Chapter 2
Introduction to electrostatics
2.1
Coulomb and Gauss’ Laws
We will restrict our discussion to the case of static electric and magnetic fields in a homogeneous, isotropic medium. In
this case the electric field satisfies the two equations, Eq. 1.59a with a time independent charge density and Eq. 1.77 with a
time independent magnetic flux density,
∇ · D (r) = ρ0 (r) ,
(1.59a)
∇ × E (r) = 0.
(1.77)
Because we are working with static fields in a homogeneous, isotropic medium the constituent equation is
D (r) = εE (r) .
(1.78)
Note : D is sometimes written :
D = ²o E + P .... SI units
D = E + 4πP in Gaussian units
in these cases ε = [1 + 4πP/E] Gaussian
(1.78b)
The solution of Eq. 1.59 is
D (r) =
1
4π
with ∇ · D0 (r) = 0
Z ZZ
ρ0 (r0 ) (r − r0 )
|r −
3
r0 |
d3 r0 + D0 (r) , SI units
(1.79)
If we are seeking the contribution of the charge density, ρ0 (r) , to the electric displacement vector then D0 (r) = 0. The
given charge density generates the electric field
E (r) =
1
4πε
ZZ Z
ρ0 (r0 ) (r − r0 ) 3 0
d r SI units
|r − r0 |3
18
(1.80)
Section 2.2
2.2
The electric or scalar potential
The electric or scalar potential
Faraday’s law with static fields, Eq. 1.77, is automatically
satisfied by any electric field E(r) which is given by
E (r) = −∇φ (r)
(1.81)
The function φ (r) is the scalar potential for the electric field. It is also possible to obtain the difference in the values of the
scalar potential at two points by integrating the tangent component of the electric field along any path connecting the two
points
−
Z
path
ra →rb
E (r) · d` =
=
=
Z
path
ra →rb
Z
path
ra →rb
Z
path
ra →rb
∇φ (r) · d`
(1.82)
·
¸
∂φ (r)
∂φ (r)
∂φ (r)
+ dy
+ dz
dx
∂x
∂y
∂z
dφ (r) = φ (rb ) − φ (ra )
The result obtained in Eq. 1.82 is independent of the path taken between the points ra and rb . It follows that the integral
of the tangential component along a closed path is zero,
I
I
E (r) · d` = dφ (r) = 0.
(1.83)
This last result actually follows from the requirement that ∇ × E (r) = 0 and the application of Stoke’s theorem.
To obtain the scalar potential due to the charge density ρ0 (r) we note that


1

∇ q
2
2
2
0
0
0
(x − x ) + (y − y ) + (z − z )
(1.84)
(x − x0 ) i + (y − y 0 ) j+ (z − z 0 ) k
= −h
i3/2
(x − x0 )2 + (y − y 0 )2 + (z − z 0 )2
= −
r − r0
3.
|r − r0 |
Comparing the expression on the right hand side of Eq. 1.84 to the integrand in Eq. 1.80 we find that can write that the scalar
potential due to the charge density ρo (r) is
1
φ (r) =
4πε
Z ZZ
ρo (r0 ) 3 0
d r + φ0 ,
|r − r0 |
(1.85)
where φ0 is a constant which fixes the (arbitrary) location of the zero for the scalar potential. Since the observed quantity is
the electric force and, therefore, the electric field, only the difference in the values of the scalar potential at any two different
points is significant. (See Eq. 1.82)
19
Section 2.3
2.3
Surface charges and charge dipoles
Surface charges and charge dipoles A surface charge is a charge density which is
‘restricted to lie on a surface’.
It is characterized by the equation defining the surface, say F (r) = 0, (or
F (x, y, z) = 0)and a surface charge density, σ (r) , with dimensions of Coulombs/m2 in SI units and statcoulombs per
square centimeter in gaussian units.
Restricting a volume integral to a surface with a delta function in the integrand
The charge density associated with a surface charge will be the product of two terms. One term will contain a delta
function which restricts the density to the surface, δ (F (r)) ∗ f (r). This term should have dimensions of inverse length,
m−1 in our units. and the other will be the surface charge density, σ (r) . The function f (r) should be determined such that
ZZZ
3
H (r) δ (F (r)) ∗ f (r) d r =
ZZ
H (r) dS (r)
(1.86)
F (r)=0
where H (r) is any ‘smooth’ function and the surface integral on the right hand side is restricted to the surface F (r) = 0.
Before attacking this problem we should note that, in one dimension,
Z
∞
−∞
a (x) δ (b (x)) |b0 (x)| dx =
Z
a (x) δ (b (x)) db (x)
(1.87)
= a (x0 ) , with b (x0 ) = 0
Since the delta function will restrict the integral to the surface we need only consider the region close to the surface. In this
region let the coordinates system consist of a coordinate axis perpendicular to the surface at each point and two orthogonal
axes which are tangent to the surface at each point. The unit vector which is perpendicular to the surface at each point is
b = +∇F (r) / |∇F (r)| or n
b = −∇F (r) / |∇F (r)| .
either n
b] · dξb
n. In addition if r0 is a point
If we let ξb
n be the displacement of a point from the surface at r0 then d3 r = [dS (r0 ) n
on the surface then
·
¸
∂F (r0 + ξb
n)
0
.
(1.88)
∇F (r ) =
∂ξ
ξ=0
When performing the volume integration we carry out the integration perpendicular to the surface at each point r0 and then
multiply the result by the differential surface element at r0 , dS (r0 ) . We have in this way reduced our problem to a set of one
dimensional problems, one for each point r0 on the surface. Viewing the problem in this way suggests that the appropriate
expression for the function f (r) is
b · ∇0 F (r0 + ξb
f (r0 + ξb
n) = n
n)
¯
¯ 0
n)¯
= ¯∇ F (r0 + ξb
(1.89)
ρ (r) = σ (r) δ (F (r)) |∇F (r)|
(1.90)
It follows then that the volume charge density associated with a surface charge density, σ (r) , on the surface defined by
F (r) = 0 is
20
Section 2.3
Surface charges and charge dipoles
R
is as follows, where δ(F (r))dF = 1:
·Z 0
¸
σ (r)
δ(F (r))dF dS (r)
(1.90b)
Note: A simpler way to obtain this result
ZZ
ZZ
σ (r) dS (r) =
F (r)=0
ZZ
=
ZZ
=
ZZ
=
ZZ
=
F (r)=0
σ (r)
F (r)=0
σ (r)
F (r)=0
σ (r)
F (r)=0
σ (r)
F (r)=0
·Z
·Z
Z
Z
0
0
0
0
0
δ(F (r))dr ·∇F
δ(F (r))dr0 ·
¸
dS (r)
¸
∇F
|∇F | dS (r)
|∇F |
δ(F (r))|∇F | dr0 ·n̂s dS (r)
δ(F (r))|∇F | dr0 · dS (r)
where the primed integration symbol indicates that the integration must be taken along a direction, dr0 which
is along the normal to the surface.
=
ZZ
ZZ
σ (r)
F (r)=0
σ (r) dS (r) =
F (r)=0
ZZ Z
Z
0
δ(F (r))|∇F | dV 0
V contains F (r)=0
(1.90c)
σ (r) δ(F (r)) |∇F | dV
(1.90d)
Note that the final volume integral must be over a volume which contains all of the surface, Using this form for the charge
density allows one to manipulate the variables of integration more freely
.
