Chapter 2Introduction 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. Inthis case the electric field satisfies the two equations, Eq. 1.59a with a time independent charge density and Eq. 1.77 with atime 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 : (1.78b)D = ²oE + P .... SI unitsD = E + 4πP in Gaussian units

in these cases ε = [1 + 4πP/E] Gaussian

The solution of Eq. 1.59 is

D (r) =1

4π

ZZZρ0 (r

0) (r− r0)|r− r0|3 d3r0 +D0 (r) , SI units (1.79)

with ∇ ·D0 (r) = 0

If we are seeking the contribution of the charge density, ρ0 (r) , to the electric displacement vector thenD0 (r) = 0. Thegiven charge density generates the electric field

E (r) =1

4πε

ZZZρ0 (r

0) (r− r0)|r− r0|3 d3r0 SI units (1.80)

18

Section 2.2 The electric or scalar potential2.2 The electric or scalar potential Faraday’s law with static fields, Eq. 1.77, is automaticallysatisfied 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 thescalar potential at two points by integrating the tangent component of the electric field along any path connecting the twopoints

−Zpathra→rb

E (r) · d` =

Zpathra→rb

∇φ (r) · d` (1.82)

=

Zpathra→rb

·dx

∂φ (r)

∂x+ dy

∂φ (r)

∂y+ dz

∂φ (r)

∂z

¸=

Zpathra→rb

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 integralof the tangential component along a closed path is zero,I

E (r) · d` =Idφ (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

∇ 1q

(x− x0)2 + (y − y0)2 + (z − z0)2

(1.84)

= − (x− x0) i+ (y − y0) j+(z − z0)kh(x− x0)2 + (y − y0)2 + (z − z0)2

i3/2= − r− r0

|r− r0|3 .

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 scalarpotential due to the charge density ρo (r) is

φ (r) =1

4πε

ZZZρo(r

0)|r− r0|d

3r0 + φ0, (1.85)

where φ0 is a constant which fixes the (arbitrary) location of the zero for the scalar potential. Since the observed quantity isthe electric force and, therefore, the electric field, only the difference in the values of the scalar potential at any two differentpoints is significant. (See Eq. 1.82)

19

Section 2.3 Surface charges and charge dipoles

2.3 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, (orF (x, y, z) = 0)and a surface charge density, σ (r) , with dimensions of Coulombs/m2 in SI units and statcoulombs persquare centimeter in gaussian units.

Restricting a volume integral to a surface with a delta function in the integrandThe 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

ZZZH (r) δ (F (r)) ∗ f (r) d3r =

ZZF (r)=0

H (r) dS (r) (1.86)

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 =

Za (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 orthogonalaxes which are tangent to the surface at each point. The unit vector which is perpendicular to the surface at each point iseither bn = +∇F (r) / |∇F (r)| or bn = −∇F (r) / |∇F (r)| .

If we let ξbn be the displacement of a point from the surface at r0 then d3r = [dS (r0) bn] · dξbn. In addition if r0 is a pointon the surface then

∇F (r0) =·∂F (r0 + ξbn)

∂ξ

¸ξ=0

. (1.88)

When performing the volume integration we carry out the integration perpendicular to the surface at each point r0 and thenmultiply the result by the differential surface element at r0, dS (r0) .We have in this way reduced our problem to a set of onedimensional problems, one for each point r0 on the surface. Viewing the problem in this way suggests that the appropriateexpression for the function f (r) is

f (r0 + ξbn) = bn ·∇0F (r0 + ξbn) (1.89)=

¯∇0F (r0 + ξbn)¯It follows then that the volume charge density associated with a surface charge density, σ (r) , on the surface defined byF (r) = 0 is

ρ (r) = σ (r) δ (F (r)) |∇F (r)| (1.90)

20

Section 2.3 Surface charges and charge dipoles

Note: A simpler way to obtain this result is as follows, where R δ(F (r))dF = 1:ZZF (r)=0

σ (r) dS (r) =

ZZF (r)=0

σ (r)

·Z 0δ(F (r))dF

¸dS (r) (1.90b)

=

ZZF (r)=0

σ (r)

·Z 0δ(F (r))dr0·∇F

¸dS (r)

=

ZZF (r)=0

σ (r)

·Z 0δ(F (r))dr0· ∇F|∇F | |∇F |

¸dS (r)

=

ZZF (r)=0

σ (r)

Z 0δ(F (r))|∇F | dr0·ns dS (r)

=

ZZF (r)=0

σ (r)

Z 0δ(F (r))|∇F | dr0· dS (r)

where the primed integration symbol indicates that the integration must be taken along a direction, dr0 whichis along the normal to the surface.

=

ZZF (r)=0

σ (r)

Z 0δ(F (r))|∇F | dV 0 (1.90c)

ZZF (r)=0

σ (r) dS (r) =

ZZZV contains F (r)=0

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 x2+y2

a2 + z2

b2 − 1 = 0 . The surface chargedensity is σ0. Find the potential on the z axis.

Solution: From the class notes, Eq. 1.90, the volume charge density is given by

ρ (r) = σ0δ

µx2 + y2

a2+z2

b2− 1¶|∇(x

2 + y2

a2+z2

b2− 1)| (1.91)

= σ0δ

µx2 + y2

a2+z2

b2− 1¶2(x2 + y2

a4+z2

b4)1/2

Because of the cylindrical symmetry of the charge distribution it is convenient to work in cylindrical coordinates, x = ξ cosϕand y = ξ sinϕ. In these coordinate the charge density is

21

Section 2.3 Surface charges and charge dipoles

.Fig. 3

ρ (r) = 2σ0δ

µξ2

a2+z2

b2− 1¶s

ξ2

a4+z2

b4

Taking the zero of potential at infinity, the potential on the z axis due to the surface charge is

φ (zz) =2σ04πε

ZZZ δ³ξ2

a2 +z2

b2 − 1´q

ξ2

a4 +z2

b4qξ0 2 + (z − z0)2

d3r0 SI units (1.92)

with d3r0 = ξ0dξ0 dϕ0 dz0. The integrand is independent of ϕ0 and the angle integration can be performed giving 2π. The

integrand is also only a function of ξ0 2. This suggests that we define an η = a−2ξ0 2. In this case ξ0 dξ0 = 0.5a2dη.Withthese changes the potential is given by

φ (zz) =2πσ0a2

4πε

Z ∞−∞

Z ∞0

δ¡η + b−2z0 2 − 1¢pa−2η + b−4z0 2q

a2η + (z − z0)2dη dz0 (1.93)

The delta function allows the η integration to be carried out yielding

φ (zz) =σ0a

2ε

Z b

−b

"a−2 + b−2z0 2

¡b−2 − a−2¢

(1− b−2z0 2) + a−2 (z − z0)2#1/2

dz0 (1.94)

It is convenient to express the z coordinates in units of b, z = bξ0 and z0 = bξ0, then, with β = (b/a)2 ,

