Fields and potential due to a surface electric dipole layer
A surface electric dipole layer is a neutral charge layer with an electric dipole moment per
unit area directed perpendicular to the surface. It can be modeled as two surface charge layers,
(r, ) and −(r, ), lying on each side of the surface defined by F(r) = 0. The unit vector

n = ∇F(r)/|∇F(r)| is directed from the negative surface charge density to the positive surface

charge density (it’s sufficient to replace F(r) by −F(r) in order to adjust the sense of n ). The

charge layers lie on the surfaces F(r ±n /2) and the surface dipole moment density is
 d (r)


= lim n (r, )  =  d (r)n
1.103
→0
lim (r, ) = 0
→0
lim (r, ) =  d (r)
→0
Since the surface is neutral (total charge = 0), in the limit that  → 0 with (r, ) fixed
,

n(r)  ( 1 E 1 (r) −  2 E 2 (r))= o for electric dipole surface layer
From Gauss’ law one can determine the electric field contributions, E + and E − , and from
Eq. (1.99) the potential field contributions,  + and,  − , from each individual layer . The latter
contributions are shown in the schematics below. The total E 1 and  1 (above the dipole layer)
and total E 2 and  2 (below the dipole layer) are given by E + + E − and  + +  − .in each region
Schematic of the E field contributions
(r, )n 1 /2 ↑
++++++++++
− (r, )n 1 /2 ↓
− (r, )n 2 /2 ↓
−−−−−−−−−
(r, )n 2 /2 ↑
where n 1 = n 2 and the negative sign on the field due to the lower layer comes from the
negative charge ”surface layer”.
1.104
Schematic of the  field contributions
− |z|(r, )/2
++++++++++
− |z|(r, )/2
|z|(r, )/2
−−−−−−−−−
|z|(r, )/2
Thus the total electric fields are:
E 1,total (r, ) = (r, )n 1 /2 − (r, )n 2 /2 = 0 above the dipole layer
E 2,total (r, ) = −(r, )n 1 /2 + (r, )n 2 /2 = 0 below the dipole layer
giving
n 1  [E 1,total − E 2,total ] = 0
Between the two layers for finite  the electric field is constant, directed downward and equal
to,
E 3,total (r, ) = −(r, )n 1 /2 − (r, )n 2 /2 = −(r, )n 1 /.
The total potential above the dipole layer (where |z 2 | =  + |z 1 |) is
 1,total (r, ) =  + (r, ) +  − (r, ) = −|z 1 |(r, )/2 + |z 2 |(r, )/2 = (r, )/2
Below the dipole layer (where |z 1 | =  + |z 2 |)
. 2,total (r, ) = |z 2 |(r, )/2 − |z 1 |(r, )/2 = −(r, )/2,
and for finite  between the two charge layers where |z 2 | =  − |z 1 | the total potential is
 3,total (r, ) = −|z 1 |(r, )/2 + |z 2 |(r, )/2 = (r, )/2 − |z 1 |(r, )/
= (r, )/2 + z 1 (r, )/.
Note that  is continuous at each individual planar charged layer:
 3,total = (r, )/2 =  1,total at the + layer when z 1 = 0
#
 3,total = −(r, )/2 =  2,total at the - layer when z 1 = −
#
Finally the discontinuity in the potential above and below the surface dipole layer is
 1 (r) −  2 (r) = lim[(r, )/2 − (−(r, )/2)] = lim 1 (r, ) = 1  d (r)
1.106
0
0
To formally obtain the potential due to a surface dipole layer we follow the standard approach.
The potential will be the sum of the potential of the positively charged surface layer and the
potential of the negatively charged surface layer as follows. After the limit  → 0 is taken a
point r ′ on the planar dipole layer will be on the surface S ′ .
(r) = lim
0
1
4
∫∫

n, 
dS ′ − 1
′
1 
4
r − (r + 2 n)
 r′ +
1
2
∫∫
 r′ −
1
2
′
r − (r −
n), 
1
2
n)
dS ′
recall : df(r ′ ) =dr ′ ∇ ′ f =f(r ′ + dr ′ ) − f(r ′ )
(r) = lim
0
1
4
= lim 1
0 4
=
1
4
∫∫
(r ′ , )
+
|r − r ′ |