2.3.1
Example: uniformly charged ellipsoidal surface
The system is a uniformly charged ellipsoidal surface (Fig. ??) defined by
density is σ0. Find the potential on the z axis.
x2 +y 2
a2
+
z2
b2
− 1 = 0 . The surface charge
Solution: From the class notes, Eq. 1.90, the volume charge density is given by
¶
x2 + y 2 z 2
x2 + y2 z 2
+
−
1
|∇(
+ 2 − 1)|
2
2
a
b
a2
b
µ 2
¶
2
2
2
2
2
x +y
z
x +y
z
= σ0 δ
+ 2 − 1 2(
+ 4 )1/2
a2
b
a4
b
ρ (r) = σ0 δ
µ
(1.91)
Because of the cylindrical symmetry of the charge distribution it is convenient to work in cylindrical coordinates, x = ξ cos ϕ
and y = ξ sin ϕ. In these coordinate the charge density is
21
Section 2.3
Surface charges and charge dipoles
.Fig. 3
µ
ρ (r) = 2σ0 δ
¶
ξ2
z2
+ 2 −1
a2
b
s
z2
ξ2
+
a4
b4
Taking the zero of potential at infinity, the potential on the z axis due to the surface charge is
ZZZ δ
2σ0
φ (zẑ) =
4πε
´q 2
2
ξ
+ zb2 − 1
a4 +
q
ξ 0 2 + (z − z 0 )2
³
ξ2
a2
z2
b4
d3 r0
SI units
(1.92)
with d3 r0 = ξ 0 dξ 0 dϕ0 dz 0 . The integrand is independent of ϕ0 and the angle integration can be performed giving 2π. The
integrand is also only a function of ξ 0 2 . This suggests that we define an η = a−2 ξ 0 2 . In this case ξ 0 dξ 0 = 0.5 a2 dη. With
these changes the potential is given by
2πσ 0 a2
φ (zẑ) =
4πε
Z
∞
−∞
Z
∞
0
¡
¢p
δ η + b−2 z 0 2 − 1
a−2 η + b−4 z 0 2
q
dη dz 0
2
a2 η + (z − z 0 )
(1.93)
The delta function allows the η integration to be carried out yielding
σ0a
φ (zẑ) =
2ε
Z
b
−b
"
¡
¢ #1/2
a−2 + b−2 z 0 2 b−2 − a−2
2
(1 − b−2 z 0 2 ) + a−2 (z − z 0 )
dz 0
(1.94)
2
It is convenient to express the z coordinates in units of b, z = bξ 0 and z 0 = bξ 0 , then, with β = (b/a) ,
σ0a
φ (ξ 0 bẑ) =
2ε
Z
1
−1
"
β + ξ 02 (1 − β)
¡
¢
¡
¢2
1 − ξ 02 + β ξ 0 − ξ 0
22
#1/2
dξ 0
(1.95)
Section 2.3
Surface charges and charge dipoles
.Fig. 4
φ (zẑ)
, is plotted below in
The integral in Eq. 1.95 can be done numerically and the normalized potential, Φ(a, b, zẑ) =
σ0 a/ε
a
Fig. 4 for three cases. b = 4a, a and . Note that when a = b the ellipsoid of revolution becomes a charged sphere.
4
Q
The potential due to a spherical shell of radius a with total charge, Q, (and uniform surface charge density, σ 0 =
)
4πa2
can be obtained using Gauss’ law. Because of the spherical symmetry E has a constant magnitude on a sphere of radius r
ZZ Z
ZZ Z
∇ · E dV =
ρ/εdV
ZZ
E · dS = Qenclosed /ε
|E|4πr2
= 4πa2 σ0 /ε
= 0
if r ≥ a
if r < a
Thus
Q
4πεr2
= 0
|E| =
if r ≥ a
if r < a
From E = −∇φ one finds
φspherical shell (r) =
=
σ 0 a2
Q
=
4πεr
ε r
σ0
Q
=
a
4πεa
ε
if r ≥ a
if r ≤ a
Thus the potential is constant inside the uniformly charged spherical shell and the electric field is zero. But inside the
ellipsoidal shell (see Fig. 4) one can not assume that E has a constant magnitude on a spherical shell and Gauss’ law is not
partiularly useful. As seen in Fig. 4 the potential inside the ellipsoidal shell is not constant, and the electric field is not zero.
23
Section 2.3
Surface charges and charge dipoles
For general information the potential, divided by σ 0 b/4ε (at z = 0, .b, 2b ) as a function of b/a is shown in Fig. 5 . The
small values of b/a correspond to a ‘pancake’ shaped ellipse and the potentials at the origin, the middle of the ‘pancake
surface’, and just above the ‘pancake’ surface are nearly the same. Large values of b/a give a ‘needle’ shaped ellipse. (Note
that the curves have some strange behavior since the normalization potential is proportional to b.)
.Fig. 5
2.3.2
The potential and fields due to a surface charge density It is possible to have physical systems with
charge distributions which lie approximately along a surface. Unless one probes within the distribution one can approximate
the fields and potentials by those of a surface charge density. The potential due to a surface charge density, σ (r) ,is formally
given by
1
φ (r) =
4πε
ZZ
surf ace
σ (r0 )
dS 0 + φ0 .
|r − r0 |
SI units
(1.96)
For finite surface charge densities the potential will be continuous at the surface.
2.3.3 Example: uniformly charged plane Let the system be a uniform plane sheet of charge, surface charge
density σ 0 , lying in the z = 0 plane. Determine the value of the potential at (x, y, z) (also denoted by r) with the condition
that the potential vanish at the surface.
Solution: The differential surface area for this problem is dS 0 = dx0 dy0 . This can be used in Eq. 1.96 to obtain the
24
Section 2.3
Surface charges and charge dipoles
required integral
φ (x, y, z) =
=
=
Z
Z
∞
−∞
∞
Z
∞
−∞
Z
∞
σ 0 /4πε
q
dx0 dy 0 + φ0
2
2
0
0
2
(x − x ) + (y − y ) + z
(1.97)
σ 0 /4πε
q
d(x − x0 )d(y − y0 ) + φ0
2
2
−∞ −∞
0
0
2
(x − x ) + (y − y ) + z
Z ∞
£
¤−1/2
σ0
r2 + z 2
2πrdr + φ0 .
4πε 0
The integral, which has the potential equal to zero at infinity, is divergent To handle this difficulty we note that the
potential is independent of x and y. We will assume that the surface has a finite radius R centered on the z axis. Then the
potential at x = y = 0 is
φ (0, 0, z) =
=
Z
¤−1/2
σ0 R £ 2
r + z2
rdr + φ0 (R).
2ε 0
i
σ0 hp 2
R + z 2 − |z| + φ0 (R)
2ε
(1.98)
The condition that the potential is zero at z = 0 requires that φ0 (R) = − σ2ε0 R. With this value of φ0 (R) we can evaluate the
potential in the limit that R goes to infinity,
i
σ 0 hp 2
R + z 2 − |z| − R
R→∞ 2ε
−σ 0
=
|z| = φ (x, y, z)
2ε
φ (0, 0, z) =
lim
(1.99)
This potential is normally obtained with less effort by starting with the electric field.Note that φ (x, y, z) <0 and that the
resulting electric field is along the z axis, as expected.
E(z) = −∇φ = ẑ
σ0
σ0
= n̂
= E1 (z)
2ε
2ε
(1.99b)
On the other side of the surface charge, along -n̂
E2 (z) = −∇φ2 = −ẑ
σ0
σ0
= −n̂
2ε
2ε
(1.99c)
This gives a surface discontinuity of the electric field shown in Eq. 1.102 below.
+———————————————————————————————————————————
The field due to an arbitrary surface charge density is formally given by
E (r) = −∇φ (r) =
1
4πε
25
ZZ
σ (r0 ) [r − r0 ]
dS.
|r − r0 |3
(1.100)
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 ′ ;

 ndS ′
(r ′ − r)  ′
 ndS ;
|r ′ −r| 3
= −1
4
(r ′ , )
|r − r ′ |
−
(r ′ , )
−
|r − r ′ |
−1 
n  ∇′
2
(r ′ , )
|r − r ′ |
(r ′ , ) is a constant, independent of r ′
Eq. 1.107
since ∇ ′
1
|r − r ′ |
= −∇
1
|r − r ′ |
=
(r − r ′ )
|r − r ′ | 3
Eq. 1.107
(r) = −1
4
∫∫  d (r ′ ) d ′s .
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
Use Eq. 1.107
cos  ′ (r ′ − r)
q
′ ′2
 r̂ r sin  ′ d ′ d ′ ; r ′ = R
(rẑ , ) = −
4a ∫∫ S
|r ′ −r| 3
′
=−
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
Section 2.4
Laplace and Poisson Equations
.
2.4
Laplace and Poisson Equations
As noted, Faraday’s law for static fields is satisfied by any electric field which can be written as
E (r) = −∇φ (r) .
The function φ (r) has been identified as the electric or scalar potential associated with the electric field E (r) . Since
D (r) = εE (r) , in order to satisfy Gauss’ law for the displacement vector the scalar potential must be a solution to
Poisson’s Equation,
1
∇2 φ (r) = − ρ (r) .
ε
(1.106)
A special case of Poisson’s Equation is obtained for ρ (r) = 0. The result is known as Laplace’s Equation,.
∇2 φ (r) = 0.
