19.5.1 Point charge q with two dielectrica. Method of images
B. Schnizer
[email protected]
2013 - 02 - 08
z
/|\
|
|
e1
. q, q1
F1
___________________ |_d_________________ x
| -d
.
e2
F2
q2
|
A point charge q is at a distance d in front of an interface, z = 0, separating two dielectric half spaces.
This problem is equivalent to a problem with dielectrica corresponding to the halfspace under consideration
but with two additional image charges q1 and q2 as drawn above. This solution is computed below in the
first part of this notebook. Thereeafter the same game is played with q located at z = - d.
Cylindrical coordinates r =
x2 + y2 , f, z .
Primary charge q in G1, z = d > 0
ü The potentials
R1 = Sqrt@r ^ 2 + Hz - dL ^ 2D
R2 = Sqrt@r ^ 2 + Hz + dL ^ 2D
H- d + zL2 + r2
Hd + zL2 + r2
sF1 = 1 ê H 4 p e1 L q ê R1 +
q
4p
sF2 =
H- d + zL2 + r2 e1
4p
q2
+
4p
1 ê H 4 p e2 L q1 ê R1
q1
H- d + zL2 + r2 e2
q2 ê H 4 p R2 e1 L
Hd + zL2 + r2 e1
2
K19charge2dielectric.nb
ü Field matching at the interface z = 0
bc1 = H e1 D@sF1 , zD ê. z -> 0 L ==
dq
2 3ê2
4 p Id2 + r M
d q2
-
2 3ê2
4 p Id2 + r M
e1
e2 D@sF2 , zD ê. z -> 0
d q1
e1 ã
4 p Id2 + r2 M
e1
3ê2
bc2 = HsF1 ê. z -> 0L == H sF2 ê. z -> 0L
q
4p
q2
+
d2 + r2 e1
d2 + r2 e1
4p
q1
ã
4p
d2 + r2 e2
so = Map@Factor, Solve@8bc1, bc2<, 8q1 , q2 <D êê Flatten êê Together, 82<D
:q1 Ø
2 q e2
e1 + e2
, q2 Ø
q He1 - e2 L
e1 + e2
>
qr2 = q2 ê q ê. so
e1 - e2
e1 + e2
F1 = sF1
q
4p
q2
+
H- d + zL2 + r2 e1
4p
Hd + zL2 + r2 e1
F2 = sF2
q1
4p
H- d + zL2 + r2 e2
ü The potential of the original and of the image charges
F1 = sF1 ê. so
q
4p
q He1 - e2 L
+
H- d + zL2 + r2 e1
4p
F2 = sF2 ê. so
q
2p
H- d + zL2 + r2 He1 + e2 L
ü Checking continuity
Limit@F1 , d Ø 0D
q
z2 + r2 H2 p e1 + 2 p e2 L
Limit@F2 , d Ø 0D
q
z2 + r2 H2 p e1 + 2 p e2 L
Hd + zL2 + r2 e1 He1 + e2 L
K19charge2dielectric.nb
% == %%
True
ü Parts Gi1 of the Green's function
G11 = F1 ê. q Ø 1
1
4p
e1 - e2
+
H- d + zL2 + r2 e1
Hd + zL2 + r2 e1 He1 + e2 L
4p
G21 = F2 ê. q Ø 1
1
2p
H- d + zL2 + r2 He1 + e2 L
Primary charge q in G2, z = - d < 0
z
/|\
|
|
e1
.
F1
q1
___________________ |_d_________________ x
| -d
. q,
e2
q2
|
ü The potentials
R1 = Sqrt@r ^ 2 + Hz - dL ^ 2D
R2 = Sqrt@r ^ 2 + Hz + dL ^ 2D
H- d + zL2 + r2
Hd + zL2 + r2
sF1 = 1 ê H 4 p e2 L q2 ê R2
q2
4p
Hd + zL2 + r2 e2
sF2 = q ê H 4 p e2 L ê R2 + 1 ê H 4 p e1 L q1 ê R1
q1
4p
H- d + zL2 + r2 e1
q
+
4p
Hd + zL2 + r2 e2
F2
3
4
K19charge2dielectric.nb
ü Field matching at the interface z = 0
bc1 = H e1 D@sF1 , zD ê. z -> 0 L ==
d q2 e1
-
2 3ê2
4 p Id2 + r M
e2 D@sF2 , zD ê. z -> 0
d q1
ã
2 3ê2
4 p Id2 + r M
e2
dq
e1
4 p Id2 + r2 M
3ê2
e2
e2
bc2 = HsF1 ê. z -> 0L == H sF2 ê. z -> 0L
q2
4p
q1
ã
d2 + r2 e2
d2 + r2 e1
4p
q
+
4p
d2 + r2 e2
so = Map@Factor, Solve@8bc1, bc2<, 8q1 , q2 <D êê Flatten êê Together, 82<D
:q1 Ø
q e1 H- e1 + e2 L
e2 He1 + e2 L
, q2 Ø
2 q e2
e1 + e2
>
qr2 = q2 ê q ê. so
2 e2
e1 + e2
F1 = sF1
q2
4p
Hd + zL2 + r2 e2
F2 = sF2
q1
4p
q
+
H- d + zL2 + r2 e1
4p
Hd + zL2 + r2 e2
ü The potential of the original and of the image charges
F1 = sF1 ê. so
q
2p
Hd + zL2 + r2 He1 + e2 L
F2 = sF2 ê. so
q
4p
q H- e1 + e2 L
+
Hd + zL2 + r2 e2
4p
ü Checking continuity
Limit@F1 , d Ø 0D
q
z2 + r2 H2 p e1 + 2 p e2 L
Limit@F2 , d Ø 0D
q
z2 + r2 H2 p e1 + 2 p e2 L
H- d + zL2 + r2 e2 He1 + e2 L
K19charge2dielectric.nb
% == %%
True
ü Parts G2i of the Green's function
G12 = F1 ê. q Ø 1
1
2p
Hd + zL2 + r2 He1 + e2 L
G22 = F2 ê. q Ø 1
1
4p
- e1 + e2
+
Hd + zL2 + r2 e2
4p
H- d + zL2 + r2 e2 He1 + e2 L
The resulting four pieces of the Green's function
G11
1
4p
e1 - e2
+
H- d + zL2 + r2 e1
4p
Hd + zL2 + r2 e1 He1 + e2 L
G12
1
2p
Hd + zL2 + r2 He1 + e2 L
G21
1
2p
H- d + zL2 + r2 He1 + e2 L
G22
1
4p
Hd + zL2 + r2 e2
- e1 + e2
+
4p
H- d + zL2 + r2 e2 He1 + e2 L
5