φ (ξ0bz) =σ0a

2ε

Z 1

−1

"β + ξ02 (1− β)¡

1− ξ02¢+ β

¡ξ0 − ξ0

¢2#1/2

dξ0 (1.95)

22

Section 2.3 Surface charges and charge dipoles

.Fig. 4

The integral in Eq. 1.95 can be done numerically and the normalized potential, Φ(a, b, zz) =φ (zz)

σ0a/ε, is plotted below in

Fig. 4 for three cases. b = 4a, a anda

4. Note that when a = b the ellipsoid of revolution becomes a charged sphere.

The potential due to a spherical shell of radius a with total charge, Q, (and uniform surface charge density, σ0 =Q

4πa2)

can be obtained using Gauss’ law. Because of the spherical symmetry E has a constant magnitude on a sphere of radius rZZZ∇ ·E dV =

ZZZρ/εdVZZ

E · dS = Qenclosed/ε

|E|4πr2 = 4πa2σ0/ε if r ≥ a= 0 if r < a

Thus

|E| =Q

4πεr2if r ≥ a

= 0 if r < a

From E = −∇φ one finds

φspherical shell (r) =Q

4πεr=

σ0ε

a2

rif r ≥ a

=Q

4πεa=

σ0εa 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 notpartiularly 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 σ0b/4ε (at z = 0, .b, 2b ) as a function of b/a is shown in Fig. 5 . Thesmall values of b/a correspond to a ‘pancake’ shaped ellipse and the potentials at the origin, the middle of the ‘pancakesurface’, and just above the ‘pancake’ surface are nearly the same. Large values of b/a give a ‘needle’ shaped ellipse. (Notethat 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 approximatethe fields and potentials by those of a surface charge density. The potential due to a surface charge density, σ (r) ,is formallygiven by

φ (r) =1

4πε

ZZsurface

σ (r0)|r− r0|dS

0 + φ0. 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 chargedensity σ0, lying in the z = 0 plane. Determine the value of the potential at (x, y, z) (also denoted by r) with the conditionthat the potential vanish at the surface.

Solution: The differential surface area for this problem is dS0 = dx0dy0. This can be used in Eq. 1.96 to obtain the24

Section 2.3 Surface charges and charge dipoles

required integral

φ (x, y, z) =

Z ∞−∞

Z ∞−∞

σ0/4πεq(x− x0)2 + (y − y0)2 + z2

dx0dy0 + φ0 (1.97)

=

Z ∞−∞

Z ∞−∞

σ0/4πεq(x− x0)2 + (y − y0)2 + z2

d(x− x0)d(y − y0) + φ0

=σ04πε

Z ∞0

£r2 + z2

¤−1/22πrdr + φ0.

The integral, which has the potential equal to zero at infinity, is divergent To handle this difficulty we note that thepotential is independent of x and y.We will assume that the surface has a finite radius R centered on the z axis. Then thepotential at x = y = 0 is

φ (0, 0, z) =σ02ε

Z R

0

£r2 + z2

¤−1/2rdr + φ0(R). (1.98)

=σ02ε

hpR2 + z2 − |z|

i+ φ0 (R)

The condition that the potential is zero at z = 0 requires that φ0 (R) = −σ02εR.With this value of φ0 (R) we can evaluate the

potential in the limit that R goes to infinity,

φ (0, 0, z) = limR→∞

σ02ε

hpR2 + z2 − |z|−R

i(1.99)

=−σ02ε

|z| = φ (x, y, z)

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 = zσ02ε= n

σ02ε= E1(z) (1.99b)

On the other side of the surface charge, along -n

E2(z) = −∇φ2 = −zσ02ε= −nσ0

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

ZZσ (r0) [r− r0]|r− r0|3 dS. (1.100)

25

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 perunit 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 vectorn = ∇F(r)/|∇F(r)| is directed from the negative surface charge density to the positive surfacecharge density (it’s sufficient to replace F(r) by −F(r) in order to adjust the sense of n ). Thecharge layers lie on the surfaces F(r ±n/2) and the surface dipole moment density is

d(r) =→0lim n(r,) = d(r)n

→0lim (r,) = 0

→0lim (r,) = d(r)

1.103

Since the surface is neutral (total charge = 0), in the limit that → 0 with (r,) fixed,

n(r) (1E1(r) − 2E2(r))= o for electric dipole surface layer 1.104

From Gauss’ law one can determine the electric field contributions, E+ and E− , and fromEq. (1.99) the potential field contributions, + and, −, from each individual layer . The lattercontributions are shown in the schematics below. The total E1and 1(above the dipole layer)and total E2and 2 (below the dipole layer) are given by E+ + E− and + + −.in each region

Schematic of the E field contributions

(r,)n1/2 ↑+ + + + + + + + + +

− (r,)n1/2 ↓

− (r,)n2/2 ↓− − − − − − − − −

(r,)n2/2 ↑

where n1 = n2 and the negative sign on the field due to the lower layer comes from thenegative charge ”surface layer”.

Schematic of the field contributions− |z|(r,)/2+ + + + + + + + + +− |z|(r,)/2

|z|(r,)/2− − − − − − − − −|z|(r,)/2

Thus the total electric fields are:E1,total(r,) = (r,)n1/2 − (r,)n2/2 = 0 above the dipole layer

E2,total(r,) = −(r,)n1/2 + (r,)n2/2 = 0 below the dipole layergiving

n1 [E1,total − E2,total] = 0Between the two layers for finite the electric field is constant, directed downward and equalto,

E3,total(r,) = −(r,)n1/2 − (r,)n2/2 = −(r,)n1/.The total potential above the dipole layer (where |z2 | = + |z1 |) is

1,total(r,) = +(r,) + −(r,) = −|z1 |(r,)/2 + |z2 |(r,)/2 = (r,)/2Below the dipole layer (where |z1 | = + |z2 |)

.2,total(r,) = |z2 |(r,)/2 − |z1 |(r,)/2 = −(r,)/2,and for finite between the two charge layers where |z2 | = − |z1 | the total potential is

3,total(r,) = −|z1 |(r,)/2 + |z2 |(r,)/2 = (r,)/2 − |z1 |(r,)/= (r,)/2 + z1(r,)/.

Note that is continuous at each individual planar charged layer:3,total = (r,)/2 = 1,total at the + layer when z1 = 03,total = −(r,)/2 = 2,total at the - layer when z1 = −

# #

Finally the discontinuity in the potential above and below the surface dipole layer is1(r) − 2(r) = lim

0[(r,)/2 − (−(r,)/2)] = lim

01 (r,) = 1

d(r) 1.106

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 thepotential of the negatively charged surface layer as follows. After the limit → 0 is taken apoint r ′ on the planar dipole layer will be on the surface S′.

