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Boundary Conditions on
r r r r
E, H , D, B
Chapter 9
General Boundary Conditions: Summary of Results
Vector form
r
r
1. e$n ¥ E 1 - E 2 = 0
d
r
¥ dH
i
r
r
H
js
=
1
2
Scalar form
Description
Et1 − Et 2 = 0
r
Tangential electric field, E , is
continuous.
Ht1 − Ht 2 = js
The discontinuity of the tangential H field equals the surface current.
r
r
3. e$n ◊ D1 - D 2 = r s
Dn1 - Dn2 = r s
r
r
4. e$n ◊ B1 - B 2 = 0
Bn1 − Bn2 = 0
The
r discontinuity of the normal
D equals the surface charge
density.
r
The normal component of B is
continuous.
2. e$n
d
d
i
i
i
Derivation of these most general forms is in the text, Chapter 4-9 and 5-7.
Study them carefully. In what follows in these notes, we give the derivations
of the same four boundary conditions for two important special cases. That
is, (a) dielectric to perfect conductor boundary, and (b) general medium to
general medium boundary with no surface charges and surface current densities.
There are two ways to solve electromagnetic problems:
1. Find out charge and current distributions everywhere in space and solve
Maxwell’s Equations everywhere.
2. Solve Maxwell’s Equations in a limited region of interest, subject to
“boundary conditions” on the boundaries defining this region. Boundary
condition means the value of the fields just at the boundary surface.
The second method is used most often. It is especially useful when the
boundaries are conductors. Boundary conditions at perfect conductors are
fairly simple. We usually approximate good conductors as perfect conductors in this class.
Boundary Conditions at a Surface of Perfect Conductor
Chapter 4-9 Text
r
r
Fields inside conductor are zero: E2 = 0 , H 2 = 0 .
1. Boundary condition for tangential component of electric field:
Dielectric (perfect, or lossy)
1
a
b
2
d
c
z
b
a
r r
E ◊ dl +
s =•
r
r
E2 = 0 , H2 = 0
Perfect Conductor
z
c
b
r r
E ◊ dl +
z
d
c
r r ar r
d r r
E ◊ dl + E ◊ dl = B ◊ ds
d
dt abcd
z
z
z z
Area enclosed is infinitely small. Therefore RHS → 0 .
“ ad ” and “ cb ” are infinitely small. Therefore,
T
E 2 (inside conductor) is zero. Therefore,
This leaves:
z
b
a
z
d
c
b
and
a
a
d
Æ0.
Æ 0.
r r
E ◊ dl = 0
r
If ab distance is finite, but small, then E is constant, and we have:
r
Et ⋅ ∆ = 0
E t = component of E along ab ;
i.e., tangential component
Thus, Et = 0 .
In vector form, this can be written:
r
e$n ¥ E = 0 , where e$n is unit normal.
A very important corollary of this boundary condition is that, since E t = 0 at
the surface of a perfect conductor,
z
b
a
r r
E ◊ dl = 0
between any point on or inside the conductor. Therefore, a perfect conductor is an equipotential surface.
2. Boundary condition for tangential component of magnetic field:
Text, 5-7
e$n
a
b
d
xxxxxxxxxxxxxxxxxxxxxxxxxxxx
r
c
js
z
r r
H ◊ dl =
d
∆ is small,
but finite
d Æ0
r r d
r r
j ◊ ds +
D◊ ds
dt abcd
abcd
z
z
The second term on the right hand side goes to zero, since the area is infinitely small.
However, the first term on the right hand side does
r not go to zero. The
reason is that, although the area is rinfinitely small, j can be infinitely large.
This means that a surface current js can exist on the boundary, normal to
the area abcd. We have already discussed this before.
r
d Æ0
U
|V
r
j Æ • |W
r r
js = j ◊ d
r
js finite
Total current enclosed in the area abcd then is ∆ ⋅ js . The direction of js
must be into the paper if Ht on ab is in the a → b direction. Then, the left
hand side becomes (by the same reasoning as in 1):
z
b
a
r r
H ◊ dl = Ht D
Thus, Ht ∆ = js ∆
r
r
This also gives
correct direction
∴ Ht = js , or e$n ¥ H = js
A
be$ g is outward normal
n
( e$n is frequently written more simply as n$ .
