6. Electrostatic Boundary Conditions

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ABSTRACT In electric field analysis, it is important to know the changes in field quantities as one crosses a boundary between a conductor and a dielectric or between two different dielectric media. In practical configurations comprising multi-dielectric media, this is of paramount importance, as quantities defining the field on the two sides of the boundary between two dielectrics often undergo significant changes. Boundary conditions are fundamental equations involving electric field quantities, which describe the changes in electrostatic field specifically in relation to boundary surfaces. In electrostatic field, another major aspect is the presence of charges on the boundaries, either free charges, as in the case of conductor boundaries, or surface charges, as in the case of dielectric–dielectric boundaries. These charges play the most significant role in the changes that the field quantities undergo at the boundary.

6.1 Introduction

In real life, any conductor is always surrounded by at least one dielectric. It should be kept in mind that air is also a dielectric, which is present almost everywhere. Therefore, even if there is no solid or liquid or any other gaseous dielectric around a conductor, it will in all probability be surrounded by air. Therefore, there will be boundaries between a conductor and a dielectric in practice. Moreover, except for very few cases like single-core cables having only one dielectric or transmission line conductors surrounded by air at mid-span, dielectric materials are arranged either in series or in parallel between two conductors having a particular potential difference. For example, if one takes the case of an outdoor porcelain insulator, it may appear that there is only one dielectric, that is, porcelain, involved. But the porcelain insulator will be surrounded by air and hence it becomes a two dielectric configuration. As a result there will be boundaries between two different dielectric media in practical configurations. Changes in some electric field quantities in direction and/or magnitude occur at such boundaries. The equations that describe such field behaviours by relating electric field quantities on two sides of a boundary surface are known as boundary conditions. The transition of electric field from one medium to another medium through a boundary surface is governed by the boundary conditions.

6.2 Boundary Conditions between a Perfect Conductor and a Dielectric

A perfect conductor is defined as a material within which the charges are able to move freely. In electrostatics, it is considered that the charges have attained the equilibrium positions and are fixed in space. Theoretically, consider that the charges are initially distributed uniformly throughout the volume of a perfect conductor. Such distributed charges should be of same polarity within a conductor of one particular value of electric potential, because if there are charges of opposite polarity within the volume of the conductor, then such charges will immediately recombine with each other as they are free to move. Hence, the charges of same polarity that are present in the volume of the conductor will exert repulsive forces on each other. Because the charges are able to move without any hindrance, the charges will disperse in such a direction, so that the distance between the charges will increase. In the process all the charges will arrive at the surface of the conductor. But the conductor being surrounded by a dielectric, the charges are unable to move further and the charges will be fixed in space on the surface of the conductor. Consequently, any Gaussian surface within a perfect conductor will enclose zero charge and hence, electric field within a perfect conductor will be zero.

6.2.1 Boundary Condition for Normal Component of Electric Flux Density

Images

FIGURE 6.1
Pertaining to boundary condition for electric flux density at conductor–dielectric boundary.

Consider a coin-like closed volume of cylindrical shape, as shown in Figure 6.1. Such a volume is often termed as a Gaussian pillbox. The pillbox has a finite surface area ΔA and an infinitesimally small height δ, such that half of the pillbox is within the conductor and the other half is within the dielectric medium as shown. The top and bottom surfaces of the pillbox are parallel to the conductor–dielectric interface.

Application of Gauss’s law to this pillbox yields

$\int_{S u r f a c e o f p i l l b o x} \vec{D} \cdot d \vec{s} = Change enclosed by the pillbox (6.1)$

For the right-hand side (RHS) of Equation 6.1 the following need to be considered: (1) the volume of the pillbox is infinitesimally small and (2) half of the volume of the pillbox within the perfect conductor does not contain any volume charge density, as the charges reside on the surface of the conductor and the other half of the volume of the pillbox within the dielectric also does not contain any volume charge density, if ideal dielectric is assumed. But the surface charge density on the conductor has a finite value and the area of the pillbox is also finite.