(r) = lim0

14 ∫∫

r ′ + 12 n,

r − (r ′ + 12 n)

dS′ − 14 ∫∫

r ′ − 12 n),

r − (r ′ − 12 n)

dS′

recall : df(r ′) =dr ′∇ ′f =f(r ′ + dr ′) − f(r ′)

(r) = lim0

14 ∫∫

(r ′,)|r − r ′ |

+ 12

n ∇ ′ (r ′,)|r − r ′ |

− (r ′,)|r − r ′ |

− −12 n ∇ ′ (r ′,)

|r − r ′ |dS′

= lim0

14 ∫∫ (r ′,)n ∇ ′ 1

|r − r ′ |dS′; (r ′,) is a constant, independent of r ′

= 14 ∫∫ d(r ′ )∇ ′ 1

|r − r ′ | ndS′

= −14 ∫∫ d(r ′ ) (r ′ − r)

|r ′−r|3 ndS′; since ∇ ′ 1

|r − r ′ |= −∇ 1

|r − r ′ |= (r − r ′)

|r − r ′ |3

= −14 ∫∫ d(r ′ )d′

(r) = −14 ∫∫ d(r ′ )ds

′ . 1.108

In this relationship ds is the solid angle subtended by the differential surface element of thedipole layer at r ′, specifically as viewed from the observation point r when in theconfiguration shown in Fig. 6 below. Formally, the solid angle subtended by a surface, S, isthe projection of the surface onto a unit sphere centered at the observation point, r:

S = ∫∫Sexposed

(r ′ − r)|r ′−r|3

ns dS′.= ∫∫Sexposedds. 1.109

S is defined by f(x,y, z) = 0 and has normals, n = ±∇f/|∇f|. The ns = n in the solid angleexpression of Eq. 1.109 is always directed out of the volume enclosed by the surface and theunit sphere upon which the surface is being projected. Using this prescription S is alwayspositive. When calculating solid angles one often only integrates over that part of the surfacewhich is visible to the observer.

In Eq. 1.107, however, n always points from − toward the charge layer (wheredepending on the problem could later be interpreted as negative) and one must integrate overthe entire surface upon which the dipole surface charge density resides. Eq. 1.107 gives thecorrect sign for all r independent of configuration. The discontinuity in (r) arisesautomatically from the change in sign of the differential d′ as one traverses the dipole layer.(In ds

′ one adjusts the sign artifically to give a positive solid angle for all configurations.)

Fig. 6aThe relationship between the parameters is shown in the above figure.

Example 1: Find the electrostatic field inside and outside a constant dipole surfacecharge density, d = q/a, residing on a sphere of radius R. Let the sphere be centered at theorigin. From Eq. 1.107 (r) is given by

(r) = − 14 ∫∫ d(r ′ ) (r ′ − r)

|r ′−r|3 n dS′

= − q4a ∫∫S

(r ′ − r)|r ′−r|3

n dS′

= − q4a ∫∫S

d′

where the sphere is defined by r′ = R. All solid angles viewed from inside the sphere are 4and correspond to integrating over the entire sphere, so

(r) = − qa for r < R

Outside the sphere, at r > R:

(rz) = q4a ∫∫ r ′ ∇ ′ 1

|r − r ′ |dS′; note that r > r′

= q4a ∫∫ ∫ (r′ − R)r ′ ∇ ′ ∑

l,m

42l+1

r′lrl+1 Ylm

∗ ( ′, ′)Ylm(,) r′2 sin ′d ′d ′dr′

= q4a ∑l,m

42l+1 Ylm(,) ∫ (r′ − R)r′2 ∂

∂r′r′l

rl+1 dr′ ∫∫ 4 Ylm∗ ( ′, ′)Y00( ′, ′) sin ′d ′d ′

= q4a ∑l,m

42l+1 Ylm(,) ∫ (r′ − R)r′2 lr′l−1

rl+1 dr′ 4 l0m0

= q4a

41 Y00(,) ∫ (r′ − R)r′2 0

rl+1 dr′ 4

= 0 for r > RNote that the above result of (rz) = 0 is only valid if r > r′. [One could use the aboveprocedure to derive the result for r < r′ In this case the partial derivative with respect to r′

gives an overall factor of −(l + 1)or −1 for l = 0 and the final result is −qa . We will discuss

these techniques later.] This result is consistent with spherical symmetry, Gauss’s law and atotal charge enclosed = 0. The electric field is zero everywhere. (Note that outside the spherethe solid angle approach is not helpful. When determining the solid angle from outside thesphere part of the sphere is obscured and one integrates over only the visible part of S. Thepotential calculation, on the other hand, calls for integrating over the entire sphere! )

Fig. 6b Sphere with d = q/aThe normalized potential is divided by q

a and uses variables (r, = 0) where r =zz.= rcosz

Example 2: Find the electrostatic potential along the symmetry axis inside and outsidea sphere of radius, R, centered at the origin, with dipole surface charge density,d = [q/a]cos On the symmetry axis r =rz, and

(rz, ) = − q4a ∫∫S

cos ′(r ′ − r)|r ′−r|3

r ′ r′2 sin ′d ′d ′; r′ = R

= − q2a ∫0

cos ′(r ′ − r)|r ′−r|3

r ′ r′2 sin ′d ′

= − q2a ∫0

cos ′(R3 − rcos ′R2)[R2 + r2 − 2rcos ′R]3 sin ′d ′; r = R

where r r ′ = cos ′ and the symmetry gives 2 for the ′ integration. The electric field alongz is

E(r) z = − d(r)dz

Fig. 6c Sphere with d = q/a cosNote that is discontinuous and E n is continuous across the dipole layer. The normalizedpotential is divided by q

a and uses variables (r, = 0) where r = zz.= rcosz.Example 3 An infinite plane with a constant dipole surface charge density, d = q/a

(r) = − q4a ∫∫S

(r ′ − r)|r ′−r|3

n dS′

We let z′ = 0 define the plane, and let r =zz. Then r r ′ = 0 and n = z. Thus

(rz. ) = − q4a ∫∫S

−r z|r ′−r|3

dx ′dy ′

= qz4a ∫∫S

1[′2 + z2]3/2

′d′d ′

= qz24a ∫0

∞ 1[′2 + z2]3/2

12 d(′2)

= qz4a ∫0

∞d(−2[′2 + z2]−1/2)

= qz2a [z2]−1/2 = q

2az|z|

The electric field is zero everywhere. This unique results occurs only when the plane isinfinite as the next example indicates.

Example 4 Assume the infinite plane is reduced to a finite disk of radius R andcalculate the potential along the symmetry axis.

(zz. ) = − q4a ∫∫S

−r z|r ′−r|3

dx ′dy ′

= qz4a ∫∫S

1[′2 + z2]3/2

′d′d ′

= qz24a ∫0

R2 1[′2 + z2]3/2

12 d(′2)

= qz4a ∫′=0

′=Rd(−2[′2 + z2]−1/2)

= qz2a [ 1

|z| − 1[R2 + z2]1/2 ]

In the plot below the normalized potential is divided by qa and uses variables (r, = 0)

where r = zz.= rcosz

Just above disk’s center the potential is the same as for an infinite plane, q2a

z|z| . The

magnitude, however can be adjusted with an appropriate o added to the potential.