∫∫ (r ′ , )n  ∇ ′
∫∫  d (r ′ )∇ ′
1
|r − r ′ |
1 
n  ∇′
2
1
|r − r ′ |
∫∫  d (r ′ )
= −1
4
∫∫  d (r ′ ) d ′
dS ′ ;
−
(r ′ , )
−
|r − r ′ |
−1 
n  ∇′
2
(r ′ , )
|r − r ′ |
(r ′ , ) is a constant, independent of r ′

 ndS ′
(r ′ − r)  ′
 ndS ;
|r ′ −r| 3
= −1
4
(r ′ , )
|r − r ′ |
since ∇ ′
(r) = −1
4
1
|r − r ′ |
∫∫  d (r ′ ) d ′s .
= −∇
1
|r − r ′ |
=
(r − r ′ )
|r − r ′ | 3
1.108
In this relationship d s is the solid angle subtended by the differential surface element of the
dipole layer at r ′ , specifically as viewed from the observation point r when in the
configuration shown in Fig. 6 below. Formally, the solid angle subtended by a surface, S, is
the projection of the surface onto a unit sphere centered at the observation point, r:
(r ′ − r)
 n̂ s dS ′ . = ∫∫
d s .
1.109
 S = ∫∫
S exp osed |r ′ −r| 3
S exp osed
S is defined by f(x, y, z) = 0 and has normals, n̂ = ±∇f/|∇f|. The n̂ s = n̂ in the solid angle
expression of Eq. 1.109 is always directed out of the volume enclosed by the surface and the
unit sphere upon which the surface is being projected. Using this prescription  S is always
positive. When calculating solid angles one often only integrates over that part of the surface
which is visible to the observer.
In Eq. 1.107, however, n̂ always points from − toward the  charge layer (where
depending on the problem  could later be interpreted as negative) and one must integrate over
the entire surface upon which the dipole surface charge density resides. Eq. 1.107 gives the
correct sign for all r independent of configuration. The discontinuity in (r) arises
automatically from the change in sign of the differential d ′ as one traverses the dipole layer.
(In d ′s one adjusts the sign artifically to give a positive solid angle for all configurations.)
dS ′
Fig. 6a
The relationship between the parameters is shown in the above figure.
Example 1: Find the electrostatic field inside and outside a constant dipole surface
charge density,  d = q/a, residing on a sphere of radius R. Let the sphere be centered at the
origin. From Eq. 1.107 (r) is given by
(r ′ − r)
 n̂ dS ′
(r) = − 1 ∫∫  d (r ′ )
4
|r ′ −r| 3
(r ′ − r)
q
 n̂ dS ′
4a ∫∫ S |r ′ −r| 3
q
=−
d ′
4a ∫∫ S
where the sphere is defined by r ′ = R. All solid angles viewed from inside the sphere are 4
and correspond to integrating over the entire sphere, so
q
(r) = − a for r < R
Outside the sphere, at r > R:
q
1
(rẑ ) =
r′  ∇′
dS ′ ; note that r > r ′
4a ∫∫
|r − r ′ |
=−
2
q
4 r ′l
(r ′ − R)r ′  ∇ ′ ∑ 2l+1
Y ∗ ( ′ ,  ′ )Y lm (, ) r ′ sin  ′ d ′ d ′ dr ′
r l+1 lm
4a ∫∫ ∫
l,m
′l
q
4
=
Y (, ) ∫ (r ′ − R)r ′2 ∂r∂ ′ rrl+1 dr ′ ∫∫ 4 Y ∗lm ( ′ ,  ′ )Y 00 ( ′ ,  ′ ) sin  ′ d ′ d ′
2l+1 lm
4a ∑
l,m
′l−1
q
4
=
Y lm (, ) ∫ (r ′ − R)r ′2 lrr l+1 dr ′ 4  l0  m0
∑
2l+1
4a l,m
q 4
0
=
Y (, ) ∫ (r ′ − R)r ′2 r l+1
dr ′ 4
4a 1 00
=0
for r > R
Note that the above result of (rẑ ) = 0 is only valid if r > r ′ . [One could use the above
procedure to derive the result for r < r ′ In this case the partial derivative with respect to r ′
=
−q
gives an overall factor of −(l + 1)or −1 for l = 0 and the final result is a . We will discuss
these techniques later.] This result is consistent with spherical symmetry, Gauss’s law and a
total charge enclosed = 0. The electric field is zero everywhere. (Note that outside the sphere
the solid angle approach is not helpful. When determining the solid angle from outside the
sphere part of the sphere is obscured and one integrates over only the visible part of S. The
potential calculation, on the other hand, calls for integrating over the entire sphere! )
Fig. 6b Sphere with  d = q/a
q
The normalized potential is divided by a and uses variables (r,  = 0) where r =
zẑ . = r cos ẑ
Example 2: Find the electrostatic potential along the symmetry axis inside and outside
a sphere of radius, R, centered at the origin, with dipole surface charge density,
 d = [q/a] cos  On the symmetry axis r =rẑ , and
cos  ′ (r ′ − r)
q
 r̂ ′ r ′2 sin  ′ d ′ d ′ ; r ′ = R
(rẑ , ) = −
3
′
4a ∫∫ S
|r −r|
′
=−
q
2a

