# 4.8 Electric Boundary Conditions

Even though the electric field may be continuous in adjoining dissimilar media, it may well be discontinuous at the boundary between them. Boundary conditions specify how the components of fields tangential and normal to an interface between two media relate across the interface Here we derive a general set of boundary conditions for E, D, and J that is applicable at the interface between any two dissimilar media—be they two dielectrics or a conductor and a dielectric. Of course, any of the dielectrics may be free space. Even though these boundary conditions are derived assuming electrostatic conditions, they remain valid for time-varying electric fields as well. Figure 4-20 shows an interface between medium 1 with permittivity ε 1 and medium 2 with permittivity ε 2. In the general case, the interface may contain a surface charge density ρ s (unrelated to the dielectric polarization charge density).

# Figure 4-20 Interface between two dielectric media.

Figure 4 20 Full Alternative Text

To derive the boundary conditions for the tangential components of E and D, we consider the closed rectangular loop a b c d a shown in Fig. 4-20 and apply the conservative property of the electric field expressed by Eq. (4.40), which states that the line integral of the electrostatic field around a closed path is always zero. By letting Δ h→0, the contributions to the line integral by segments b c and d a vanish. Hence,

∮C E⋅d l=∫a b E 1⋅ℓ ˆ 1 d l+∫c d E 2⋅ℓ ˆ 2 d l=0,(4.97)

where ℓ ˆ 1 and ℓ ˆ 2 are unit vectors along segments a b and c d and E 1 and E 2 are the electric fields in media 1 and 2. Next, we decompose E 1 and E 2 into components tangential and normal to the boundary (Fig. 4-20),

E 1=E 1 t+E 1 n, (4.98a)
E 2=E 2 t+E 2 n. (4.98b)

Noting that ℓ^1=−ℓ^2, it follows that

(E 1−E 2) .ℓ ˆ 1=0.(4.99)

In other words, the component of E 1 along ℓ ˆ 1 equals that of E 2 along ℓ^1 for all ℓ ˆ 1 tangential to the boundary. Hence,

Upon decomposing D 1 and D 2 into tangential and normal components (in the manner of Eq. (4.98)) and noting that D 1 t=ε 1 E 1 t and D 2 t=ε 2 E 2 t, the boundary condition on the tangential component of the electric flux density is

Next, we apply Gauss’s law, as expressed by Eq. (4.29), to determine boundary conditions on the normal components of E and D. According to Gauss’s law, the total outward flux of D through the three surfaces of the small cylinder shown in Fig. 4-20 must equal the total charge enclosed in the cylinder. By letting the cylinder’s height Δ h→0, the contribution to the total flux through the side surface goes to zero. Also, even if each of the two media happens to contain free charge densities, the only charge remaining in the collapsed cylinder is that distributed on the boundary. Thus, Q=ρ s Δ s, and

∮S D⋅d s=∫top D 1⋅n ˆ 2 d s+∫bottom D 2⋅n ˆ 1 d s=ρ s Δ s,(4.102)

where n ˆ 1 and n ˆ 2 are the outward normal unit vectors of the bottom and top surfaces, respectively. It is important to remember that the normal unit vector at the surface of any medium is always defined to be in the outward direction away from that medium. Since n^1=−n^2, Eq. (4.102) simplifies to

If D 1 n and D 2 n denote as the normal components of D 1 and D 2 along n^2, we have

The corresponding boundary condition for E is

or equivalently

In summary, (1) the conservative property of E,

∇×E=0⇔∮C E⋅d l=0,(4.106)

led to the result that E has a continuous tangential component across a boundary, and (2) the divergence property of D,

∇⋅D=ρ v⇔∮S D⋅d s=Q,(4.107)

led to the result that the normal component of D changes by ρ s across the boundary. A summary of the conditions that apply at the boundary between different types of media is given in Table 4-3.

# Table 4-3 Boundary conditions for the electric fields.

Table 4 3 Full Alternative Text

# Example 4-13: Application of Boundary Conditions

The x–y plane is a charge-free boundary separating two dielectric media with permittivities ε 1 and ε 2, as shown in Fig. 4-21. If the electric field in medium 1 is

E 1 x+y ˆ E 1 y+z ˆ E 1 z,

and E 1=x ˆ, find (a) the electric field E 2 in medium 2 and (b) the angles θ 1 and θ 2.