.

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) . SinceD (r) = εE (r) , in order to satisfy Gauss’ law for the displacement vector the scalar potential must be a solution toPoisson’s Equation,

∇2φ (r) = −1ερ (r) . (1.106)

A special case of Poisson’s Equation is obtained for ρ (r) = 0. The result is known as Laplace’s Equation,.

∇2φ (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 functionsThere 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) =

−14π |r− r0| (1.108)

∇2g0 (r, r0) = δ(3) (r− r0) . (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) aformal solution of Eq. 1.106 is given by

φ (r) =1

ε

ZZZρ (r0) g (r, r0) d3r0. (1.110)

However, this solution will not generally satisfy the boundary conditions placed on φ (r) . In the next section we derive amore formally correct solution which contains the expression in Eq. 1.110

32

Section 2.4 Laplace and Poisson Equations

2.4.1 Green’s theorem and Green’s function solution for φ(r)

Green’s theorem, uses the divergence theorem to relate solutions of Poisson’s equation. Assume that we have twofunctions, φ (r) and g (r, ro) , which satisfy Poisson equations as follows in volume, V, with surface, S:

∇2ϕ (r) = −ρ(r)/ε (1.111)

∇2g (r, ro) = δ(r − ro) (1.112)

The divergence theorem states that

ZZZvolume V

∇ · [ϕ (r)∇g (r, ro)− g (r, ro)∇φ (r)] d3r =ZZ

boundingsurfaces, S

[φ (r)∇g (r, ro)− g (r, ro)∇φ (r)] · dS

(1.113)

The left hand side can be written as follows:

ZZZvolume

[φ (r)∇2g (r, ro)− g (r, ro)∇2φ (r)]d3r +ZZZ

volume

[∇φ (r) ·∇g(r, ro)−∇g (r, ro) ·∇φ (r)]d3r

=

ZZZvolume

φ (r) δ (r− ro) d3r −ZZZ

volume

g (r, ro) [−ρ (r)]d3r since the second term above is 0

= φ (ro)−ZZZ

volume

g (r, ro) [−ρ (r) /ε]d3r if ro is in the volume!

So if ro is in the volume, V, the left hand side substituted into Eq. 1.113 gives:

φ (ro) =

ZZZvolume

g (r, ro) [−ρ (r) /ε]d3r +ZZ

boundingsurface

[φ (r)∇g (r, ro)− g (r, ro)∇φ (r)] · n (r) dS (1.114)

This is a Green’s function solution of Eq.1.111 . Outside the volume, where ro is not in V,

ZZZvolume

φ (r) δ (r− ro) d3r = 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) · n (r) at thesurface is surface charge layer (because ∇φ (r) · n (r) = −E · n (r) and the surface might be replaced with an effectivediscontinuity in E ) and φ (r) n (r) at the surface is a surface dipole layer (because the surface might be replaced with a aneffective discontinuity in φ (r)). But the latter is not crucial to what we derive in this section. Here we have assumed thatn (r) is the bounding surface unit outward normal vector. In the next section we will find that specifying both the potentialand the normal gradient of the potential on the bounding surface will ‘over specify the problem’. In this case the problemwill only have a solution when the specified potential and normal derivative are compatible.

33

Section 2.4 Laplace and Poisson Equations

2.4.2 Uniqueness of the solutions to Poisson’s Equation Let φ1 (r) and φ2 (r) betwo solutions to Eq. 1.111 and let both satisfy the same boundary conditions (still of an unspecified nature) on the surfacebounding the volume. It follows that

χ (r) = φ1 (r)− φ2 (r) (1.115)

satisfies Laplace’s Equation,

∇2χ (r) = ∇2φ1 (r)−∇2φ2 (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 obtainZZZ

volume

∇ · [χ (r)∇χ (r)]d3r =ZZ

boundingsurface

χ (r)∇χ (r) · dS (1.116)

=

ZZZvolume

[χ (r)∇2χ (r) + |∇χ (r)|2]d3r =ZZ

boundingsurface

χ (r)∇χ (r) · dS

ZZZvolume

|∇χ (r)|2 d3r =ZZ

boundingsurface

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 volumeintegral 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 onthe boundary gives a unique boundary value problem satisfied by only one potential function. This particular problem inwhich 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 thenormal derivative of χ (r) will vanish on the boundary. Again the right hand side of Eq. 1.116 will vanish and therefore, asbefore, φ1 (r)− φ2 (r) will be constant. In this case it’s only the normal derivative which vanishes on the boundary and thegradient of any constant is zero. It follows that φ1 (r) = φ2 (r) + φ0 and the solution to this boundary value problem isonly 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 somesections of the boundary and the value of the potential is specified on other, different, sections of the boundary. In this casethe 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 ofthe potential on the boundary. Similarly the solution of a Neumann problem will give the potential on the boundary (up to anadditive constant). We find then that Eq. 1.114, which requires the values of the potential and its normal derivative on thebounding surface, does not provide a convenient solution to Poisson’s equation unless one can eliminate one of the surfaceintegrals 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 thatboth the potential and it’s normal derivative appear in the integrand of the surface integral. However, we have not imposedany boundary conditions on the Green’s function. If we could find a function g (r, r0) ,satisfying Eq.1.112 whose normalderivative on the bounding surface vanished, only the potential on the surface would be required. In the other case, if thefunction g (r, r0) vanished on the bounding surface, only the normal derivative of the potential on the surface would berequired. The apparent difficulty in applying Eq. 1.116 to obtain the solution of Poisson’s Equation is circumvented byobtaining 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

4π |r− r0| + F (r, r0) (1.117)

with

∇2F (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 therequired 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) =−1ε

ZZZvolume

GD (r, r0) ρ (r) d3r +

ZZboundingsurface

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

4π |r− r0| for r = rS (1.121)

where rS is on the boundary of the volume. Note that this does not mean that FD (r, r0) = 14π|r−r0| . Rather FD (r, r0) is a

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 functionA characteristic of the Green’s function is that GD (r, r0) = 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 boundary35

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 findZZZvolume

∇ · [GD (r, r0)∇GD (r, r00)−GD (r, r00)∇GD (r, r0)] d3r =ZZZvolume

GD (r, r0) δ (r− r0o) d3r −ZZZ

volume

GD (r, r00) δ (r− ro) d3r

= GD (r0o, ro)−GD (ro, r0o)

=

ZZboundingsurface

[GD (r, r0)∇GD (r, r00)−GD (r, r00)∇GD (r, r0)] · n (r) dS

= 0

and

GD (r00, r0) = GD (r0, r

00) . or (1.122)

GD (r0, r) = GD (r, r

0) since r00, r0 are arbitrary variables.