∫0
cos  ′ (r ′ − r)
 r̂ ′ r ′2 sin  ′ d ′
3
′
|r −r|
=−
q
2a

cos  ′ (R 3 − r cos  ′ R 2 )
sin  ′ d ′ ;
[R 2 + r 2 − 2r cos  ′ R] 3
∫0
r = R
where r̂  r̂ ′ = cos  and the symmetry gives 2 for the  ′ integration. The electric field along
ẑ is
d(r)
E(r)  ẑ = −
dz
Fig. 6c Sphere with  d = q/a cos
Note that  is discontinuous and E  n̂ is continuous across the dipole layer. The normalized
q
potential is divided by a and uses variables (r,  = 0) where r = zẑ . = r cos ẑ .
Example 3 An infinite plane with a constant dipole surface charge density,  d = q/a
(r ′ − r)
q
 n̂ dS ′
(r) = −
4a ∫∫ S |r ′ −r| 3
We let z ′ = 0 define the plane, and let r =zẑ . Then r  r ′ = 0 and n̂ = ẑ . Thus
q
−r  ẑ dx ′ dy ′
4a ∫∫ S |r ′ −r| 3
qz
1
=
 ′ d ′ d ′
4a ∫∫ S [ ′2 + z 2 ] 3/2
qz2 ∞
1
1 d( ′2 )
=
4a ∫ 0 [ ′2 + z 2 ] 3/2 2
qz ∞
d(−2[ ′2 + z 2 ] −1/2 )
=
4a ∫ 0
qz 2 −1/2
q z
=
[z ]
=
2a
2a |z|
The electric field is zero everywhere. This unique results occurs only when the plane is
infinite as the next example indicates.
(rẑ . ) = −
Example 4
Assume the infinite plane is reduced to a finite disk of radius R and
calculate the potential along the symmetry axis.
q
−r  ẑ dx ′ dy ′
4a ∫∫ S |r ′ −r| 3
qz
1
=
 ′ d ′ d ′
4a ∫∫ S [ ′2 + z 2 ] 3/2
qz2 R 2
1
1 d( ′2 )
=
4a ∫ 0 [ ′2 + z 2 ] 3/2 2
qz  ′ =R
d(−2[ ′2 + z 2 ] −1/2 )
=
4a ∫  ′ =0
qz 1
1
[
−
]
=
2
2a |z|
[R + z 2 ] 1/2
q
In the plot below the normalized potential is divided by a and uses variables (r,  = 0)
where r = zẑ . = r cos ẑ
(zẑ . ) = −
z . The
|z|
magnitude, however can be adjusted with an appropriate  o added to the potential.
Just above disk’s center the potential is the same as for an infinite plane,
.
q
2a