advanced27 min studyLesson 238 in Physics

Match the right field components at a dielectric interface

Use displacement for free charge and electric field for total charge.

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What you’ll understand

  • Derive tangential electric and normal displacement boundary conditions.
  • Separate free interface charge from bound polarization charge.
  • Solve a two-layer capacitor with a consistent voltage constraint.

A dielectric interface usually bends an electric field without making either vector continuous in every component. The tangential electric field is continuous in electrostatics, while the normal displacement jumps by the free surface charge. Bound charge is already incorporated through polarization. Keeping these roles distinct lets one solve layered capacitors and refracted field patterns without accidentally discarding the physical charge that causes the normal electric field to change.

A loop and a pillbox ask different questions

Take a thin rectangular loop crossing an interface. In the electrostatic limit its circulation vanishes; shrinking its height leaves equality of the tangential electric fields. Next place a thin pillbox across the same interface and apply the displacement version of Gauss’s law. Its flux jump equals the free surface charge. Choose one normal, pointing from material 1 into material 2, and keep it fixed for both limiting fields.

For homogeneous isotropic linear materials, displacement equals absolute permittivity times electric field. Thus without free surface charge the normal displacement is continuous, while the normal electric field generally changes inversely with permittivity. Relative permittivity is dimensionless and must be multiplied by vacuum permittivity in SI formulas. Potential is continuous across an ordinary interface with finite fields, even when its normal derivative jumps. An ideal dipole sheet would require a separate potential-jump model.

n×(E2−E1)=0,n⋅(D2−D1)=σf,Di=ϵiEi\mathbf n\times(\mathbf E_2-\mathbf E_1)=0,\qquad \mathbf n\cdot(\mathbf D_2-\mathbf D_1)=\sigma_f,\qquad \mathbf D_i=\epsilon_i\mathbf E_i
n points from 1 to 2, εi is absolute permittivity in F/m, E is in V/m, D in C/m², and σf is free surface charge in C/m². The fields are static.↔ Scroll wide equations horizontally.

Bound charge has not disappeared

The displacement is defined as vacuum permittivity times electric field plus polarization. Its divergence counts free charge because the divergence of polarization accounts for bound charge with the opposite sign. If there is no free surface charge, a discontinuity of normal electric field can still occur because of bound surface charge. Declaring that every surface charge is zero whenever normal displacement is continuous confuses two different conservation statements.

At an interface the net bound surface charge is the normal component of polarization in material 1 minus that in material 2. Combining it with free surface charge recovers the vacuum-permittivity jump condition for electric field. This is a useful audit: compute the electric-field discontinuity using permittivities, then check that the resulting total charge equals free plus bound contributions. Interface sign errors often become obvious when both descriptions are compared.

D=ϵ0E+P,σb=(P1−P2)⋅n,ϵ0(E2n−E1n)=σf+σb\mathbf D=\epsilon_0\mathbf E+\mathbf P,\qquad \sigma_b=(\mathbf P_1-\mathbf P_2)\cdot\mathbf n,\qquad \epsilon_0(E_{2n}-E_{1n})=\sigma_f+\sigma_b
P is polarization in C/m². σb denotes the combined bound sheet at the interface with the fixed 1→2 normal, not either material’s independently oriented surface formula.↔ Scroll wide equations horizontally.

Local matching needs a global condition

A two-layer parallel-plate capacitor has constant normal displacement when the internal interface has no free charge. Each layer’s field is that common displacement divided by its own permittivity. The total imposed voltage is the sum of the field times thickness in each layer. Interface conditions alone do not determine the field magnitude: the external plate voltage or plate charge supplies the missing global constraint. Fringing is neglected only when plate dimensions greatly exceed layer thicknesses.

Use the vector lab to inspect components rather than assume full-vector continuity. Reset and set angle to 0° and magnitude to 2. Reset and set angle to 60° and magnitude to 2; predict the x projection falling from 2 to 1 while the y projection grows. These are independent vector examples, not a material-interface protocol. The lab neither enforces displacement matching nor contains a permittivity control, so material calculations must use the equations here.

ΔV=Dn(d1ϵ1+d2ϵ2),CA=(d1ϵ1+d2ϵ2)−1\Delta V=D_n\left(\frac{d_1}{\epsilon_1}+\frac{d_2}{\epsilon_2}\right),\qquad \frac CA=\left(\frac{d_1}{\epsilon_1}+\frac{d_2}{\epsilon_2}\right)^{-1}
ΔV is the positive voltage drop along the field, di are layer thicknesses in m, and C/A is capacitance per area in F/m². The ideal planar layers have no internal free sheet charge.↔ Scroll wide equations horizontally.

Audit a field crossing

A flat interface has ε₁=2ε₀, ε₂=4ε₀, no free surface charge, and normal from 1 to 2. Just inside material 1, E₁n=6 V/m and E₁t=3 V/m. Find E₂ and the net bound interface charge.

  1. Normal displacement continuity gives 2ε₀×6=4ε₀E₂n. Therefore E₂n=3 V/m. Tangential field continuity independently gives E₂t=3 V/m; neither rule sets the full electric vectors equal.

  2. The normal polarizations are P₁n=(ε₁−ε₀)E₁n=6ε₀ V/m and P₂n=(ε₂−ε₀)E₂n=9ε₀ V/m. The interface’s net bound sheet is P₁n−P₂n=−3ε₀ V/m.

  3. With ε₀≈8.85×10⁻¹² F/m this charge is −2.66×10⁻¹¹ C/m². Check the total-charge law: ε₀(E₂n−E₁n)=ε₀(3−6), the same negative value. No free charge was needed to produce the field jump.

The result

The second field has normal and tangential components both 3 V/m. The bound surface charge is approximately −2.66×10⁻¹¹ C/m².

Explore the model

Interactive lab

Interactive lab · vectors

Inspect normal and tangential components

Reset and set angle=0°, magnitude=2. Reset and set angle=60°, magnitude=2; predict x changing from 2 to 1. Choose +x as a normal and y as tangent for this projection exercise.

Inspect normal and tangential componentsPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThese are two independently selected vectors. The lab does not impose dielectric interface laws, compute displacement, or vary permittivity.

Check your understanding

At a static interface without free surface charge, which pair is continuous?

Try it yourself

Two planar layers each have thickness 1 mm, with ε₁=2ε₀ and ε₂=4ε₀. A 6 V drop is imposed across them without internal free sheet charge. Find each field and capacitance per area.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use one common Dn and add the two voltage drops; the lower-permittivity layer takes the larger share.

Compare with the solution

The series denominator is 0.001/(2ε₀)+0.001/(4ε₀)=0.00075/ε₀ m²/F. Thus Dn=8000ε₀ V/m, E₁=4000 V/m and E₂=2000 V/m. Their drops are 4 V and 2 V. Capacitance per area is ε₀/0.00075 m≈1.18×10⁻⁸ F/m² under the stated planar approximation.

Take it with you

  • Tangential E and normal D obey distinct interface laws.
  • Zero free charge does not imply zero bound charge.
  • A global voltage or charge condition fixes the field scale in a layered problem.

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