2.5.2 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

∇2FN (r, r0) = 0 in V.

satisfies a slightly more complicated boundary condition. If we apply the divergence theorem to

∇2GN (r, r0) = δ(3) (r− r0) (1.123)ZZZ∇ ·∇GN (r, r0) d3r =

ZZZδ(3) (r− r0) d3r

we find the Neumann restriction: ZZboundingsurface

∇GN (r, r0) · n (r) dS=1 (1.124a)

It follows that we are not permitted to set∇GN (r, r0) · n (r) = 0 on the boundary of the volume. The next simplest choice

is to satisfy Eq.1.123 with a simple boundary condition onGN :

simple boundary condition: ∇GN (r, r0) · n (r) =1

Stot(1.124b)

The solution to the Neumann problem is then given by

φ (r0) =1

ε

ZZZvolume

GN (r, r0) [−ρ (r)]d3r + φavg +

ZZboundingsurface

GN (r, r0)∇φ (r) · n (r) dS (1.125)

36

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 =

RRφ (r) dSRRdS

(1.126)

where S is the area of the boundary

Stot =

ZZboundingsurface

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

[∇GN (r, r0)] · n (r) =·∇FN (r, r0) +∇

µ −14π |r− r0|

¶¸· n (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.andlet

∇02G0N (r0, r0o) = δ(r0 − r0o) (1.128)

G0N (r0, r0o) is a solution to the Neumann problem, with ρ(r0)/ε = δ(r0−r0o) and from Eq. 1.125,

G0N (r0, r00) =

ZZZvolume

GN (r0, r0) δ(r

0−r0o)d3r0 + (1.129)

G0avg(r00) +

ZZboundingsurface

GN (r0s, r0)∇0G0N (r0s, r00) · dS0

The result is

G0N (r0, r00) = GN (r

00, r0) +G

0ave(r

00)−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 Neumanngreen’s function and subtract its average value on the boundary we will obtain a green’s function which is symmetric in theinterchange of the parameters.

37

Section 2.6 Solutions∇2Φ = 0 using GD and GN for Poisson’s Equation

2.5.3 The functions F (r, r0) The simple green’s function g (r, r0) = −14π|r−r0| can be identified as the

potential due to a ‘unit’ negative point charge located at r0. The function F (r, r0) must satisfy Laplace’s equation in thevolume, V, of interest. If we identify it as a potential its sources will be charges which lie outside our volume. We willfind that viewing F (r, r0) as the potential due to charges outside the region of interest will provide us with a technique forgenerating this function.

2.6 Solutions∇2Φ = 0 using GD and GN for Poisson’s Equation This section has introduced the Green’sfunctions for Poisson’s equation with Dirichlet or Neumann boundary conditions given. Since Laplace’s equation is aspecial case of Poisson’s equation with the source term, ρ (r) , equal to zero, the Green’s functions can be used to provide asolution to the Dirichlet problems for Laplace’s equation, φL−D (r) ,

φL−D (ro) =

ZZZ0 ·GD (r0, r0) d3x0 + (1.132)ZZ

boundingsurface

φ (rs)L−D∇GD (rs, r0) · dS−ZZ

GD (rs, r0)∇φ(rs) · dS

=

ZZboundingsurface

φ (rs)L−D∇GD (rs, r0) · dS

where the second term and third terms are zero because ∇2Φ = 0 and GD (rs, r0) = 0, respectively. A solution to theNeumann problem for Laplace’s equation,φL−N (r)

φL−N (r) =

ZZZ0 ·GN (r0, r0) d3x0 + (1.133)ZZ

φ (rs)L−N∇GN (rs, r0) · dS−ZZ

boundingsurface

GN (rs, r0)∇φL−N (rs) · dS

= −ZZ

boundingsurface

GN (rs, r0)∇φL−N (rs) · dS + φavg

= −ZZ

boundingsurface

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 Electrostatic energy density, capacitance

2.7 Electrostatic energy density, capacitance

2.7.1 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)] (1.134)

= −E (r, t) · [J (r, t) + ∂

∂tD (r, t)] +

·H (r, t) ·− ∂

∂tB (r, t)

¸.

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 toassociate E (r, t)×H (r, t) with S (r, t), the energy current density carried in the electromagnetic field. S (r, t) is calledthe Poynting vector.

S (r, t) = E (r, t)×H (r, t) (1.135)

and to associate the term

∂u (r, t)

∂t=

·H (r, t) · ∂

∂tB (r, t) +E (r, t) · ∂

∂tD (r, t)

¸(1.136)

with the time rate of change of the energy density stored in the ‘electromagnetic fields’.where

u (r, t) =1

2[E(r, t) ·D(r, t) +H (r, t) ·B (r, t)] (1.137)

Then Eq. 1.134 gives

∇ · S(r, t) = −∂u (r, t)∂t

−E(r, t) · J(r, t (1.138)

and ZZS(r, t) · dA+

ZZZE(r, t) · J(r, t)d3r = −

ZZZ∂u (r, t)

∂td3r (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 thefields are varied ‘adiabatically’ we have a unique value for the energy stored by the electromagnetic fields. We will thereforeignore for the present some subtle or not so subtle problems which arise due to frequency dependent relationships betweenEandD.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∂thas a contribution from an

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 Electrostatic energy density, capacitance

w (r) =ε

2|∇φ (r)|2 SIunits (1.140)

The electromagnetic energy of a system is given by integrating the energy density over the volume of the system

W =ε

2

ZZZvolume

|∇φ (r)|2 d3r ≥ 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 theindividual charge densities

ρ (r) =Xi

ρi (r) . (1.142)

Each charge density will contribute, −∇φi (r) , to the total electric field at each point where∇2φi (r) = −ρi (r) /ε. In this

case

−∇φ (r) = −Xi

∇φi (r) (1.143)

In terms of the fields of the individual charge elements the electromagnetic energy of the system is

W =ε

2

ZZZvolume

Xi

∇φi (r) ·Xj

∇φj (r) d3r (1.144)

=ε

2

Xi

ZZZvolume

|∇φi (r)|2 d3r

+ε

22Xi,ji< j

ZZZvolume

∇φi (r) ·∇φj (r) d3r

The term

Wi =ε

2

ZZZvolume

|∇φi (r)|2 d3r 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

ZZZvolume

∇φi (r) ·∇φj (r) d3r, interaction energy (1.146)

40

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.

ExampleConsider 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 -σ0 (a)What is the self energy ofeach spherical shell? ~(b)What is the interaction energy of the spherical shells?

