At a boundary with surface free charge density σ_free, which relation describes the normal components of the D field across the boundary?

At a boundary with surface free charge density σ_free, which relation describes the normal components of the D field across the boundary?

• A. D1_n − D2_n = σ_free
• B. D1_n − D2_n = − σ_free
• C. D1_n = D2_n
• D. D1_n + D2_n = σ_free

Answer: (A)

The normal component of the displacement field D can have a jump across a boundary when free charge sits on that boundary. This comes from Gauss’s law in the form ∮ D · dA = Q_free_enc. If you take a tiny pillbox that straddles the boundary, the only significant flux is through the top and bottom faces. The flux through the top face is D1_n times the area, and the flux through the bottom face is D2_n times the area, with signs set by the outward normals. The enclosed free charge is σ_free times the area. Equating flux to enclosed charge gives D1_n A − D2_n A = σ_free A, so D1_n − D2_n = σ_free. If there were no free surface charge, the jump would be zero, so D1_n = D2_n. The other relations don’t follow from this boundary argument: adding the two normal components or introducing a negative sign would not match the flux balance for the boundary charge.

The normal component of the displacement field D can have a jump across a boundary when free charge sits on that boundary. This comes from Gauss’s law in the form ∮ D · dA = Q_free_enc. If you take a tiny pillbox that straddles the boundary, the only significant flux is through the top and bottom faces. The flux through the top face is D1_n times the area, and the flux through the bottom face is D2_n times the area, with signs set by the outward normals. The enclosed free charge is σ_free times the area. Equating flux to enclosed charge gives D1_n A − D2_n A = σ_free A, so D1_n − D2_n = σ_free.

If there were no free surface charge, the jump would be zero, so D1_n = D2_n. The other relations don’t follow from this boundary argument: adding the two normal components or introducing a negative sign would not match the flux balance for the boundary charge.