Hence, the RHS of Equation 6.1

$= σ Δ A (6.2)$

where:

σ is the surface charge density on the conductor

The integral on the left-hand side (LHS) of Equation 6.1 could be expanded as follows:

$\int_{Surface of pillbox} \vec{D} \cdot d \vec{s} = \int_{Top surface} \vec{D} \cdot d \vec{s} + \int_{Bottom s urface} \vec{D} \cdot d \vec{s} + \int_{Wall surface} \vec{D} \cdot d \vec{s} (6.3)$

As the height of the pillbox is infinitesimally small, the integral over the wall surfaces is negligible. The integral over the bottom surface is also zero as the field within the perfect conductor is zero. Hence, Equation 6.3 could be rewritten as

$\int_{Surface of pillbox} \vec{D} \cdot d \vec{s} = \int_{Top surface} \vec{D} \cdot d \vec{s} = D_{n} Δ A (6.4)$

where:

D_n is the normal component of electric flux density

Hence, from Equations 6.2 and 6.4,

$D_{n} Δ A = σ Δ A, or, D_{n} = σ (6.5)$

Bringing in the unit normal vector, Equation 6.5 could be written as

${\hat{u}}_{n} \cdot \vec{D} = σ (6.6)$

6.2.2 Boundary Condition for Tangential Component of Electric Field Intensity

Consider an infinitesimally small closed rectangular loop abcda as shown in Figure 6.2, of which the length segments ab and cd are parallel to the conductor–dielectric boundary and the length segments bc and da are normal to the boundary. The length of the loop parallel to the boundary is Δl and is finite, but the length of the loop normal to the boundary is δ, which is negligibly small.

As E-field is conservative in nature, the integral of $\vec{E} \cdot d \vec{l}$ over the loop contour abcda will be zero, that is,

$∮_{abcda} \vec{E} \cdot d \vec{l} = \int_{a}^{b} \vec{E} \cdot d \vec{l} + \int_{b}^{c} \vec{E} \cdot d \vec{l} + \int_{c}^{d} \vec{E} \cdot d \vec{l} + \int_{d}^{a} \vec{E} \cdot d \vec{l} = 0 (6.7)$

As stated, the lengths bc and da are negligibly small. Hence,

$\int_{b}^{c} \vec{E} \cdot d \vec{l} = \int_{d}^{a} \vec{E} \cdot d \vec{l} \approx 0$

Again, the field within the perfect conductor is zero. Hence,

$\int_{c}^{d} \vec{E} \cdot d \vec{l} = 0$

Therefore, from Equation 6.7, $\int_{a}^{b} \vec{E} \cdot d \vec{l} = 0$

The length ab is Δl, which is small but finite. Then considering $\vec{E}$ to be constant over the small length Δl, it may be written that

$E_{t} Δ l = 0, or, E_{t} = 0 (6.8)$

Images

FIGURE 6.2
Pertaining to boundary condition for electric field intensity at conductor–dielectric boundary.

where:

E_t is the component of electric field intensity along the length ab, which is the tangential component of electric field intensity

Bringing in the unit normal vector, Equation 6.8 could be written as

${\hat{u}}_{n} \times \vec{E} = 0 (6.9)$

6.2.3 Field Just Off the Conductor Surface

From Equation 6.8, on the conductor surface E_t = 0. In other words, the electric field acts in the direction normal to the conductor surface. Then, from Equation 6.5, the normal component of electric field intensity could be written as

$E_{n} = \frac{σ}{ε} = \frac{σ}{ε_{r} ε_{0}} (6.10)$

As discussed in Section 6.2.1, the electric field intensity as given by Equation 6.10 is for the top surface of the pillbox. The height of the pillbox is infinitesimally small, but is not zero. Hence, the top surface of the pillbox, as shown in Figure 6.1, is not exactly on the conductor surface. As a result, the value of electric field intensity as obtained from Equation 6.10 is stated to be electric field intensity within the dielectric medium just off the conductor surface.