(1.107)
This is the equation satisfied by the scalar potential in a charge free region. The general problem is to obtain the solution,
φ (r) , of Eq. 1.106 in a region of space with φ (r) satisfying specified boundary conditions on the boundary of the region
.
Green’s functions
There is a class of solutions, {g (r, r0 )} , to Poisson’s Equation with the source term given by δ (3) (r − r0 ) . These
solutions are called ‘Green’s functions’. One possible solution for g (r, r0 ) which we have already seen is
g0 (r, r0 ) =
−1
4π |r − r0 |
∇2 g0 (r, r0 ) = δ (3) (r − r0 ) .
(1.108)
(1.109)
This g0 (r, r0 ) vanishes as |r| → ∞ for all finite |r0 | . If g (r, r0 ) is known (and has the correct boundary conditions) a
formal solution of Eq. 1.106 is given by
1
φ (r) =
ε
ZZ Z
ρ (r0 ) g (r, r0 ) d3 r0 .
(1.110)
However, this solution will not generally satisfy the boundary conditions placed on φ (r) . In the next section we derive a
more formally correct solution which contains the expression in Eq. 1.110
32
Section 2.4
Laplace and Poisson Equations
Green’s theorem and Green’s function solution for φ(r)
2.4.1
Green’s theorem, uses the divergence theorem to relate solutions of Poisson’s equation. Assume that we have two
functions, φ (r) and g (r, ro ) , which satisfy Poisson equations as follows in volume, V, with surface, S:
∇2 ϕ (r) = −ρ(r)/ε
(1.111)
∇2 g (r, ro ) = δ(r − ro )
(1.112)
The divergence theorem states that
Z ZZ
volume V
∇ · [ϕ (r) ∇g (r, ro ) − g (r, ro ) ∇φ (r)] d3 r =
ZZ
bounding
surf aces, S
[φ (r) ∇g (r, ro ) − g (r, ro ) ∇φ (r)] · dS
(1.113)
The left hand side can be written as follows:
ZZ Z
ZZ Z
[φ (r) ∇2 g (r, ro ) − g (r, ro ) ∇2 φ (r)]d3 r +
[∇φ (r) · ∇g(r, ro ) − ∇g (r, ro ) · ∇φ (r)]d3 r
volume
volume
ZZ Z
ZZ Z
3
φ (r) δ (r − ro ) d r −
g (r, ro ) [−ρ (r)]d3 r
since the second term above is 0
=
volume
volume
ZZZ
g (r, ro ) [−ρ (r) /ε]d3 r if ro is in the volume!
= φ (ro ) −
volume
So if ro is in the volume, V, the left hand side substituted into Eq. 1.113 gives:
φ (ro ) =
ZZZ
g (r, ro ) [−ρ (r) /ε]d3 r +
volume
ZZ
bounding
surf ace
[φ (r) ∇g (r, ro ) − g (r, ro ) ∇φ (r)] · n̂ (r) dS
(1.114)
This is a Green’s function solution of Eq.1.111 . Outside the volume, where ro is not in V,
ZZ Z
volume
φ (r) δ (r − ro ) d3 r = 0. for ro not in V
and we obtain no solution for φ (ro ) . The surface terms in Eq. 1.114 suggest the interpretation that ∇φ (r) · n̂ (r) at the
surface is surface charge layer (because ∇φ (r) · n̂ (r) = −E · n̂ (r) and the surface might be replaced with an effective
discontinuity in E ) and φ (r) n̂ (r) at the surface is a surface dipole layer (because the surface might be replaced with a an
effective discontinuity in φ (r)). But the latter is not crucial to what we derive in this section. Here we have assumed that
n̂ (r) is the bounding surface unit outward normal vector. In the next section we will find that specifying both the potential
and the normal gradient of the potential on the bounding surface will ‘over specify the problem’. In this case the problem
will only have a solution when the specified potential and normal derivative are compatible.
33
Section 2.4
2.4.2
Laplace and Poisson Equations
Uniqueness of the solutions to Poisson’s Equation
Let φ1 (r) and φ2 (r) be
two solutions to Eq. 1.111 and let both satisfy the same boundary conditions (still of an unspecified nature) on the surface
bounding the volume. It follows that
(1.115)
χ (r) = φ1 (r) − φ2 (r)
satisfies Laplace’s Equation,
∇2 χ (r) = ∇2 φ1 (r) − ∇2 φ2 (r)
= ρ(r)/ε − ρ(r)/ε
= 0
with ‘zero boundary conditions’. If we now apply the divergence theorem to χ (r) ∇χ (r) we obtain
ZZ Z
=
ZZZ
volume
volume
ZZZ
∇ · [χ (r) ∇χ (r)]d3 r =
ZZ
bounding
surf ace
[χ (r) ∇2 χ (r) + |∇χ (r)|2 ]d3 r =
2
volume
3
|∇χ (r)| d r =
ZZ
ZZ
bounding
surf ace
χ (r) ∇χ (r) · dS
bounding
surf ace
(1.116)
χ (r) ∇χ (r) · dS
χ (r) ∇χ (r) · dS
Using Eq. 1.116 we now investigate the boundary conditions satisfied by φ1 (r) and φ2 (r) ..
(1)
First, let us assume that the values of the potential, but not its normal derivative, are given on the bounding surface.
In this case χ (r) = 0 on the boundary and the right hand side of Eq.1.116 vanishes. Since the integrand of the volume
integral cannot be negative it must be identically zero or ∇χ (r) = 0. This requires that φ1 (r) − φ2 (r) be constant.
Since they are equal on the boundary the appropriate constant is zero and φ1 (r) = φ2 (r) . Specifying the potential on
the boundary gives a unique boundary value problem satisfied by only one potential function. This particular problem in
which the function on the boundary is specified is called a ‘Dirichlet problem’ and the boundary conditions are called
‘Dirichlet boundary conditions’.
(2)
Next we assume that the values of the normal derivative of the potential on the boundary is given.. In this case the
normal derivative of χ (r) will vanish on the boundary. Again the right hand side of Eq. 1.116 will vanish and therefore, as
before, φ1 (r) − φ2 (r) will be constant. In this case it’s only the normal derivative which vanishes on the boundary and the
gradient of any constant is zero. It follows that φ1 (r) = φ2 (r) + φ0 and the solution to this boundary value problem is
only unique up to an additive constant. The problems in which the normal derivative is given on the boundary are known as
‘Neumann problems’ and the boundary conditions are called ‘Neumann boundary conditions’.
(3)
One can also contemplate problems in which the value of the normal derivative of the potential is specified on some
sections of the boundary and the value of the potential is specified on other, different, sections of the boundary. In this case
the potential is said to satisfy ‘mixed boundary conditions’.
34
Section 2.5
The Green’s functions for Dirichlet and Neumann problems
Since the Dirichlet problem has unique solutions, specifying the potential on the boundary fixes the normal derivative of
the potential on the boundary. Similarly the solution of a Neumann problem will give the potential on the boundary (up to an
additive constant). We find then that Eq. 1.114, which requires the values of the potential and its normal derivative on the
bounding surface, does not provide a convenient solution to Poisson’s equation unless one can eliminate one of the surface
integrals by imposing boundary conditions on g (r, ro ) .
2.5 The Green’s functions for Dirichlet and Neumann problems The difficulty with Eq.1.114 is that
both the potential and it’s normal derivative appear in the integrand of the surface integral. However, we have not imposed
any boundary conditions on the Green’s function. If we could find a function g (r, r0 ) ,satisfying Eq.1.112 whose normal
derivative on the bounding surface vanished, only the potential on the surface would be required. In the other case, if the
function g (r, r0 ) vanished on the bounding surface, only the normal derivative of the potential on the surface would be
required. The apparent difficulty in applying Eq. 1.116 to obtain the solution of Poisson’s Equation is circumvented by
obtaining the appropriate g (r, r0 ) .
One approach is to write g (r, r0 ) in the region of interest as the sum of a solution to the inhomogeneous equation
(Eq.1.112) and the homogeneous equation (Laplace’s equation). That is,
g (r, r0 ) =
−1
+ F (r, r0 )
4π |r − r0 |
(1.117)
with
∇2 F (r, r0 ) = 0. inside V
(1.118)
V is the region for which the solution, φ (r), is defined. The function F (r, r0 ) is determined so that g (r, r0 ) will satisfy the
required boundary conditions.
2.5.1
The Dirichlet problem For Dirichlet problems we require that g (r, r0) = GD (r, r0 ) with
GD (r, r0 ) = 0 for r on the boundary.
