Match the right field components at a dielectric interface
Use displacement for free charge and electric field for total charge.
On this page
The question for today
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.
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.
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.
Work it through
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.
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.
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.
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 second field has normal and tangential components both 3 V/m. The bound surface charge is approximately −2.66×10⁻¹¹ C/m².
The living notebook
Explore the model
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.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
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.
Pause & reason
Check your understanding
Your turn
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.
Keep the idea
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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