PROBLEM 6.1

A charged conductor is surrounded by air. Calculate the maximum charge density that the conductor can hold at standard temperature and pressure (STP).

Solution:

The conductor can hold the maximum charge density for which the electric field intensity just off the surface is equal to the breakdown strength of air at STP. This is due to the fact that any further increase in the value of charge density on the conductor surface will cause breakdown of air and the charges will be drained from the conductor.

Breakdown strength of air at STP = 30 kV/cm = 30 × 10³ × 10² V/m = 3 × 10⁶ V/m.

Let the maximum charge density that the conductor can hold be σ_m.

Now, ε_r for air = 1

Therefore, σ_m/ε₀ = 3 × 10⁶

or, σ_m = 3 × 10⁶ × 8.854 × 10⁻¹² = 26.5 μC/m².

6.3 Boundary Conditions between Two Different Dielectric Media

In the case of practical configurations comprising multiple homogeneous dielectric media, there could be many dielectric–dielectric boundaries. On such dielectric–dielectric boundaries, there could be charges depending on the dissimilarities of the dielectric media that are present on the two sides of the boundaries. These surface charges will serve as source of electric field acting in opposite directions on the two sides of the boundary. Consequently, electric field quantities get changed in direction as well as magnitude on the two sides of the boundary.

6.3.1 Boundary Condition for Normal Component of Electric Flux Density

For the boundary between two dielectric media having permittivities ε₁ and ε₂, consider a cylindrical coin-like Gaussian pillbox, as shown in Figure 6.3. The height of the pillbox is infinitesimally small. As in Equation 6.1 application of Gauss’s law to this pillbox yields

$\int_{S u r f a c e o f p i l l b o x} \vec{D} \cdot d \vec{s} = Charge enclosed by the pillbox$

Subdividing the integral on the LHS into contributions from the top, bottom and wall surfaces of the pillbox and subdividing the wall surface into two halves, one in dielectric 1 and the other in dielectric 2,

$\begin{matrix} \int_{Surface of pillbox} \vec{D} \cdot d \vec{s} = & \int_{Top surface dielectric 2} {\vec{D}}_{2} \cdot d \vec{s} + \int_{Bottom surface dielectric 1} {\vec{D}}_{1} \cdot d \vec{s} \\ + \int_{Wall surface dielectric 1} {\vec{D}}_{1} \cdot d \vec{s} + \int_{Wall surface dielectric 2} {\vec{D}}_{2} \cdot d \vec{s} \end{matrix}$

Images

FIGURE 6.3
Pertaining to boundary condition for electric flux density at dielectric–dielectric boundary.

As the height of the pillbox is vanishingly small, the contribution of the integral over the wall surfaces will be negligible. Thus

$\int_{Surface of pillbox} \vec{D} \cdot d \vec{s} = \int_{Top surface dielectric 2} {\vec{D}}_{2} \cdot d \vec{s} + \int_{Bottom surface dielectric 1} {\vec{D}}_{1} \cdot d \vec{s} (6.11)$

As the top and bottom surfaces are small, ${\vec{D}}_{1}$ and ${\vec{D}}_{2}$ could be assumed to be constant over the bottom and top surfaces, respectively. Hence,

$\int_{Top surface dielectric 2} {\vec{D}}_{2} \cdot d \vec{s} = {\vec{D}}_{2} \cdot Δ \vec{A} = ({\vec{D}}_{2} \cdot {\hat{u}}_{n 2}) Δ A (6.12)$

$and \int_{Top surface dielectric 1} {\vec{D}}_{1} \cdot d \vec{s} = {\vec{D}}_{1} \cdot Δ \vec{A} = ({\vec{D}}_{1} \cdot {\hat{u}}_{n 1}) Δ A (6.13)$