Solution: (a) The magnitude of the charge on each shell is q = 4πa2σ0. The electric field of spherical shell vanishesinside the shell while outside the shell the electric field is

E (r− r±) =±q (r− r±)4πε |r− r±|3

SI units

=±q (r− r±)ε |r− r±|3

in Gaussian units

= −∇φ±(r).To calculate the self energy of each shell we evaluate, Eq.1.145 ,

W0 =εq2

2(4πε)2

ZZZ|r−r±|>a

d3r

|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 =εq2

2(4πε)2

ZZZ|r0|>a

d3r0

|r0|4 =4π εq2

2(4πε)2

Z ∞a

1

r02dr0

=q2

2 · 4πε·−1r0

¸∞a

W0 =q2

2 · 4πεa. Self energy of each shell in SI units (1.147)

=q2

2εaGaussian units

Aside: the charge radius of the electronEquation 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 electronwould be (ε = 1)

We =e2

2ain Gaussian units

41

Section 2.7 Electrostatic energy density, capacitance

The total energy of an electron at rest ismec2. 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

a =e2

2mec2=

(4.8× 10−10statC)22× (9.1× 10−28g)× (3× 1010cm/s)2

= 1. 4× 10−13cm !!

If this were correct the electron radius would be the same order of magnitude as the observed proton radius. But the upperlimit on the radius of the electron is several orders of magnitude below the radius of the proton. In fact the results of allexperiments 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) =q

4πεrfor r > a

φ (r) =q

4πεafor r ≤ a

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

ZZZ∇φ (r− r+) ·∇φ (r− r−) d3r

Using the identity,

∇ · [φ (r− r+)∇φ (r− r−)] = ∇φ (r− r+) ·∇φ (r− r−) + φ (r− r+)∇2φ (r− r−)

Wint = −εZZZ

allspace

φ (r− r+)∇2φ (r− r−) d3r + ε

ZZsurfaceat∞

φ (r− r+)∇φ (r− r−) · dS

Now let r0 =r− r−

= −εZZZ

allspace

φ (r0−r+ + r−)∇2φ (r0) d3r0 + ε

ZZsurfaceat r0s−→∞

φ (r0s−r+ + r−)∇φ (r0s) · dS0

Since the product of the field and the potential vanish at infinity as r0−3s whereas dS0only increases like r02s the surfaceintegral does not contribute to the interaction energy. To evaluate the spatial integration we note that∇2φ (r0) refers to thenegative shell and (see Jackson problem 1.3a):

∇2φ (r0) = − σ0εδ (r0 − a) . SI units

In the example the distance between the centers of the spheres is greater than the diameter of the spheres. Therefore the42

Section 2.7 Electrostatic energy density, capacitance

spheres do not overlap and the interaction energy is given by

Wint = −εZZZ

allspace

φ (r0−r+ + r−)∇2φ (r0) d3r0

= −εZZZ

allspace

q

4πε|r0−2a| [−σ0εδ (r0 − a)]d3r0

Letting 2a =Rz

Wint = −q σ0ε

1

4π

Z π

0

Z ∞0

Z 2π

0

δ (r − a)[r2 +R2 − 2rR cos θ]1/2

dγ r2dr sin θ dθ

The integration over the angle γ and the radial direction r are readily carried out leaving (ξ = cos θ)

Wint = −q σ0ε

2π

4π

Z π

0

a2

[a2 +R2 − 2aR cos θ]1/2sin θ dθ

Wint = −q£

q4πa2

¤a2

2ε

Z +1

−1

dξ

[a2 +R2 − 2aRξ]1/2

= − q2

2 · 4πεZ ξ=+1

ξ=−1− 1

aRd£a2 +R2 − 2aRξ¤1/2

=q2

2 · 4πε[R− a]− [R+ a]

aR

= − q2

4πεRinteraction energy of two shells

The interaction energy is equal to that of two charged point particles, with charges ±q, separated by a distance R. Thetotal energy stored in the electrostatic field is

W =q2

4πεa− q2

4πεR> 0 total energy for the two shells

As R decreases the stored energy decreases.

dW = −|dR| ∂∂R

·− q2

4πεR

¸= − q2

4πεR2|dR| = −F |dR|

Therefore the force between the shells is attractive with a magnitude of q2/[4πεR2] and does work dW

W + dW =q2

4πεa− q2

4πεR

·1 +

dR

R

¸

43

Section 2.8 Conductors and Capacitance

2.8 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 havea charge Qi and let all other conductors have zero net charge. The charge on the ith conductor will be distributed over itssurface with a surface charge density σii (r) which is proportional toQi. It also will induce a surface charge density on eachof the other conductors σji (r) which will be proportional to Qi. Of course

Qi =

ZZi

σii (r) dS (1.148)

0 =

ZZj

σji (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 chargedensities which would be on the conductor with only one conductor charged. [This follows from the uniqueness of thesolution to the boundary value problem. In this case the potential of each conductor can not vary over the conductor. This issatisfied if either the ith conductor has a charge Qi or the jth conductor has a chargeQj . If the potentials obtained with eachof these conditions are added the result will also satisfy the boundary conditions].

σi (r) =Xj

σij (r) .

The jth term in the sum is proportional to Qj . The potential of the ith conductor is given by

Vi =NXk=1

ZZk

σk (r)

|ri−r|dS (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 =NXj=1

pijQj . (1.150)

The proportionality constants {pij} will depend on the geometry of the system of conductors but not the charges on theconductors. ( Note that pij 6= 1

Qj

RRjσj(r)|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 onthe conductors in terms of the potentials of the conductors

Qi =NXj=1

CijVj (1.151)

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 ithconductor is obtained by ‘grounding’ all the other conductors (forcing Vj = 0 for j 6= i) andmeasuring the ratio

Cii=QiVi, V j= 0 for j 6= i. (1.152)

The potential energy for a system of charged conductors is

W =1

2

NXi=1

QiVi=1

2

NXi,j=1

CijViVj (1.153)

This equation must give the energy stored by the electric fields of the system and thereforeW ≥ 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 itis 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 morefunctions. In classical mechanics, for example, we seek rk(t) for a system of particles. A common technique is to minimizethe action integral, I =

RL( rk(t), rk(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 suchas

I(φ) =1

2

ZZZvolume

∇φ ·∇φ d3r.

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] =ZZZ

F (r,ψ (r) ,∇ψ (r))d3r (1.154)

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 ψδ∇ψ =∇[δψ] andδI = I [ψ,∇ψ]− I [ψ + δψ,∇ψ + δ∇ψ].

45

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,

I [ψ + δψ,∇ψ + δ∇ψ] = exp[δψδ

δψ+ δ∇ψ · δ

δ∇ψ ] I [ψ,∇ψ] (1.156)

= [1 + [δψδ

δψ+ δ∇ψ · δ

δ∇ψ ] + [δψδ

δψ+ δ∇ψ · δ

δ∇ψ ]2/2! + ....] I + ...

= I [ψ,∇ψ] + [δψ δ

δψ+ δ∇ψ · δ

δ∇ψ ]I + [δψδ

δψ+ δ∇ψ · δ

δ∇ψ ]2I/2! + ...

