfy a simpler law (∇·D = ρ_free) that makes problems tractable. This is a common pattern in physics: introduce a combination of fields that satisfies a simpler equation, even if the combination has no independent physical meaning.