Now considering the unit normal to be directed into dielectric 2, ${\hat{u}}_{n} = {\hat{u}}_{n 2} = - {\hat{u}}_{n 1}$

Therefore, from Equation 6.11,

$\int_{S urface of pillbox} \vec{D} \cdot d \vec{s} = ({\vec{D}}_{2} \cdot {\hat{u}}_{n} - {\vec{D}}_{1} \cdot {\hat{u}}_{n}) Δ A (6.14)$

Let the surface charge density on the boundary be ρ_s and the volume charge densities in the two halves of the pillbox be ρ_v1 and ρ_v2. Then the net charge enclosed by the pillbox is given by

$Charge enclosed by the pillbox = ρ_{s} Δ A + ρ_{v 1} Δ A \frac{δ}{2} + ρ_{v 2} Δ A \frac{δ}{2} (6.15)$

As the height of the pillbox, δ, is vanishingly small, the contribution of the terms involving volume charge densities will be negligible. Hence, Equation 6.15 becomes

$Charge enclosed by the pillbox = ρ_{s} Δ A (6.16)$

Therefore, from Equations 6.14 and 6.16, $({\vec{D}}_{2} \cdot {\hat{u}}_{n} - {\vec{D}}_{1} \cdot {\hat{u}}_{n}) = ρ_{s}$

$or, D_{2 n} - D_{1 n} = ρ_{s} (6.17)$

as ${\vec{D}}_{2} \cdot {\hat{u}}_{n} = D_{2 n}$ and ${\vec{D}}_{1} \cdot {\hat{u}}_{n} = D_{1 n},$ , which are normal components of electric flux density in dielectrics 2 and 1, respectively.

Equation 6.17 is valid in general. For example, consider that the medium 1 is a perfect conductor. Then D₁ is zero. Then D_2n is equal to the surface charge density, which is in accordance with Equation 6.5.

6.3.2 Boundary Condition for Tangential Component of Electric Field Intensity

Considering the infinitesimally small closed rectangular contour abcdefa, as shown in Figure 6.4 and applying the principle of conservative E-field along this closed contour

$\oint_{abcdefa} \vec{E} \cdot d \vec{l} = \int_{a}^{b} {\vec{E}}_{2} \cdot d \vec{l} + \int_{b}^{c} {\vec{E}}_{2} \cdot d \vec{l} + \int_{c}^{d} {\vec{E}}_{1} \cdot d \vec{l} + \int_{d}^{e} {\vec{E}}_{1} \cdot d \vec{l} + \int_{e}^{f} {\vec{E}}_{1} \cdot d \vec{l} + \int_{f}^{a} {\vec{E}}_{2} \cdot d \vec{l} = 0 (6.18)$

But the length of the segments bc, cd, ef and fa are negligibly small and hence the integrals over these length elements contribute insignificantly. Thus from Equation 6.18

$\int_{a}^{b} {\vec{E}}_{2} \cdot d \vec{l} + \int_{d}^{e} {\vec{E}}_{1} \cdot d \vec{l} = 0$

Then considering ${\vec{E}}_{2}$ and ${\vec{E}}_{1}$ , to be constant over the small lengths ab and de and noting that $d {\vec{l}}_{a b} = Δ l = - d {\vec{l}}_{d e}$ , it may be written that

$\begin{matrix} \int_{a}^{b} {\vec{E}}_{2} \cdot d \vec{l} + \int_{d}^{e} {\vec{E}}_{1} \cdot d \vec{l} = E_{2 t} Δ l - E_{1 t} Δ l = 0 \\ or, E_{2 t} = E_{1 t} \end{matrix} (6.19)$

where:

E_2t and E_1t are tangential components of E₂ and E₁, respectively, along the boundary

Images

FIGURE 6.4
Pertaining to boundary condition for electric field intensity at dielectric–dielectric boundary.