(1.119)
In this case we find that
φD (r0 ) =
−1
ε
ZZ Z
GD (r, r0 ) ρ (r) d3 r +
volume
ZZ
bounding
surf ace
φ (rs ) ∇GD (rs , r0 ) · dS
(1.120)
which only requires the values of the potential on the boundary. The function GD (r, r0 ) is the ‘Dirichlet green’s function’.
The function FD (r, r0 ) satisfies
FD (r, r0 ) =
1
for r = rS
4π |r − r0 |
(1.121)
1
. Rather FD (r, r0 ) is a
where rS is on the boundary of the volume. Note that this does not mean that FD (r, r0 ) = 4π|r−r
0|
function which reduces to the latter expression on the surface. In particular, Eq. 1.118 must be satisfied in V.
Symmetry of the Green’s function
0
A characteristic of the Green’s function is that GD (r, r ) = GD (r0 , r) . This result can be obtained by considering two
Dirichlet Green’s functions GD (r, r0 ) and GD (r, r00 ) each satisfying the condition that they vanish for r on the boundary
35
Section 2.5
The Green’s functions for Dirichlet and Neumann problems
surface. If we now apply the divergence theorem to GD (r, r0 ) ∇GD (r, r00 ) − GD (r, r00 ) ∇GD (r, r0 )we find
ZZZ
∇ · [GD (r, r0 ) ∇GD (r, r00 ) − GD (r, r00 ) ∇GD (r, r0 )] d3 r =
volume
ZZZ
ZZ Z
GD (r, r0 ) δ (r − r0o ) d3 r −
GD (r, r00 ) δ (r − ro ) d3 r
volume
volume
= GD (r0o , ro ) − GD (ro , r0o )
ZZ
[GD (r, r0 ) ∇GD (r, r00 ) − GD (r, r00 ) ∇GD (r, r0 )] · n̂ (r) dS
=
bounding
surface
= 0
and
GD (r00 , r0 ) = GD (r0 , r00 ) . or
GD (r0 , r) = GD (r, r0 ) since r00 , r0 are arbitrary variables.
2.5.2
(1.122)
The Neumann problem Since the solutions to the Neumann problems are only unique up to an
additive constant the Green’s function for the Neumann problem,
GN (r, r0 ) = g (r, r0 ) + FN (r, r0 )
with
∇2 FN (r, r0 ) = 0 in V.
satisfies a slightly more complicated boundary condition. If we apply the divergence theorem to
we find the Neumann restriction:
Z ZZ
δ (3) (r − r0 )
ZZ Z
=
δ (3) (r − r0 ) d3 r
∇2 GN (r, r0 ) =
∇ · ∇GN (r, r0 ) d3 r
ZZ
bounding
surf ace
(1.123)
(1.124a)
∇GN (r, r0 ) · n̂ (r) dS=1
It follows that we are not permitted to set ∇GN (r, r0 ) · n̂ (r) = 0 on the boundary of the volume. The next simplest choice
is to satisfy Eq.1.123 with a simple boundary condition on GN :
simple boundary condition: ∇GN (r, r0 ) · n̂ (r) =
1
Stot
(1.124b)
The solution to the Neumann problem is then given by
1
φ (r0 ) =
ε
ZZ Z
3
GN (r, r0 ) [−ρ (r)]d r + φavg +
volume
36
ZZ
bounding
surface
GN (r, r0 ) ∇φ (r) · n̂ (r) dS
(1.125)
Section 2.5
The Green’s functions for Dirichlet and Neumann problems
with φavg the average value of the potential on the bounding surface,
φavg =
RR
where S is the area of the boundary
Stot =
ZZ
φ (r) dS
RR
dS
bounding
surf ace
(1.126)
dS.
Since the Neumann problem only defines the potential up to an additive constant we could arbitrarily take φavg = 0 without
losing any information. The function FN (r, r0 ) satisfies the condition that
·
µ
[∇GN (r, r0 )] · n̂ (r) = ∇FN (r, r0 ) + ∇
let
−1
4π |r − r0 |
¶¸
· n̂ (r) =
1
.
Stot
(1.127)
To investigate the relationship between GN (r, r0 ) and GN (r0 , r) which we artificially label G0N (r0 , r) with a prime.and
∇02 G0N (r0 , r0o ) = δ(r0 − ro0 )
(1.128)
G0N (r0 , r0o ) is a solution to the Neumann problem, with ρ(r0 )/ε = δ(r0 −r0o ) and from Eq. 1.125,
Z ZZ
G0N (r0 , r00 ) =
ZZ
0
0
Gavg (r0 ) +
volume
bounding
surf ace
GN (r0 , r0 ) δ(r0 −r0o )d3 r0 +
(1.129)
GN (r0s , r0 ) ∇0 G0N (r0s , r00 ) · dS0
The result is
G0N (r0 , r00 ) = GN (r00 , r0 ) + G0ave (r00 ) − Gave (r0 )
(1.130)
Dropping the primes on the G symbols,
GN (r0 , r00 ) − hGN (., r00 )iavg = GN (r00 , r0 ) − hGN (., r0 )iavg .
(1.131)
Since the solutions of the Neumann problem are only unique up to an additive constant if we take any Neumann
green’s function and subtract its average value on the boundary we will obtain a green’s function which is symmetric in the
interchange of the parameters.
37
Solutions ∇2 Φ = 0 using GD and GN for Poisson’s Equation
Section 2.6
2.5.3
The functions F (r, r0)
−1
The simple green’s function g (r, r0 ) = 4π|r−r
0 | can be identified as the
0
0
potential due to a ‘unit’ negative point charge located at r . The function F (r, r ) must satisfy Laplace’s equation in the
volume, V, of interest. If we identify it as a potential its sources will be charges which lie outside our volume. We will
find that viewing F (r, r0 ) as the potential due to charges outside the region of interest will provide us with a technique for
generating this function.
2.6
Solutions ∇2 Φ = 0 using GD and GN for Poisson’s Equation This section has introduced the Green’s
functions for Poisson’s equation with Dirichlet or Neumann boundary conditions given. Since Laplace’s equation is a
special case of Poisson’s equation with the source term, ρ (r) , equal to zero, the Green’s functions can be used to provide a
solution to the Dirichlet problems for Laplace’s equation, φL−D (r) ,
ZZ Z
φL−D (ro ) =
ZZ
bounding
surf ace
ZZ
=
0 · GD (r0 , r0 ) d3 x0 +
(1.132)
φ (rs )L−D ∇GD (rs , r0 ) · dS −
bounding
surf ace
ZZ
φ (rs )L−D ∇GD (rs , r0 ) · dS
GD (rs , r0 ) ∇φ(rs ) · dS
Given the solution on the
boundary, S, one can find a
Green's function solution to
Laplace's equation.
where the second term and third terms are zero because ∇2 Φ = 0 and GD (rs , r0 ) = 0, respectively. A solution to the
Neumann problem for Laplace’s equation,φL−N (r)
φL−N (r) =
Z ZZ
ZZ
=
−
= −
0 · GN (r0 , r0 ) d3 x0 +
(1.133)
φ (rs )L−N ∇GN (rs , r0 ) · dS −
ZZ
ZZ
bounding
surf ace
bounding
surf ace
ZZ
bounding
surf ace
GN (rs , r0 ) ∇φL−N (rs ) · dS
GN (rs , r0 ) ∇φL−N (rs ) · dS + φavg
GN (rs , r0 ) ∇φL−N (rs ) · dS
as we can set φavg = 0 on the bounding surface because the Neumann problem gives the solution to within a constant.
38
Section 2.7
2.7
2.7.1
Electrostatic energy density, capacitance
Electrostatic energy density, capacitance
The electrostatic energy density Starting with Maxwell’s equations we can obtain the relationship
∇ · [E (r, t) × H (r, t)] = −E (r, t) · [∇ × [H (r, t)] + H (r, t) · [∇ × [E (r, t)]
·
¸
∂
∂
= −E (r, t) · [J (r, t) + D (r, t)] + H (r, t) · − B (r, t) .
∂t
∂t
(1.134)
We can identify E (r, t) · J (r, t) as the rate at which the E field does work on the mechanical system per unit area, or
equivalently, the power density transfer between the fields and the mechanical system. With this identification we are led to
associate E (r, t) × H (r, t) with S (r, t), the energy current density carried in the electromagnetic field. S (r, t) is called
the Poynting vector.