To first order in δψ we have:

δI =

ZZZvol

[δF

δψδψ (r) +∇[δψ] · δF

δ∇ψ ]d3r + higher order terms.... (1.157)

where

∇[δψ] · δF

δ∇ψ ≡3Xi=1

∂

∂xiδψ

δF

δ[∂ψ/xi].

We can replace δFδψ by ∂F

∂ψ when F is a continuous function of ψ in the volume.

δI =

ZZZvol

[∂F

∂ψδψ (r) +∇[δψ] · ∂F

∂∇ψ ]d3r + higher order terms.... (1.158)

=

ZZZvol

[∂F

∂ψδψ (r) +∇ · (δψ ∂F

∂∇ψ )− δψ∇ ·∂F

∂∇ψ ]d3r + higher order terms...

Using the divergence theorem,

δI =

ZZZvol

[∂F

∂ψ−∇ · ∂F

∂∇ψ ]δψd3r +

ZZZvol

∇ · (δψ ∂F

∂∇ψ )d3r + higher order terms (1.159)

=

ZZZvol

[∂F

∂ψ−∇ · ∂F

∂∇ψ ]δψd3r +

ZZboundingsurface

δψ∂F

∂∇ψ · n dS + .higher order terms....

The functional I will have an extremum when δI = 0 or,

ZZZvol

"[∂F

∂ψ−

3Xi=1

∂

∂xi

∂F (r,ψ (r) ,∇ψ (r))∂ (∂ψ/∂xi)

#δψd3r +

ZZboundingsurface

"3Xi=1

δF

δ (∂ψ/∂xi)n · xi

#δψ dS = 0 (1.160)

for all ‘permissible’ δψ (r) . Generally the values of the function ψ (r) are specified on the boundary of the volume. In this46

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 onlyrestriction on the variation in ψ(r) then the requirement that the first order term vanishes constrains ψ (r) to be a solution ofthe following ”Euler-Lagrange” equation.

∂F

∂ψ−

3Xi=1

∂

∂xi

∂F (r,ψ (r) , ∂ψ/∂xi)

∂ (∂ψ/∂xi)= 0 (1.161)

To determine whether the extremum is a minimum, maximum, or stationary point the second order term in δψ must becalculated.

2.9.2 Variations with equality constraints and fixed boundary conditions In an extension of thevariational calculations the function ψ (r) is required to satisfy one or more constraints in addition to the boundaryconditions. Generally these have a form,

Cα [ψ] =

ZZZKα (r,ψ (r) ,∇ψ (r)) d3r = 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

ZZZvol

"δKα

δψ−

3Xi=1

∂

∂xi

δKα

δ (∂ψ/∂xi)

#δψd3r +

ZZboundingsurface

3Xi=1

·δKα

δ (∂ψ/∂xi)xi

¸· n δψdS = 0 (1.163)

Because ψ (r) must satisfy fixed boundary conditions δψ (r) is zero on the surface and the surface integral vanishes. Theequations which must be satisfied are

ZZZvol

"δF (r,ψ (r) ,∇ψ (r))

δψ−

3Xi=1

∂

∂xi

δF (r,ψ (r) ,∇ψ (r))δ (∂ψ/∂xi)

#δψ (r) d3r = 0 (1.164a)

and

ZZZvol

"δKα(r,ψ (r) ,∇ψ (r))

δψ−

3Xi=1

∂

∂ri

δKα(r,ψ (r) ,∇ψ (r))δ (∂ψ/∂ri)

#δψ (r) d3r = 0. (1.164b)

We can consider the47

Section 2.9 Variational calculations

uα (r) =δKα

δψ−

3Xi=1

∂

∂xi

δK

δ (∂ψ/∂xi)(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 isorthogonal to the uα (r) . (Note that Eq. 1.164b looks like an inner product condition, (uα, δψ) = 0 as does Eq.1.164a, ( δFδψ −

P3i=1

∂∂xi

δFδ(∂ψ/∂xi)

, δψ) = 0.) In this case Eq. 1.164a requires that the function multiplying δψ (r) lie in thesubspace spanned by the uα (r) or

δF

δψ−

3Xi=1

∂

∂xi

δF

δ (∂ψ/∂xi)=Xα

λαuα (r) (1.166)

The ‘expansion coefficients’ λα are constants, called Lagrange multipliers, which are determined such that the constraintequations are satisfied.

2.9.3 Approximation using the functional If the function describing the property of thesystem provides an extremum for Eq. 1.159 an approximation to the function can be generated. In the simplest case letψ (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 isotherwise arbitrary. The functional equation is now a function of β

W (β) =

ZZZF (r,w (β, r) ,∇w (β, r))d3r. (1.167)

Assume further that,

W (β) =ε

2

ZZZ|∇w (β, r) |2d3r.

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, sayb0, for whichW 0(β) = 0 restricted to functions of the form w (β, r) (See also Eq. 1.166.). IfW 00 (bo) is consistent with thetype of extremum required ,w (b0, r) is an approximation to ψ (r) .

2.9.4 Estimation of capacitance using a variational approach In anelectrostatic problem the potential of each conductor is fixed. One then guesses a trial potential which satisfies the boundaryconditions. This trial potential should depend on one or more parameters which are varied so as to obtain the minimum valueof the energy for the system. In the case that the trial potential =1 on the ith conductor and zero on all other conductors thecapacitance, Cii, is approximated by the estimated energy. That is, for two conductors,

W =1

2

2Xi,j=1

CijViVj

=1

2C11V

21 =

1

2C11 if V1 = 1 and V2 = 0

=1

2C22V

22 =

1

2C22 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 isW =

1

2QV . (Hint: solve the boundary value problem for φ(r) everywhere and calculateW from Eq. 1.141)

49

Section 2.9 Variational calculations

Example of simple variational method: coaxial cylindersConsider 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− ρ

2b+

ρ2

2b)] +

2ρ

3c− 5ρ

2

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 obtainW = 12Cbb. The estimated energy of the coaxial system

per unit length, L, in Gaussian units is

W (β) =²08πL

Z ZZ|∇φ|2 ρdρdϕdz ²0 = 1 in Gaussian units

=1

8πL

ZZ ·ρ∂φ

∂ρ+ ϕ

1

ρ

∂φ

∂ϕ+ z

∂φ

∂z

¸··ρ∂φ

∂ρ+ ϕ

1

ρ

∂φ

∂ϕ+ z

∂φ

∂z

¸ρdρdϕdz

=1

8πL

Z ¯∂φ

∂ρ

¯2ρdρ · 2πL

W (β) =a2

4

Z c

b

| β2b[−1 + 2ρ] + 2

3c(1− 5ρ)|2ρdρ

This has an extremum (minimum) for

W 0(β) =a2

2

Z c

b

¯β

2b[−1 + 2ρ] + 2

3c(1− 5ρ)

¯1

2b[−1 + 2ρ] ρdρ = 0

β

Z c

b

¯1

2b[−1 + 2ρ]