This is the notation our text uses.)
3. Boundary condition on normal component of electric field:
small, but finite, area ∆A
U|
V|
W
z
r r
D◊ ds =
surface of box
z
infinitesimal thickness d
r dv
volume of box
If there is only a volume charge density, the right hand side would be zero,
since the volume of the box is infinitely small.
However, there can be a surface charge density on the surface of the conductor. Because of that:
right hand side = DA ◊ r s
left hand side =
z
r r
D◊ ds +
side surface
z
top
r r
D ◊ ds +
z
r r
D◊ ds
bottom
r r
The integral of D ◊ d s over the side surface is zero, since its height is infinitesimal. The integral over the bottom surface is zero because the field is
zero inside.
\LHS =
z
r
positive, if D
is pointing away
from conductor
r r
D ◊ ds = Dn D A
top
Thus, D A r s = D ADn
r
Dn = r s , or e$n ◊ D = r s
r
Note that ρ s is positive for D pointing outward, as it should be (Field lines
originate on positive charges).
4. Boundary condition on normal component of magnetic field:
We apply
z
r r
B ◊ ds = 0 to the same pillbox as in 3.
Then, obviously, since RHS = 0 , we get:
r
Bn = 0 , or e$ n ◊ B = 0
Summary of B.C. At the surface of a perfect conductor:
Et = 0
Ht = js
Dn = r s
Bn = 0
r
e$ n ¥ E = 0
r r
e$ n ¥ H = j s
r
e$ n ◊ D = r s
r
e$ n ◊ B = 0
Tangential component of E = 0 .
r
Tangential component of H = surface current density.
r
Normal component of D = surface charge density.
r
Normal component of B = 0 .
Boundary conditions at boundary between two dielectrics (or two general media). No surface charges or currents: r s = 0 , js = 0
We use exactly the same methods as we did in the previous sections. The
main difference are that:
a. There are fields on both sides of the boundary, since only inside perfect conductors are the field’s zero.
b. Under normal conditions, there is no surface charge on the boundary.
c. There is no surface current on boundary, since surface currents can
exist only on a perfect conductor surface.
1. Tangential Electric Field Component:
∆
1
a
b
2
d
c
z
b
a
r r
E ◊ dl +
z
r r
E ◊ dl = 0
d
c
r
E t1
r
Et2
r
r
E t1 D - E t 2 D = 0
r
r
r
r
\ E t1 = E t 2 , or e$ n ¥ E 1 - E 2 = 0
d
i
2. Tangential Magnetic Field Component:
Since js = 0 , we get the same result as in 1.
r
r
Ht1 = Ht 2 , or e$n ¥ H 1 - H 2 = 0
d
i
3. Normal Electric Field Component:
Since r s = 0 ,
z
r r
D ◊ ds +
z
r r
D◊ ds = 0
bottom
top
dr
r
Dn1 = Dn2 , or e$n ◊ D1 - D 2 = 0
i
4. Normal Magnetic Field Component:
Same as 3.
r
r
Bn1 = Bn2 , or e$n ◊ B1 - B 2 = 0
d
i
5. Examples Useful for Remembering 2. and 3.
ΟΟΟΟΟΟΟΟ
××××××××
++++++++
−−−−−−−−
1
2
3
1
2
3
H=0
H = js
H=0
parallel plate
currents
E=0
E = rs / e
E=0
parallel plate
capacitor
More on Boundary Conditions
A. It can be easily shown
r that an equivalent form of Boundary Condition 1.
(that is, tangential E continuous: E t1 = E t 2 ) is:
V1 = V2 That is, potential is continuous at an interface.
r
B. Consider e$n ◊ D = r s , which is the boundary condition for the normal
component of the electric displacement at the interface between a perfect conductor and a dielectric.
r
r
r
D = eE
E = - —V
∂V
Then: En = - e$n ◊ — V = ∂n
\ rs = - e
where
∂V
,
∂n
∂V
means the normal derivative (that is, derivative in the direc∂n
tion of the normal out of the conducting surface). Example shortly!
C. If the value of the potential V is given on a conducting surface, that too is
called a boundary condition (to Laplace’s Equation).