S (r, t) = E (r, t) × H (r, t)
units of S:
energy/sec/area
= power/area
= Watts/square meter
(1.135)
and to associate the term
·
¸
∂
∂
∂u (r, t)
= H (r, t) · B (r, t) + E (r, t) · D (r, t)
∂t
∂t
∂t
(1.136)
with the time rate of change of the energy density stored in the ‘electromagnetic fields’.where
u (r, t) =
1
[E(r, t) · D(r, t) + H (r, t) · B (r, t)]
2
(1.137)
Then Eq. 1.134 gives
∇ · S(r, t) = −
∂u (r, t)
− E(r, t) · J(r, t
∂t
ZZ Z
3
(1.138)
and
ZZ
S(r, t) · dA +
E(r, t) · J(r, t)d r = −
Z ZZ
∂u (r, t) 3
d r
∂t
(1.139)
Equation 1.139 states that the rate at which energy leaves a region plus the rate at which field energy is converted to
mechanical energy equals the rate at which energy is transferred to the fields and the material 1 . We can expect that if the
fields are varied ‘adiabatically’ we have a unique value for the energy stored by the electromagnetic fields. We will therefore
ignore for the present some subtle or not so subtle problems which arise due to frequency dependent relationships between E
and D. We will claim that the energy density stored in the electrostatic electric field is
1 We note that there is an ad hoc separation of hte charge currents. Part of the current is explicitly given by J(r,t) while ∂u has a contribution from an
∂t
implicit current. In addition the magnetic field energy includes an implicit energy current density which transmits energy across the boundary of the region.
39
Section 2.7
w (r) =
Electrostatic energy density, capacitance
ε
|∇φ (r)|2 SIunits
2
(1.140)
The electromagnetic energy of a system is given by integrating the energy density over the volume of the system
ε
W =
2
ZZ Z
2
volume
|∇φ (r)| d3 r ≥ 0.
(1.141)
Let ρ (r) denote the charge density for the system. If the charge density is composed of charge densities for fixed elements,
charged point particles, charged surfaces, etc., then it is convenient to express the total charge density as a sum of the
individual charge densities
ρ (r) =
X
(1.142)
ρi (r) .
i
Each charge density will contribute, −∇φi (r) , to the total electric field at each point where ∇2 φi (r) = −ρi (r) /ε. In this
case
−∇φ (r) = −
X
i
(1.143)
∇φi (r)
In terms of the fields of the individual charge elements the electromagnetic energy of the system is
W
=
ε
2
ZZ Z
volume
X
i
∇φi (r) ·
X
j
∇φj (r) d3 r
(1.144)
ZZZ
εX
|∇φi (r)|2 d3 r
=
2 i
volume
ZZZ
ε X
∇φi (r) · ∇φj (r) d3 r
+ 2
2 i,j
volume
i< j
The term
Wi =
ε
2
ZZZ
2
volume
|∇φi (r)| d3 r
self energy
(1.145)
is the electromagnetic ‘self energy’ of the ith charge element. As long as the charge density of the element does not change
this ‘self energy’ will be fixed. The second term,
Wij = ε
ZZZ
volume
∇φi (r) · ∇φj (r) d3 r,
40
interaction energy
(1.146)
Section 2.7
Electrostatic energy density, capacitance
with i 6= j provides the interaction energy between the charged elements. The significance of the various terms is illustrated
by the following example.
Example
Consider two identical non-conducting spherical shells. Let each have a radius a and let their centers be separated by a
distance 2a. The surface charge of one shell is σ0 and the surface charge of the other is -σ 0 (a)What is the self energy of
each spherical shell? ~(b)What is the interaction energy of the spherical shells?
Solution: (a) The magnitude of the charge on each shell is q = 4πa2 σ0 . The electric field of spherical shell vanishes
inside the shell while outside the shell the electric field is
E (r − r± ) =
=
±q (r − r± )
4πε |r − r± |3
±q (r − r± )
ε |r − r± |3
= −∇φ± (r).
SI units
r+ and r- locate the
center of the spherical
shells.
in Gaussian units
To calculate the self energy of each shell we evaluate, Eq.1.145 ,
W0 =
εq 2
2(4πε)2
ZZ Z
|r−r± | > a
d3 r
.
|r − r± |4
For each shell we can let r0 = r − r± and the self energy of each spherical shell is found to be
W0
=
=
W0
=
=
ZZ Z
Z ∞
εq 2
4π εq 2
d3 r0
1 0
=
dr
4
2
02
0
2(4πε)2
2(4πε)
r
0
a
|r | >a |r |
· ¸∞
Why does one integrate
q2
−1
from a to infinity, rather
2 · 4πε r0 a
than zero to infinity?
q2
. Self energy of each shell in SI units
2 · 4πεa
q2
Gaussian units
2εa
(1.147)
Aside: the charge radius of the electron
Equation 1.147 can be used to introduce a classic physics problem which persists in all models. Suppose we consider the
electron to be a spherical charged shell. In this case the electromagnetic energy stored in the electric field of the electron
would be (ε = 1)
We =
e2
in Gaussian units
2a
41
Section 2.7
Electrostatic energy density, capacitance
The total energy of an electron at rest is me c2 . The question now is what would be the radius of the electron if all its energy
was that stored by the electric field? The answer is
(4.8 × 10−10 statC)2
e2
=
2me c2
2 × (9.1 × 10−28 g) × (3 × 1010 cm/s)2
= 1. 4 × 10−13 cm !!
a =
If this were correct the electron radius would be the same order of magnitude as the observed proton radius. But the upper
limit on the radius of the electron is several orders of magnitude below the radius of the proton. In fact the results of all
experiments at this time are consistent with the electron being a point particle! What happened to its electromagnetic energy?
Solution(b) The potential due to a uniform spherical shell of charge centered at the origin is
φ (r) =
φ (r) =
q
for r > a
4πεr
q
for r ≤ a
4πεa
where q is the total charge of the shell and a is the radius of the shell. In terms of this potential the interaction energy stored
by the electric fields is, Eq. 1.146,
Wint = ε
Z ZZ
∇φ (r − r+ ) · ∇φ (r − r− ) d3 r
Using the identity,
∇ · [φ (r − r+ ) ∇φ (r − r− )] = ∇φ (r − r+ ) · ∇φ (r − r− ) + φ (r − r+ ) ∇2 φ (r − r− )
Wint = −ε
Z ZZ
Now let r0 =r − r−
= −ε
ZZ Z
all
space
φ (r − r+ ) ∇2 φ (r − r− ) d3 r + ε
ZZ
surf ace
at ∞
φ (r − r+charge
) ∇φ (rdensity
− r− ) for
· dSshell of charge
Note: r- is at the center of the shell.
all
space
φ (r0 −r+ + r− ) ∇2 φ (r0 ) d3 r0 + ε
ZZ
surf ace
at r0s −→∞
φ (r0s −r+ + r− ) ∇φ (r0s ) · dS0
Since the product of the field and the potential vanish at infinity as rs0−3 whereas dS0 only increases like rs02 the surface
integral does not contribute to the interaction energy. To evaluate the spatial integration we note that ∇2 φ (r0 ) refers to the
negative shell and (see Jackson problem 1.3a):
∇2 φ (r0 ) = −
σ0
δ (r0 − a) . SI units
ε
In the example the distance between the centers of the spheres is greater than the diameter of the spheres. Therefore the
42
Section 2.7
Electrostatic energy density, capacitance
spheres do not overlap and the interaction energy is given by
ZZ Z
φ (r0 −r+ + r− ) ∇2 φ (r0 ) d3 r0
Wint = −ε
ZZ Z
= −ε
all
space
all
space
σ0
q
[− δ (r0 − a)]d3 r0
0
4πε|r −2a|
ε
Letting 2a =Rẑ
Wint = −
q σ0 1
ε 4π
Z
π
0
Z
0
∞ Z 2π
0
[r2
δ (r − a)
+ R2
1/2
− 2rR cos θ]
dγ r2 dr sin θ dθ
The integration over the angle γ and the radial direction r are readily carried out leaving (ξ = cos θ)
Z
q σ 0 2π π
a2
sin θ dθ
Wint = −
ε 4π 0 [a2 + R2 − 2aR cos θ]1/2
Wint
= −
= −
q
£
q
4πa2
2ε
q2
2 · 4πε
2
¤
a2
Z
Z
+1
−1
ξ=+1
ξ=−1
−
dξ
[a2
+ R2
− 2aRξ]1/2
¤1/2
1 £ 2
d a + R2 − 2aRξ
aR
q
[R − a] − [R + a]
2 · 4πε
aR
q2
= −
interaction energy of two shells
4πεR
=
The interaction energy is equal to that of two charged point particles, with charges ±q, separated by a distance R. The
total energy stored in the electrostatic field is
W =
q2
q2
−
> 0 total energy for the two shells
4πεa 4πεR
As R decreases the stored energy decreases.