Equation 6.19 is also valid in general. For example, if the medium 1 is considered to be a perfect conductor, then E₁ is zero. Then E_2t is also zero, which is in accordance with Equation 6.8.

PROBLEM 6.2

For a two-dielectric arrangement comprising transformer oil (ε_r1 = 2) as dielectric 1 and mica (ε_r2 = 6) as dielectric 2, it is given that ${\vec{E}}_{1} = 6 i + 4 j + 9 \hat{k} k V / c m$ . Find ${\vec{E}}_{2}$ : (A) considering the boundary to be charge free and (B) considering a surface charge density of +70 pC/cm² on the boundary. Given that x–y plane is the boundary between the two media.

Solution:

Because x–y plane is the boundary between transformer oil and mica, x- and y-components of $\vec{E}$ are the tangential components along the boundary and are equal on both sides of the boundary. The z-component of $\vec{D}$ is the normal component of electric flux density on the boundary.

Application of boundary condition for tangential component of electric field intensity yields

$E_{1 x} = E_{2 x} = 6 k V / c m and E_{1 y} = E_{2 y} = 4 k V / c m$

The z-component of $\vec{E}$ need to be determined from the boundary condition of D_n separately for Case A and Case B.

Case A: Boundary is charge free.

$\begin{matrix} As ρ_{s} = 0, D_{2 n} = D_{1 n}, or, ε_{r 2} ε_{0} E_{2 z} = ε_{r1} ε_{0} E_{1z} \\ As E_{1 z} = 9 k V / c m, E_{2 z} = \frac{2}{6} \times 9 = 3 k V / c m \end{matrix}$

Therefore,

${\vec{E}}_{2} = 6 \hat{i} + 4 \hat{i} + 3 \hat{k} k V / c m$

Case B: Surface charge density (ρ_s) on the boundary is +70 pC/cm².

From the boundary condition of D_2n,

$\begin{matrix} ε_{r 2} ε_{0} E_{2 z} - ε_{r1} ε_{0} E_{1z} = ρ_{s} \\ \begin{matrix} or, E_{2 z} = \frac{ε_{r 1} ε_{0} E_{1 z} + ρ_{s}}{ε_{r 2} ε_{0}} = & \frac{ε_{r 1} E_{1 z}}{ε_{r 2}} + \frac{ρ_{s}}{ε_{r 2} ε_{0}} = \frac{2 \times 9 \times 1 0^{3} \times 1 0^{2}}{6} + \frac{70 \times 1 0^{- 12} \times 1 0^{4}}{6 \times 8.854 \times 1 0^{- 12}} \\ = & 313, 176 V / m = 3.13 l k V / c m \end{matrix} \end{matrix}$

Therefore,

${\vec{E}}_{2} = 6 \hat{i} + 4 \hat{i} + 3.131 \hat{k} k V / c m$

6.3.3 Boundary Condition for Charge-Free Dielectric–Dielectric Interface

Consider an interface between two different dielectric media having permit-tivities ε₁ and ε₂, respectively, having no surface charge density on the boundary, as shown in Figure 6.5. Typically, if the dielectric media are ideal in nature, so that there is no volume as well as surface conduction and also if there is no discharge taking place within the dielectric media or on the interface between the dielectric media, then the surface charge density on the interface is zero.

From Equation 6.17,

D_2n − D_1n = 0, as ρ_s = 0

$or, D_{2 n} = D_{1 n}, or, ε_{r 2} ε_{0} E_{2 n} = ε_{r 1} ε_{0} E_{1 n}, or, ε_{r 2} E_{2} \cos θ_{2} = ε_{r 1} E_{1} \cos θ_{1} (6.20)$

Again, from Equation 6.19,

$E_{2 t} = E_{1 t}, or, E_{2} \sin θ_{2} = E_{1} \sin θ_{1} (6.21)$

From Equations 6.20 and 6.21,

$\frac{\tan θ_{2}}{\tan θ_{1}} = \frac{ε_{r 2}}{ε_{r 1}} (6.22)$

Images

FIGURE 6.5
Pertaining to boundary condition for charge-free dielectric–dielectric boundary.

