Questions: Electric Field Inside Dielectric Materials

Question 1

A parallel-plate capacitor is filled with a dielectric of constant κ = 4. Compared to the same capacitor in vacuum with identical free charge on its plates, the electric field between the plates is:

• A. Four times stronger, because the dielectric material amplifies the applied field
• B. Unchanged — the dielectric affects capacitance but not the electric field inside
• C. Reduced by a factor of 4, because polarization produces bound charges that oppose the external field
• D. Reduced by a factor of 2, because the dielectric splits the field symmetrically between two surfaces

Answer: (C)

The most common misconception is that the dielectric leaves the field unchanged. In fact, polarization creates bound surface charges whose field OPPOSES the external field, reducing E_inside to E_vacuum/κ. With κ = 4, the field is one-quarter of the vacuum value. The capacitance increases by κ precisely because the same charge produces a smaller E (and therefore smaller voltage V = Ed), while C = Q/V increases.

Question 2

At a boundary between two dielectrics (κ₁ = 2 and κ₂ = 5) with no free surface charge, which statement correctly describes the boundary conditions?

• A. Both the normal component of E and the normal component of D are continuous across the boundary
• B. The normal component of D is continuous; the normal component of E is discontinuous by factor κ₁/κ₂
• C. The tangential component of D is continuous; the normal component of E is also continuous
• D. Both E and D are fully continuous across the boundary whenever free charges are absent

Answer: (B)

With no free surface charge, ∇·D = ρ_free = 0 at the boundary, so the normal component of D is continuous (D₁ₙ = D₂ₙ). But since D = ε₀κE, if D_n is continuous and κ changes, E_n must be discontinuous by factor κ₁/κ₂. Separately, from ∇×E = 0 in electrostatics, the TANGENTIAL component of E is always continuous — but the tangential component of D is discontinuous. D and E satisfy opposite boundary conditions: D_normal continuous, E_tangential continuous.

Question 3

Inside a dielectric material, the electric field is unchanged compared to the vacuum field — the dielectric mainly modifies the displacement field D.

• T. True
• F. False

Answer: False

This is explicitly identified as a misconception. The E field IS reduced inside the dielectric by the factor κ: E_inside = E_vacuum/κ. Polarization produces bound charges that create an opposing field, weakening the net E. The displacement field D = ε₀κE is introduced precisely because its divergence depends only on free charges — making it convenient for calculations — but E is the physically observable field that is genuinely modified by the material.

Question 4

The displacement field D is useful for solving dielectric problems primarily because its divergence depends only on free charges, not bound charges.

• T. True
• F. False

Answer: True

∇·D = ρ_free means Gauss's law in terms of D reads ∮D·dA = Q_free_enclosed — identical in form to the vacuum version but with D replacing ε₀E. When designing capacitors or waveguides, you typically know the free charges placed on conductors but not the bound charges induced by the field you are trying to find. Working with D sidesteps this circular dependence and makes the problem tractable.

Question 5

Why does the presence of bound charges at dielectric surfaces weaken the electric field inside the material?

Model answer: When a dielectric is placed in an external electric field, its molecules polarize — their charge distributions shift, forming aligned dipoles. At the surfaces, the net effect is exposed bound charges: positive on the face pointing toward the negative source, negative on the other. These bound surface charges produce their own electric field directed OPPOSITE to the external field. The total E inside is the superposition of the external field and this opposing bound-charge field, giving E_inside = E_external/κ — always weaker than the applied field.

This is the physical origin of the dielectric constant κ: the material's polarization response creates an internal opposing field proportional to the applied field, and κ characterizes how strongly the material polarizes. High κ means strong polarization, a large opposing bound-charge field, and a much-weakened interior E. Understanding this physical picture explains why κ > 1 always weakens (never strengthens) the field, and why it appears as a divisor in E_inside = E_vacuum/κ.