·
¸
∂
q2
q2
dW = −|dR|
|dR| = −F |dR|
−
=−
∂R
4πεR
4πεR2
Therefore the force between the shells is attractive with a magnitude of q 2 /[4πεR2 ] and does work dW
·
¸
q2
q2
dR
W + dW =
−
1+
4πεa 4πεR
R
43
Section 2.8
2.8
Conductors and Capacitance
Conductors and Capacitance A commonly encountered system consists of a set of N isolated charged
conductors. (In electrostatics the potential inside and at the surface of a conductor is constant.). Let the ith conductor have
a charge Qi and let all other conductors have zero net charge. The charge on the ith conductor will be distributed over its
surface with a surface charge density σ ii (r) which is proportional to Qi . It also will induce a surface charge density on each
of the other conductors σji (r) which will be proportional to Qi . Of course
Qi
ZZ
=
ZZ
0 =
σii (r) dS
(1.148)
i
j
σ ji (r) dS, j 6= i
If all the conductors are charged then the surface charge density on a conductor will be the sum of the surface charge
densities which would be on the conductor with only one conductor charged. [This follows from the uniqueness of the
solution to the boundary value problem. In this case the potential of each conductor can not vary over the conductor. This is
satisfied if either the ith conductor has a charge Qi or the j th conductor has a charge Qj . If the potentials obtained with each
of these conditions are added the result will also satisfy the boundary conditions].
σ i (r) =
X
σ ij (r) .
j
The j th term in the sum is proportional to Qj . The potential of the ith conductor is given by
Vi =
N ZZ
X
k=1
k
σ k (r)
dS
|ri −r|
(1.149)
where ri is any point on the ith conductor. Because the {σk (r)} , are linear functions of the {Qi } , it follows that
Vi =
N
X
pij Qj .
(1.150)
j=1
The proportionality constants {pij } will depend on the geometry of the system of conductors but not the charges on the
RR σj (r)
conductors. ( Note that pij 6= Q1j
j |ri −r| dS since the charge distribution on the surface of each conductor depends
linearly on the charges of the other conductors.) The linear relationship in Eq. 1.150 can be inverted to obtain the charges on
the conductors in terms of the potentials of the conductors
Qi =
N
X
Cij Vj
(1.151)
j=1
The coefficient Cii is the capacitance of the ith conductor in the system
.................while the {Cij , with i 6= j} are the coefficients of induction.
44
Section 2.9
Variational calculations
The capacitance of the ith conductor is obtained by ‘grounding’ all the other conductors (forcing Vj = 0 for j 6= i) and
measuring the ratio
Cii =
Qi
, V j = 0 f or j 6= i.
Vi
(1.152)
The potential energy for a system of charged conductors is
W =
N
N
1 X
1X
Qi Vi =
Cij Vi Vj
2 i=1
2 i,j=1
(1.153)
This equation must give the energy stored by the electric fields of the system and therefore W ≥ 0 for all choices of {Vi } . It
follows that Cii > 0 for all i and det [Cij ] > 0. (We have tacitly assumed that the inverse of the matrix Cij exists. In fact it
is the matrix pij . The elements of Cij form a non-singular matrix and therefore det [Cij ] 6= 0.)
If we know the Dirichlet green’s function for the region of space excluding the conductors we are able to generate the
Cij .(The surface charge density on a conductor in electrostatics equals the normal derivative of the potential at the surface.)
It is also possible by using a variational technique to obtain estimates of the Cij .
2.9
Variational calculations
2.9.1 A variational principle The behavior or properties of a system can generally be described by one or more
functions. In classical mechanics,
for example, we seek rk (t) for a system of particles. A common technique is to minimize
R
the action integral, I = L( rk (t), ṙk (t))dt. Recall that this leads to the Euler-Lagrange equations. If constraints are
imposed on the system one introduces Lagrange multipliers. In electrostatics we apply a similar technique to equations such
as
I(φ) =
1
2
Z ZZ
volume
∇φ · ∇φ d3 r.
subject to constraints. This works particularly well when the electrical potential φ is due to a system of charged conductors.
More formally, we vary the functional form for the ”system function”, ψ (r) ,via a ”functional”, I, defined as
I [ψ, ∇ψ] =
Z ZZ
F (r,ψ (r) , ∇ψ (r))d3 r
which has an extremum in for the correct functional form for ψ (r) . We shall use the following definitions:
δψ = ”the variation in the functional form for ψ
δ∇ψ = ∇[δψ] and
δI = I [ψ, ∇ψ] − I [ψ + δψ, ∇ψ + δ∇ψ].
45
(1.154)
Section 2.9
Variational calculations
Using an expansion similar to a three dimensional Taylor series expansion,
f (r + ∆r) = e∆r·∇ f = [1 + ∆r · ∇ + [∆r · ∇]2 /2! + ....[∆r · ∇]n /n! + ...]f
(1.155)
one can obtain,
δ
δ
+ δ∇ψ ·
] I [ψ, ∇ψ]
(1.156)
δψ
δ∇ψ
δ
δ
δ
δ 2
= [1 + [δψ
+ δ∇ψ ·
] + [δψ
+ δ∇ψ ·
] /2! + ....] I + ...
δψ
δ∇ψ
δψ
δ∇ψ
δ
δ
δ
δ 2
= I [ψ, ∇ψ] + [δψ
+ δ∇ψ ·
]I + [δψ
+ δ∇ψ ·
] I/2! + ...
δψ
δ∇ψ
δψ
δ∇ψ
I [ψ + δψ, ∇ψ + δ∇ψ] = exp[δψ
To first order in δψ we have:
δI =
ZZ Z
[
vol
δF 3
δF
δψ (r) + ∇[δψ] ·
]d r + higher order terms....
δψ
δ∇ψ
(1.157)
where
3
∇[δψ] ·
We can replace
δI
δF
δψ
by
X ∂
δF
δF
≡
.
δψ
δ∇ψ
∂xi
δ[∂ψ/xi ]
i=1
when F is a continuous function of ψ in the volume.
∂F
∂ψ
ZZZ
∂F 3
∂F
δψ (r) + ∇[δψ] ·
]d r + higher order terms....
∂ψ
∂∇ψ
ZZZ vol
∂F
∂F 3
∂F
δψ (r) + ∇ · (δψ
) − δψ∇ ·
]d r + higher order terms...
[
=
∂ψ
∂∇ψ
∂∇ψ
vol
=
[
(1.158)
Using the divergence theorem,
ZZZ
Z ZZ
∂F
∂F
∂F
−∇·
]δψd3 r +
)d3 r + higher order terms
∇ · (δψ
∂ψ
∂∇ψ
∂∇ψ
Z Z vol
ZZZ vol
∂F
∂F
∂F
3
−∇·
]δψd r +
· n̂ dS + .higher order terms....
[
δψ
=
bounding
∂∇ψ
∂∇ψ
vol ∂ψ
δI
=
[
(1.159)
surf ace
The functional I will have an extremum when δI = 0 or,
ZZ Z
vol
"
#
ZZ
3
∂F X ∂ ∂F (r,ψ (r) , ∇ψ (r))
−
[
δψd3 r +
∂ψ i=1 ∂xi
∂ (∂ψ/∂xi )
bounding
surf ace
"
3
X
i=1
#
δF
n̂ · x̂i δψ dS = 0
δ (∂ψ/∂xi )
(1.160)
for all ‘permissible’ δψ (r) . Generally the values of the function ψ (r) are specified on the boundary of the volume. In this
46
Section 2.9
Variational calculations
case only δψ (r) which vanish on the boundary would be permitted and the surface integral would be zero. If that is the only
restriction on the variation in ψ(r) then the requirement that the first order term vanishes constrains ψ (r) to be a solution of
the following ”Euler-Lagrange” equation.