PROBLEM 6.3

The flux lines of an electric field pass from air into glass, making an angle 30° with the normal to the plane surface separating air and glass at the air-side of the surface. The relative permittivity of glass is 5.0. The field intensity in air is 200 V/m. Calculate the electric flux density in glass and also the angle, which the flux lines make with the normal on the glass side.

Solution:

Given, ε_r1 = 1.0, ε_r2 = 5.0 and θ₁ = 30°

Therefore, $\frac{\tan θ_{2}}{\tan θ_{1}} = \frac{\tan θ_{2}}{\tan 3 0^{o}} = \frac{ε_{r 2}}{ε_{r 1}} = \frac{5}{1}, or, \tan θ_{2} = 2.887$ or tan

or, θ₂ = 70.9°.

Again, E₁ = 200 V/m and E₁ sin θ₁ = E₂ sin θ₂, so that E₂ sin70.9° = 200 × sin 30° or, E₂ = 105.8 V/m.

Hence, D₂ = ε_r2ε₀E₂ = 5 × 8 854 × 10^-12 × 105.8 = 4.68 nC/m²

Objective Type Questions

1. On any conductor–dielectric boundary, the tangential and the normal components of electric field intensity are E_t and E_n, respectively. Then

a. E_t = E_n/2

b. E_t = 2E_n

c. E_t = ∞

d. E_t = 0

2. On any dielectric–dielectric boundary, the tangential and the normal components of electric field intensity are E_t and E_n, respectively, where the suffixes 1 and 2 denote dielectrics 1 and 2, respectively. Then

a. D_n = D_t

b. D_t1 = D_t2

c. E_t = 0

d. E_t1 = E_t2

3. Surface charge density on a charged conductor is equal to

a. Tangential component of electric flux density

b. Normal component of electric flux density

c. Resultant electric flux density

d. Both (b) and (c)

4. For any dielectric–dielectric boundary, which one of the following is a valid boundary condition, when the notations have their usual meanings

a. $({\vec{D}}_{2} \cdot {\hat{u}}_{n} - {\vec{D}}_{1} \cdot {\hat{u}}_{n}) = ρ_{s}$

b. $({\vec{D}}_{2} \times {\hat{u}}_{n} - {\vec{D}}_{1} \times {\hat{u}}_{n}) = ρ_{s}$

c. $({\vec{E}}_{2} \cdot {\hat{u}}_{n} - {\vec{E}}_{1} \cdot {\hat{u}}_{n}) = ρ_{s}$

d. $({\vec{E}}_{2} \times {\hat{u}}_{n} - {\vec{E}}_{1} \times {\hat{u}}_{n}) = ρ_{s}$

5. For any dielectric–dielectric boundary, which one of the following is a valid boundary condition, when the notations have their usual meanings

a. D_2t = D_1t

b. E_2t = E_1t

c. D_n = σ

d. E_t = 0

6. For any charge-free dielectric–dielectric boundary, which one of the following is a valid boundary condition, when the notations have their usual meanings

a. D_2n − D_1n = ρ_s

b. E_2t = E_1t

c. D_2n = D_1n

d. Both (b) and (c)

7. On any conductor–dielectric boundary, the tangential and the normal components of electric field intensity are E_t and E_n, respectively, and surface charge density is σ. Then

a. D_n = σ

b. E_n = σ

c. D_t = σ

d. E_t = σ

Answers:

1) d;

2) d;

3) d;

4) a;

5) b;

6) d;

7) a

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Table of Contents for 6. Electrostatic Boundary Conditions

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PROBLEM 6.1

PROBLEM 6.2

PROBLEM 6.3

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Table of Contents for
6. Electrostatic Boundary Conditions