2.4 Solutions to Problems
27
− 4π
x+l/2
x−l/2
P
S
y
x
, y +
l
2
− P
S
y
x
, y −
l
2
dx
= −4π
∂P S
x
∂x
l +
∂P S
y
∂y
l
l + o(l).
By taking the limit of l → 0, this equation coincides with Eq. (2.8),
D z =−4π ∇ t ·P S .
2.4.2 Boundary Condition at Interface (2)
[Problem 2.2] Derive Eq. (2.10), after the discussion associated to Eq. (2.9).
We carry out surface integral of Eq. (2.5) within the small area S in the right
panel of Fig. 2.2, and apply the Stokes’ theorem to H ;
S
(∇ × H ) · (ˆ z × ˆ
t)ds =
S
1
c
∂D
∂t
+ 4π
∂P (2)
∂t
· (ˆ z × ˆ
t)ds
=
A
H · dl = llH t +
+0
−0
H z
z, t −
l
2
− H z
z, t +
l
2
dz.
(2.27)
In Eq. (2.27), the surface integral of ∂D/∂t and the line integral of H z are neglected
in the limit of infinitesimal area of S, since H z and D t are not singular there.
Consequently, Eq. (2.27) is simplified to be
llH t =
4π
c
S
∂P (2)
∂t
·
ˆ
z × ˆ
t
ds
=
4π
c
S
∂
∂t
P
S (x, y)δ(z)
·
ˆ
z × ˆ
t
ds =
4π
c
S
∂P S
∂t
× ˆ
z
· ˆ
tδ(z)ds
=
4π
c
l
∂P S
∂t
× ˆ
z
· ˆ
t.
Therefore, Eq. (2.10) is obtained.