3
∂F X ∂ ∂F (r,ψ (r) , ∂ψ/∂xi )
−
=0
∂ψ i=1 ∂xi
∂ (∂ψ/∂xi )
(1.161)
To determine whether the extremum is a minimum, maximum, or stationary point the second order term in δψ must be
calculated.
2.9.2
Variations with equality constraints and fixed boundary conditions In an extension of the
variational calculations the function ψ (r) is required to satisfy one or more constraints in addition to the boundary
conditions. Generally these have a form,
Cα [ψ] =
ZZ Z
Kα (r,ψ (r) , ∇ψ (r)) d3 r = Dα
(1.162)
where {Dα , α = 1, ..., N } is a set of N constants. This places restrictions on the δψ (r) used in the variation of ψ (r) .in
Eqs.1.156-1.160. Proceeding as before to set the δCα = 0, we have α = 1, 2, 3, ...N. equations
ZZ Z
vol
"
#
ZZ
3
δKα X ∂
δKα
3
−
δψd r +
δψ
∂xi δ (∂ψ/∂xi )
i=1
3 ·
X
bounding
surf ace i=1
¸
δKα
x̂i · n̂ δψdS = 0
δ (∂ψ/∂xi )
(1.163)
Because ψ (r) must satisfy fixed boundary conditions δψ (r) is zero on the surface and the surface integral vanishes. The
equations which must be satisfied are
ZZZ
vol
"
#
3
δF (r,ψ (r) , ∇ψ (r)) X ∂ δF (r,ψ (r) , ∇ψ (r))
−
δψ (r) d3 r = 0
δψ
∂x
δ
(∂ψ/∂x
)
i
i
i=1
(1.164a)
#
3
δKα (r,ψ (r) , ∇ψ (r)) X ∂ δKα (r,ψ (r) , ∇ψ (r))
−
δψ (r) d3 r = 0.
δψ
∂r
δ
(∂ψ/∂r
)
i
i
i=1
(1.164b)
and
Z ZZ
vol
"
We can consider the
47
Section 2.9
Variational calculations
3
uα (r) =
δKα X ∂
δK
−
δψ
∂x
δ
(∂ψ/∂x
i
i)
i=1
(1.165)
to be a set of ‘vectors’ in a function space. In this case the restriction on δψ (r) is that it lie in the subspace which is
orthogonal to the uα (r) . (Note that Eq. 1.164b looks like an inner product condition, (uα , δψ) = 0 as does Eq.1.164a
P3
∂
δF
, ( δF
i=1 ∂xi δ(∂ψ/∂xi ) , δψ) = 0.) In this case Eq. 1.164a requires that the function multiplying δψ (r) lie in the
δψ −
subspace spanned by the uα (r) or
3
X
δF
δF X ∂
−
=
λα uα (r)
δψ i=1 ∂xi δ (∂ψ/∂xi )
α
(1.166)
The ‘expansion coefficients’ λα are constants, called Lagrange multipliers, which are determined such that the constraint
equations are satisfied.
2.9.3
Approximation using the functional
If the function describing the property of the
system provides an extremum for Eq. 1.159 an approximation to the function can be generated. In the simplest case let
ψ (r) = w (β, r) where β is a free parameter and the w (β, r) is chosen to satisfy the boundary conditions on ψ (r) but is
otherwise arbitrary. The functional equation is now a function of β
ZZZ
(1.167)
W (β) =
F (r,w (β, r) , ∇w (β, r))d3 r.
Assume further that,
W (β) =
ε
2
ZZZ
|∇w (β, r) |2 d3 r.
To find the extremum of W (β) ( δW (β) = 0) subject to w (β, rs1 ) = 1,and w (β, rs2 ) = 0 we find the value of β, say
b0 , for which W 0 (β) = 0 restricted to functions of the form w (β, r) (See also Eq. 1.166.). If W 00 (bo ) is consistent with the
type of extremum required ,w (b0 , r) is an approximation to ψ (r) .
2.9.4
Estimation of capacitance using a variational approach
In an
electrostatic problem the potential of each conductor is fixed. One then guesses a trial potential which satisfies the boundary
conditions. This trial potential should depend on one or more parameters which are varied so as to obtain the minimum value
of the energy for the system. In the case that the trial potential =1 on the ith conductor and zero on all other conductors the
capacitance, Cii , is approximated by the estimated energy. That is, for two conductors,
W
=
=
=
2
1 X
Cij Vi Vj
2 i,j=1
1
1
C11 V12 = C11
2
2
1
1
C22 V22 = C22
2
2
if V1 = 1 and V2 = 0
if V2 = 1 and V1 = 0
48
Section 2.9
Variational calculations
Exercise:
Show that the electrostatic energy stored in the field of a charged conductor (with charge Q) and potential V is
1
W = QV . (Hint: solve the boundary value problem for φ(r) everywhere and calculate W from Eq. 1.141)
2
49
Section 2.9
Variational calculations
Example of simple variational method: coaxial cylinders
Consider two coaxial cylinders, one with a radius b = 0.2cm and the other with a radius c = 0.8cm = 4b. Let
(x2 + y2 ) = ρ2 and let the z axis be along the cylindrical axis. Using the potential form ,
w(β, ρ) = β[1 −
ρ2
2ρ 5ρ2
ρ
+ )] +
−
2b 2b
3c
3c
estimate the capacitance per unit length of the combination and compare the result to the actual value.
Solution :
Note that we want the function,φ(β, ρ), to be 1 at ρ = b and 0 at ρ = c.This can be done by normalizing w(ρ, z):
φ(β, ρ) = aw(β, ρ) − b where a and b are constants to be determined after we find β
= [w(β, ρ) − w(β, c)]/ [w(β, b) − w(β, c)]
Since Vb = φ(β, b) = 1, and Vc = φ(β, c) = 0 we will obtain W = 12 Cbb . The estimated energy of the coaxial system
per unit length, L, in Gaussian units is
Z ZZ
²0
|∇φ|2 ρdρdϕdz
²0 = 1 in Gaussian units
W (β) =
8πL
¸ ·
¸
ZZ ·
1 ∂φ
∂φ
1 ∂φ
∂φ
1
∂φ
∂φ
+ ϕ̂
+ ẑ
+ ϕ̂
+ ẑ
=
ρ̂
· ρ̂
ρdρdϕdz
8πL
∂ρ
ρ ∂ϕ
∂z
∂ρ
ρ ∂ϕ
∂z
Z ¯¯ ¯¯2
1
¯ ∂φ ¯ ρdρ · 2πL
=
8πL ¯ ∂ρ ¯
W (β) =
a2
4
Z
b
c
|
2
β
[−1 + 2ρ] + (1 − 5ρ)|2 ρdρ
2b
3c
This has an extremum (minimum) for
a2
W (β) =
2
0
β
Z
b
c
¯
Z c¯
¯
¯β
¯ [−1 + 2ρ] + 2 (1 − 5ρ)¯ 1 [−1 + 2ρ] ρdρ = 0
¯ 2b
¯ 2b
3c
b
¯2
¯
Z
¯
¯1
¯ [−1 + 2ρ]¯ ρdρ = −
¯
¯ 2b
c
b
¯
¯
¯
¯1
¯ [−1 + 2ρ] 2 (1 − 5ρ)¯ ρdρ
¯
¯ 2b
3c
¯
Rc¯ 1
¯ [−1 + 2ρ] 2 (1 − 5ρ)¯ ρdρ
3c
b 2b
β=−
¯2
Rc¯ 1
¯
¯ ρdρ
b 2b [−1 + 2ρ]
4b
β=−
3c
R 0.8 cm
[−1 cm + 2ρ] [2 cm − 5ρ] ρdρ
= 1.35
R 0.8 cm
|[−1cm + 2ρ]|2 ρdρ
0.2 cm
0.2 cm
50
Note that b' disappears
and a is a constant
determined from the
normalization.
Section 2.9
Variational calculations
This gives an approximation to the capacitance per unit length equal to
a2
2W (1.6) = C =
2
= 0.367
¯2
Z c ¯¯
¯
¯ β [−1 + 2ρ] + 2 (1 − 5ρ)¯ ρdρ
¯
¯ 2c
3b
b
The plot in Fig. 7 shows the W(β) (using the normalized potential) generated via mathcad. The β from this numerical
approach is approximately 1.35 (see two.cylinder.example.mcd on the web.). The value of W (1.35) = 0.183 and C =
2 · .183 = 0.367. The capacitance appears to be not too dependent on β. Figure 8 is a plot of the normalized potential as a
function of ρ.