# Figure 4-21 Application of boundary conditions at the interface between two dielectric media (Example 4-13).

Figure 4 21 Full Alternative Text

Solution: (a) Let E 2=x ˆ E 2 x+y ˆ E 2 y+z ˆ E 2 z. Our task is to find the components of E 2 in terms of the given components of E 1. The normal to the boundary is z ˆ. Hence, the x and y components of the fields are tangential to the boundary and the z components are normal to the boundary. At a charge-free interface, the tangential components of E and the normal components of D are continuous. Consequently,

E 2 x=E 1 x,E 2 y=E 1 y,

and

D 2 z=D 1 z or ε 2 E 2 z=ε 1 E 1 z.

Hence,

E 2=x ˆ E 1 x+y ˆ E 1 y+z ˆ ε 1 ε 2 E 1 z.(4.108)

(b) The tangential components of E 1 and E 2 are

E 1 t=E 1 x 2+E 1 y 2 and E 2 t=E 2 x 2+E 2 y 2.

The angles θ 1 and θ 2 are then given by

tan θ 1=E 1 t E 1 z=E 1 x 2+E 1 y 2 E 1 z,tan θ 2=E 2 t E 2 z=E 2 x 2+E 2 y 2 E 2 z=E 1 x 2+E 1 y 2(ε 1/ε 2)E 1 z,

and the two angles are related by

tan θ 2 tan θ 1=ε 2 ε 1. (4.109)

# 4.8.1 Dielectric-Conductor Boundary

Consider the case when medium 1 is a dielectric and medium 2 is a perfect conductor. In a perfect conductor, because electric fields and fluxes vanish, it follows that E 2=D 2=0, which implies that components of E 2 and D 2 tangential and normal to the interface are zero. Consequently, from Eq. (4.100) and Eq. (4.104), the fields in the dielectric medium at the boundary with the conductor satisfy

E 1 t=D 1 t=0, (4.110a)
D 1 n=ε 1 E 1 n=ρ s. (4.110b)

These two boundary conditions can be combined into

where n ˆ is a unit vector directed normally outward from the conducting surface.

Figure 4-22 shows an infinitely long conducting slab placed in a uniform electric field E 1. The media above and below the slab have permittivity ε 1. Because E 1 points away from the upper surface, it induces a positive charge density ρ s=ε 1|E 1| on the upper slab surface. On the bottom surface, E 1 points toward the surface; therefore, the induced charge density is −ρ s. The presence of these surface charges induces an electric field E i in the conductor, resulting in a total field E=E 1+E i. To satisfy the condition that E must be everywhere zero in the conductor, E i must equal −E 1.

# Figure 4-22 When a conducting slab is placed in an external electric field E 1, charges that accumulate on the conductor surfaces induce an internal electric field E i=−E 1. Consequently, the total field inside the conductor is zero.

Figure 4 22 Full Alternative Text

If we place a metallic sphere in an electrostatic field (Fig. 4-23), positive and negative charges accumulate on the upper and lower hemispheres, respectively. The presence of the sphere causes the field lines to bend to satisfy the condition expressed by Eq. (4.111); that is, E is always normal to a conductor boundary.

# Figure 4-23 Metal sphere placed in an external electric field E 0.

# 4.8.2 Conductor–Conductor Boundary

We now examine the general case of the boundary between two media—neither of which is a perfect dielectric or a perfect conductor (Fig. 4-24). Medium 1 has permittivity ε 1 and conductivity σ 1, medium 2 has ε 2 and σ 2, and the interface between them holds a surface charge density ρ s. For the electric fields, Eqs. (4.100) and (4.105b) give

E 1 t=E 2 t,ε 1 E 1 n−ε 2 E 2 n=ρ s.(4.112)

Since we are dealing with conducting media, the electric fields give rise to current densities J 1=σ 1 E 1 and J 2=σ 2 E 2. Hence,

J 1 t σ 1=J 2 t σ 2,ε 1 J 1 n σ 1−ε 2 J 2 n σ 2=ρ s.(4.113)

The tangential current components J 1 t and J 2 t represent currents flowing in the two media in a direction parallel to the boundary; hence, there is no transfer of charge between them. This is not the case for the normal components. If J 1 n≠J 2 n, then a different amount of charge arrives at the boundary than leaves it. Hence, ρ s cannot remain constant in time, which violates the condition of electrostatics requiring all fields and charges to remain constant. Consequently, the normal component of J has to be continuous across the boundary between two different media under electrostatic conditions. Upon setting J 1 n=J 2 n in Eq. (4.113), we have

# Figure 4-24 Boundary between two conducting media.

Figure 4 24 Full Alternative Text

Concept Question 4-22: What are the boundary conditions for the electric field at a conductor–dielectric boundary?

Concept Question 4-23: Under electrostatic conditions, we require J 1 n=J 2 n at the boundary between two conductors. Why?