Boundary condition for the normal component of D across an interface with free surface charge density σ_free.

Boundary condition for the normal component of D across an interface with free surface charge density σ_free.

• A. D2_perp − D1_perp = σ_free
• B. D1_perp − D2_perp = σ_free
• C. D1_perp − D2_perp = σ_free
• D. D1_perp − D2_perp = 0

Answer: (C)

The normal component of D has a jump across a boundary equal to the free surface charge present there. This comes from Gauss’s law in integral form: a small pillbox straddling the interface encloses the free surface charge, and the net flux of D through the pillbox equals the enclosed free charge. If we take the unit normal to point from medium 2 into medium 1, the flux through the pillbox leads to D1_perp − D2_perp = σ_free. Because the sign depends on how you orient the normal, the same physical boundary condition can be written as D2_perp − D1_perp = σ_free if the normal is chosen in the opposite direction. The given convention here yields D1_perp − D2_perp = σ_free, which is why that form is correct. When there is no free surface charge, the normal component is continuous: D1_perp = D2_perp.

The normal component of D has a jump across a boundary equal to the free surface charge present there. This comes from Gauss’s law in integral form: a small pillbox straddling the interface encloses the free surface charge, and the net flux of D through the pillbox equals the enclosed free charge.

If we take the unit normal to point from medium 2 into medium 1, the flux through the pillbox leads to D1_perp − D2_perp = σ_free. Because the sign depends on how you orient the normal, the same physical boundary condition can be written as D2_perp − D1_perp = σ_free if the normal is chosen in the opposite direction. The given convention here yields D1_perp − D2_perp = σ_free, which is why that form is correct. When there is no free surface charge, the normal component is continuous: D1_perp = D2_perp.