Fig. 7
Fig. 8
The exact result is C = 2 ln1|4| = 0. 361. So the variational results differ by less than 2% . We shall determine the
”exact” result for L –> ∞ later in the notes.
51
Section 2.9
Variational calculations
The exact result for the coaxial cylinders is non-trivial:
The exact result for the capacitance of two long cylindrical shells is non-trivial. But we can begin by letting the outer
c
conductor be at 0 potential and the inner potential at V then C = Q
V . The length, L, of the cylinders satisfies L << 1.
Φ(ρ, ϕ, z) =
−1
4πε
ZZ Z
σ0 δ(ρ0 − b)|∇(ρ0 − b)|
1
ρ0 dρ0 dϕ0 dz 0 + Φ0 (b, c)
|r − r0 |
where Φ0 (b, c) is a constant.and we shall let z be along the cylindrical axis, with ρ2 = x2 + y 2 . Furthermore, we shall
evaluate the potential on the x axis so that r = x = ρ cos ϕ = ρ, z = 0 and r · r0 = ρρ0 cos ϕ0
Φ(ρ, 0, 0) =
Φ(ρ, 0, 0) =
≈
=
ZZZ
−1
1
σ 0 δ(ρ0 − b)|∇(ρ0 − b)|
ρ0 dρ0 dϕ0 dz 0 + Φ0 (b, c)
4πε
|r − r0 |
ZZ
1
−Q
δ(ρ0 − b)|
ρ0 dρ0 dϕ0 + Φ0 (b, c)
1/2
2
02
02
0
4πε(2πbL)
[ρ + ρ + z − 2r · r ]
Z L/2 Z 2π
−Q
bdϕ0 dz 00
h
i1/2 + Φ0 (b, c)
4πε(2πbL) −L/2 0
1/2
2
2
02
2
02
0
ρ + b + z − 2ρ [b + z ] cos ϕ
Z L/2 Z 2π
−Q
dϕ0 dz 0
+ Φ0 (b, c) where a2 = [b2 + z 02 ]
4πε(2πL) −L/2 0 [ρ2 + a2 − 2ρa cos ϕ0 ]1/2
Unfortunately this reduces to an elliptic integral which is unmanageable. In the next chapter we shall have some methods
for handling this problem.
Using Gauss’ law one can obtain the exact result: C =
4π²0
b
2 ln[ ]
a
52
Section 2.9
Variational calculations
Using the variational technique
to find the upper and lower bounds for the capacitance:
The capacitance of a system of conductors.depends on the geometry. However we often use the potential difference and
the charges on the conductors to determine C. In Jackson problems 1.17 and 1.18 one is asked to show that upper and lower
bounds can be placed on the capacitance by considering two different variations in the functional forms of C:
·
¸
Q2 [σ]
|maximized ≤ Cactual ≤ 2W [Ψ]minimized
2W [σ]
(1.90)
where
·

¸

i2
hH
S1 σ(r)dS1
Q2 [σ]


|maximized =  H H
0 )σ(r0 )dS dS 0 
2W [σ]
σ(r)G(r,
r
1
1
S1 S1
(1.91)
max
and is more commonly written:
·
1
C[σ]
¸
min


=
H
S1
H
S1
0
0
σ(r)G(r, r )σ(r
hH
i2
σ(r)dS
1
S1

)dS1 dS10 

(1.92)
min
The right hand side of Eq. 1.90 is relatively easy to use and is often employed to determine C. The left hand condition
of Eq. 1.90 is in contrast often intractable because of the need to find G(r, r0 ). The Green’s function in Eqs. 1.91 and 1.92
must = 0: on all conductors except S1 . The solution to Problem 1.18 provides the derivation of Eq 1.92. In this solution
one considers the system for which all Qi are
· fixed ¸and all Φi = 0 but Φ1 . Subject to this condition, the surface charge
W [σ]
densities are allowed to adjust to minimize
. For two conductors, this means Φ2 = 0 (with fixed Q2 ) while
Q21 [σ]
1
. Problem 1.19 provides an example of how the actual C is always less
Φ1 varies as σ1 (rS1 ) is adjusted to minimize .
C[σ]
than the approximated C[Ψ]., and is only equal to C[Ψ] when the correct potential function is used. The minimization of
1
occurs when the charge distributed on S1 puts all the other conductors at constant potential (0)
C[σ]
53
Section 2.10
2.10
Linear charge density
Linear charge density Since a line is the intersection of two surfaces we could extend the approach used for
surface charge densities to the case of linear charge densities. In this case σ (r) would involve a delta function which would
restrict the charge to also lie on a second surface, say G (r) = 0. The result would be that
σ (r) = λ (r) δ (G (r)) |∇G (r)|
with λ (r) the linear charge density. However a linear charge density is often given along a parametrized curve,
r (s) = (x (s) , y (s) , z (s)) being the points along the curve. With this information we know that the charge at the point r0
(parameter s0 ) on the curve is
¯
¯
q
¯
¯d
¯
dq = λ (r0 ) ¯ r (s)¯¯ ds = λ (r0 ) x0 (s0 )2 + y 0 (s0 )2 + z 0 (s0 )2 ds
ds
r0
and the function which gives the variation in the electrostatic potential due to this linear charge density is
φ (R) =
Z
λ (r (s))
|R − r (s)|
q
x0 (s)2 + y0 (s)2 + z 0 (s)2 ds
54
Appendix
1.
A
Special
problems
A hyperboloid of revolution has a uniform surface charge density of 7 statcoul /cm2 . The charged surface lies between
z = −2 cm and z = +2 cm and is given by the equation
x2 + y2
− z 2 = 1 cm2 .
2
2.
3.
(a) Determine the function which gives the volume charge density corresponding to this surface charge. (b) Taking the
potential at the origin to be zero, evaluate and plot the potential along the positive x axis from x = 0 cm to x = 4 cm. 2
(c) Evaluate and plot the components of the electric field along the positive x axis from x = 0 cm to x = 4 cm.
A line charge lies along the ¡parabola z = x2 /(3
¢ cm), y = 0 cm in the region 0 cm ≤ z ≤ 3 cm. The linear charge
density is given by λ (z) = 5 statcoul / cm2 z. Along the line z = 3 cm, x = 0 cm for 0 cm ≤ y ≤ 3 cm evaluate
and plot the (a) potential and (b) the components of the electric field.
In Example ?? we found that the energy stored by the electric field of a charged spherical shell is
W0 =
q2
2εa
with q the total charge of the shell and a the radius of the shell. Since
q2
δW0
=−
δa
2εa2
4.
5.
there is a force acting at each point on the shell which would cause the shell to expand. (a) Determine and explain the
source of this force. (b) Demonstrate that the source is correct by calculating the force per unit area of the sphere and the
total work done by the force if the radius of the sphere expands by δa.
A spherical capacitor consists of two concentric conducting spherical shells. The inner shell has a radius a and the outer
a radius 2a. Using the ‘test potential’
i
h
(r
−
a)
sin 2πβ
a
, a<r<b
φ(β, r) =
sin (2πβ)
φ (β, r) = 0, r < a; φ (β, r) = 1, 2a < r
(a) determine the value of β which gives the minimum value for the energy stored in the electric field. (Show work) (b)
Compare the energy obtained by the minimization technique to the actual energy this system would have for the given
parameters. (c) Estimate the capacitance of the system and (d) compare this estimated value to the actual value for the
capacitance of this spherical capacitor. (e) In the range a < r < 2a plot the difference between the actual potential
function between the spheres and the potential function obtained with φ (β, r) . (f) In the range a < r < 2a plot the
difference between the actual radial electric field between the spheres and the radial field obtained with φ (β, r) .
A sphere of radius 1 cm sets 2 cm above the center of a large grounded conducting plane. As a model assume that
the plane is infinite in extent. Choosing a coordinate system with the sphere centered at the origin and the plane at
z = −2 cm use the function
Φ (p; r, θ, φ) = (
1 cm p r cos θ + 2 cm
)
r
1 cm cos θ + 2 cm
to obtain an upper bound on the capacitance of the sphere.
2 Note: At best the potential can be expressed in terms of tabulated integrals. Because those with experience with these tabulated integrals can often
describe the general properties of the potentials the ‘closed’ form for the integral is often obtained. This capability is associated with having solved
similar problems many times and checking the plots in a handbook of functions. In this course I will assume that the function is known if it is
written as a convergent integral with a finite integrand.
55