¯2ρdρ = −

Z c

b

¯1

2b[−1 + 2ρ] 2

3c(1− 5ρ)

¯ρdρ

β = −R cb

¯12b [−1 + 2ρ] 23c (1− 5ρ)

¯ρdρR c

b

¯12b [−1 + 2ρ]

¯2ρdρ

β = −4b3c

R 0.8 cm0.2 cm [−1 cm+ 2ρ] [2 cm− 5ρ] ρdρR 0.8 cm

0.2 cm|[−1cm+ 2ρ]|2 ρdρ

= 1.35

50

Section 2.9 Variational calculations

This gives an approximation to the capacitance per unit length equal to

2W (1.6) = C =a2

2

Z c

b

¯β

2c[−1 + 2ρ] + 2

3b(1− 5ρ)

¯2ρdρ

= 0.367

The plot in Fig. 7 shows the W(β) (using the normalized potential) generated via mathcad. The β from this numericalapproach is approximately 1.35 (see two.cylinder.example.mcd on the web.). The value ofW (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 afunction of ρ.

Fig. 7

Fig. 8

The exact result is C = 12 ln |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

conductor be at 0 potential and the inner potential at V then C = QV . The length, L, of the cylinders satisfies

cL << 1.

Φ(ρ,ϕ, z) =−14πε

ZZZσ0δ(ρ

0 − b)|∇(ρ0 − b)| 1

|r− r0| ρ0dρ0dϕ0dz0 +Φ0(b, c)

where Φ0(b, c) is a constant.and we shall let z be along the cylindrical axis, with ρ2 = x2 + y2. Furthermore, we shallevaluate the potential on the x axis so that r = x = ρ cosϕ = ρ, z = 0 and r · r0 = ρρ0 cosϕ0

Φ(ρ, 0, 0) =−14πε

ZZZσ0δ(ρ

0 − b)|∇(ρ0 − b)| 1

|r− r0| ρ0dρ0dϕ0dz0 +Φ0(b, c)

Φ(ρ, 0, 0) =−Q

4πε(2πbL)

ZZδ(ρ0 − b)| 1

[ρ2 + ρ02 + z02 − 2r · r0]1/2ρ0dρ0dϕ0 +Φ0(b, c)

≈ −Q4πε(2πbL)

Z L/2

−L/2

Z 2π

0

bdϕ0dz00hρ2 + b2 + z02 − 2ρ [b2 + z02]1/2 cosϕ0

i1/2 +Φ0(b, c)=

−Q4πε(2πL)

Z L/2

−L/2

Z 2π

0

dϕ0dz0

[ρ2 + a2 − 2ρa cosϕ0]1/2+Φ0(b, c) where a2 = [b2 + z02]

Unfortunately this reduces to an elliptic integral which is unmanageable. In the next chapter we shall have some methodsfor handling this problem.

Using Gauss’ law one can obtain the exact result: C =4π²0

2 ln[b

a]

52

Section 2.9 Variational calculations

Using the variational techniqueto 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 lowerbounds can be placed on the capacitance by considering two different variations in the functional forms of C:

·Q2[σ]

2W [σ]

¸|maximized ≤ Cactual ≤ 2W [Ψ]minimized (1.90)

where

·Q2[σ]

2W [σ]

¸|maximized =

hHS1

σ(r)dS1

i2HS1

HS1

σ(r)G(r, r0)σ(r0)dS1dS01

max

(1.91)

and is more commonly written:

·1

C[σ]

¸min=

HS1 HS1 σ(r)G(r, r0)σ(r0)dS1dS01hHS1

σ(r)dS1

i2min

(1.92)

The right hand side of Eq. 1.90 is relatively easy to use and is often employed to determine C. The left hand conditionof 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 1.92must = 0: on all conductors except S1. The solution to Problem 1.18 provides the derivation of Eq 1.92. In this solutionone considers the system for which all Qi are fixed and all Φi = 0 but Φ1. Subject to this condition, the surface charge

densities are allowed to adjust to minimize·W [σ]

Q21[σ]

¸. For two conductors, this means Φ2 = 0 (with fixed Q2 ) while

Φ1varies as σ1(rS1) is adjusted to minimize .1

C[σ]. Problem 1.19 provides an example of how the actual C is always less

than the approximated C[Ψ]., and is only equal to C[Ψ] when the correct potential function is used. The minimization of1

C[σ]occurs when the charge distributed on S1 puts all the other conductors at constant potential (0)

53

Section 2.10 Linear charge density

2.10 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 wouldrestrict the charge to also lie on a second surface, sayG (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

dq = λ (r0)

¯d

dsr (s)

¯r0

ds = λ (r0)

qx0 (s0)

2 + y0 (s0)2 + z0 (s0)

2ds

and the function which gives the variation in the electrostatic potential due to this linear charge density is

φ (R) =

Zλ (r (s))

|R− r (s)|qx0 (s)2 + y0 (s)2 + z0 (s)2ds

54

Appendix ASpecial problems

1. A hyperboloid of revolution has a uniform surface charge density of 7 statcoul /cm2. The charged surface lies betweenz = −2 cm and z = +2 cm and is given by the equation

x2 + y2

2− z2 = 1 cm2.

(a) Determine the function which gives the volume charge density corresponding to this surface charge. (b) Taking thepotential 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.

2. A line charge lies along the parabola z = x2/(3 cm), y = 0 cm in the region 0 cm ≤ z ≤ 3 cm. The linear chargedensity 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.3. 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

δW0

δa= − q2

2εa2

there is a force acting at each point on the shell which would cause the shell to expand. (a) Determine and explain thesource of this force. (b) Demonstrate that the source is correct by calculating the force per unit area of the sphere and thetotal work done by the force if the radius of the sphere expands by δa.

4. A spherical capacitor consists of two concentric conducting spherical shells. The inner shell has a radius a and the outera radius 2a. Using the ‘test potential’

φ(β, r) =sinh2πβa (r − a)

isin (2πβ)

, a < r < b

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 givenparameters. (c) Estimate the capacitance of the system and (d) compare this estimated value to the actual value for thecapacitance of this spherical capacitor. (e) In the range a < r < 2a plot the difference between the actual potentialfunction between the spheres and the potential function obtained with φ (β, r) . (f) In the range a < r < 2a plot thedifference between the actual radial electric field between the spheres and the radial field obtained with φ (β, r) .

5. A sphere of radius 1 cm sets 2 cm above the center of a large grounded conducting plane. As a model assume thatthe plane is infinite in extent. Choosing a coordinate system with the sphere centered at the origin and the plane atz = −2 cm use the function

Φ (p; r, θ,φ) = (1 cm

r)p

r cos θ + 2 cm

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 oftendescribe the general properties of the potentials the ‘closed’ form for the integral is often obtained. This capability is associated with having solvedsimilar 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 iswritten as a convergent integral with a finite integrand.

55