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Electric Flux Density

D = epsilon*E, displacement vector, flux concept.

Darshan N
Updated: 19 March 2026
12 min read

Electric Flux Density is a field quantity that combines the electric field with the properties of the medium in which it exists. Unlike the electric field intensity E, the flux density D does not change at material boundaries when there is no surface charge, making it a far more convenient quantity for working with problems involving multiple dielectric media. In GATE, D is central to Gauss's Law in integral form and to all boundary condition problems.

Electric Flux Density D and Relation to E+QrDD = ε₀ E (free space) D = ε₀εr E = ε E (medium)For point charge: D = Q / (4π r²) r̂D independent of medium | Unit: C/m²source chargefield point
Figure 1: Electric flux density D at a point due to a positive charge Q — D is independent of the medium permittivity

Core Concept Explanation

The electric flux density D, also called the electric displacement vector, is defined as D = ε E, where ε = ε₀ εr is the absolute permittivity of the medium. In free space, D = ε₀ E. The quantity D accounts for both the free charge that drives the field and the polarisation of the dielectric medium, making it the quantity whose flux through any closed surface equals only the free (not bound) charge enclosed.

For a point charge Q in any medium, D = Q / (4πr²) r̂. This expression has no permittivity in it at all, which is the key insight: D depends only on the source free charge and the geometry, not on the medium. The electric field E = D/ε does depend on the medium. This makes D the natural quantity to work with when the medium changes across a boundary.

Mathematical Expression and Gauss's Law

Gauss's Law in terms of D is expressed as the closed surface integral of D dotted with dS equals the total free charge enclosed Q_enc: closed integral of D · dS = Q_enc. This form of Gauss's Law is general and holds in any medium, including non-uniform dielectrics. The E-based form closed integral of E · dS = Q_enc / ε₀ applies only in free space.

The electric flux Ψ through a surface is defined as Ψ = integral of D · dS. The total flux leaving a closed surface equals the total free charge enclosed. For a point charge Q, integrating D over any sphere of radius r centred on Q gives D × 4πr² = Q, confirming that flux equals Q regardless of r or the medium.

Practical Understanding

When an electric field crosses a boundary between two dielectric media with permittivities ε1 and ε2, the tangential component of E is continuous while the normal component of D is continuous (when there is no surface charge). This is the boundary condition: D1n = D2n at a charge-free interface. Since D = εE, this means ε1 E1n = ε2 E2n, so the normal component of E is discontinuous at the boundary.

In a parallel plate capacitor filled with a dielectric of relative permittivity εr, the electric field is E = V/d, but the flux density is D = ε₀ εr V/d. The charge stored on the plates equals Q = D × A = ε₀ εr A V / d, which directly gives the capacitance C = ε₀ εr A / d. This is the practical consequence of the relationship between D, E, and the dielectric.

Example
Given:
Point charge Q = 5 µC = 5×10⁻⁶ C
Observation distance r = 0.3 m
Medium: dielectric with εr = 4

Why this formula applies:
Point charge in a dielectric — D depends only on Q and r, not on εr.
E depends on εr through E = D/ε.

Formula:
D = Q / (4π r²)  (independent of medium)
E = D / (ε₀ εr)

Substitution for D:
D = (5×10⁻⁶) / (4π × (0.3)²)
D = (5×10⁻⁶) / (4π × 0.09)

Calculation of D:
Denominator = 4 × 3.1416 × 0.09 = 1.131
D = (5×10⁻⁶) / 1.131 = 4.42×10⁻⁶ C/m²

Calculation of E:
ε = ε₀ × εr = 8.854×10⁻¹² × 4 = 3.54×10⁻¹¹ F/m
E = D / ε = (4.42×10⁻⁶) / (3.54×10⁻¹¹)

Final Answer:
D ≈ 4.42 µC/m²  (same in any medium at r = 0.3 m)
E ≈ 1.25×10⁵ V/m  (reduced by factor εr = 4 vs free space)
Exam Tip: For a point charge, D = Q/(4πr²) has no permittivity — it is the same in any medium. E = Q/(4πε₀εr r²) depends on the medium. GATE questions often ask for D at a point in a dielectric: always remember D is medium-independent for a given charge configuration. Using ε in the D formula is the most common error.
D Boundary Conditions at Dielectric InterfaceMedium 1: ε₁ = ε₀ εr1D₁, E₁Medium 2: ε₂ = ε₀ εr2D₂, E₂interfaceD₁nD₂nE₁tE₂tD₁n = D₂n | E₁t = E₂t(no surface charge at interface)
Figure 2: At a dielectric boundary with no surface charge, normal D is continuous and tangential E is continuous
  • D = ε₀ εr E in a dielectric medium; D = ε₀ E in free space. Units of D: C/m².
  • For a point charge: D = Q/(4πr²) r̂ — this is independent of the surrounding medium.
  • Gauss's Law: closed integral of D · dS = Q_enc (free charge only, valid in any medium).
  • Boundary condition: normal D is continuous across a charge-free interface (D₁n = D₂n).
  • Boundary condition: tangential E is continuous across any interface (E₁t = E₂t).

Quick Revision

  • D = ε E = ε₀ εr E; unit is C/m².
  • For point charge: D = Q/(4πr²) — no permittivity, medium-independent.
  • Gauss's Law in integral form: ∮ D · dS = Q_enc (free charges only).
  • Electric flux: Ψ = ∮ D · dS = Q_enc.
  • Boundary conditions: D₁n = D₂n (normal, no surface charge); E₁t = E₂t (tangential, always).
  • Trap: D for a point charge has no ε in it — adding ε in the denominator of D is wrong.
  • Parallel plate capacitor with dielectric: C = ε₀ εr A/d derived from D = ε₀ εr E.

Electric Flux Density

Test your command of the displacement vector D and its independence from permittivity.

Question 1 of 3

Q1.In a medium with relative permittivity εr = 4, the electric field E = 100 V/m. The electric flux density D is: