Chapter 2

Electrostatic Energy and GreensTheorem

2.1 Electrostatic energy

The work required to bring the charge qi from innity to the point xi in the presence ofexisting charges q1; : : : ; qi1 is

Wi =1

40qi

i1Xj=1

qjjxi xjj : (2.1)

The total work required to assemble N charges is then:

WN =1

40

NXi=2

qii1Xj=1

qjjxi xjj : (2.2)

Each pairwise interaction appears once in this double sum. If we sum over i and j indepen-dently, but exclude i = j, each pair is counted twice. Thus we can also write

WN =1

80

NXi;j=1

0 qiqjjxi xjj ; (2.3)

where the prime on the sum reminds us to omit the (innite) terms with i = j.Going to a continuum distribution, we have no way to exclude the interaction of an

innitesimal charge element with itself. So we leave it in and write:

W =1

80

Z Z (x)(x0)jx x0j d

3x d3x0: (2.4)

For a smooth three dimensional charge distribution, the interaction of a charge element withthe nearby charges is

1

80(x) d3x

ZV

(x0)jx x0j d

3x0 V 2=32(x) d3x (2.5)

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and vanishes for V ! 0. So the self-interaction does not contribute and the formula (2.4)is correct, unless (x) contains a point charge, or even a line charge. At the macroscopiclevel the charge distributions always have a nite extent, really, so there is no problem. Atthe fundamental level, there is a problem if the elementary particles (leptons, quarks) aretruly point charges. Relativistic QED (quantum electrodynamics) deals with this dicultyby postulating a simple negative term that cancels the innite self-energy in W .

2.1.1 The eld energy

One reason why we like Eq. (2.4) is that it can be transformed to

W =02

Zjrj2 d3x = 0

2

ZjEj2 d3x; (2.6)

which shows that the work done in assembling the charges is stored as eld energy. To getthis expression we rewrite Eq. (2.4) as

W =1

2

Z(x)(x) d3x; (2.7)

substitute (x) = 0r2(x) and use r2 = r (r) jrj2 to obtain

W =02

Zjrj2 d3x 0

2

ZS

@

@nda: (2.8)

So it seems that Eq. (2.6) is obtained only if the integral is over all space and there are nocharges at innity, or in special cases such as grounded conductors ( = 0) on the boundaryS. In practice, however, one can usually nd an S such that the boundary term vanishes oris unimportant.

Equation (2.6) is a fundamental result. It is very useful to think that there really isan energy density

w =02jEj2 (2.9)

that resides in the eld and is greater where the eld is stronger. Note that the eld energy isalways positive. A consequence of this is that an assembly of charges cannot be in equilibriumunder electrostatic forces alone.

We already noted that for point charges (2.3) is not the same as (2.4) or (2.6). Fortwo unlike charges, (2.3) is negative, while (2.6) is positive (and innite). Jackson discussesthis point in greater detail.

2.2 Boundary value problems

A number of relations can be obtained by ddling around with the eld energy expressionand related integrals. They are very useful for discussing the properties of the E eld in aconned geometry. Typically, the boundaries are made of conductors, each held at a knownpotential, or carrying a known total charge. The problem is to nd the charge density onthe surface of each conductor and the eld E in the space between them.

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2.2.1 Greens rst identity

Greens rst identity is

ZV(r2 +r r ) d3x =

IS@

@nda: (2.10)

This is obtained simply by integrating the identity

r (r ) = r2 +r r ; (2.11)and using the divergence theorem. It doesnt seem very interesting, but it has many uses.An important corollary is the following.

Uniqueness theorem

The solution of Poissons equation is unique if is given on the boundary (Dirichlet boundarycondition). It is unique up to an additive constant if @=@n is given on the boundary(Neumann boundary condition). It is also unique if the boundary conditions are Dirichlet onpart of the boundary, Neumann on the rest.

This is proven by assuming that there are two distinct solutions, 1 and 2, withr21 = =0 and r22 = =0, and showing that U = 1 2 must vanish. Use (2.10)with = = U : Z

V(Ur2U +rU rU) d3x =

ISU@U

@nda; (2.12)

and note that r2U = 0 in all V and that either U = 0 (Dirichlet) or @U=@n = 0 (Neumann)on S. It follows that jrU j vanishes and U is constant in V . The constant must be zero inthe Dirichlet case, or for mixed boundary conditions, but is arbitrary in the pure Neumanncase.

By a modication of this argument one can prove that, in charge-free space, thefunctional W [] =

R jrj2 d3x is minimal, subject to the appropriate boundary conditions,when coincides with the potential , the solution of Laplaces equation r2 = 0.

2.2.2 Greens theorem and integral equations

Write down Greens rst identity for ; and ; :

ZV(r2 +r r ) d3x =

IS@

@nda; (2.13)Z

V( r2+r r) d3x =

IS @

@nda; (2.14)

and subtract side by side. The result is Greens second identity, or Greens theorem:

ZV(r2 r2) d3x =

IS

@

@n @

@n

!da: (2.15)

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This gives an important formula if we choose

=1

R=

1

jx x0j ; (2.16)

so that r2 = 4(3)(x x0), and we identify with the potential . If x is inside V wend

4(x0)ZV

1

Rr2 d3x =

IS

@

@n

1

R 1R

@

@n

!da: (2.17)

Recalling that r2 = =0 and interchanging the names of x and x0, we arrive at our nalresult:

(x) =1

40

ZV

(x0)R

d3x0 +1

4

IS

1

R

@

@n0 (x0) @

@n01

R

!da0: (2.18)

If x lies outside the volume V , then the left hand side of Eq. (2.18) is zero.There is a lot of mathematical physics in this equation:

If V is all of space and there are no charges at innity, it simply gives back Coulombslaw,

(x) =1

40

ZV1

(x0)R

d3x0: (2.19)

If @=@n is given on the boundary S (Neumann), we obtain an integral equation for on S. This equation may be easier to solve than the original Poisson equation. Once(x) is known for x on S, it can be found everywhere by integration (in principle, ifnot in practice).

If is given on the boundary S (Dirichlet), we can similarly get an integral equationfor @=@n on S and then nd (x) everywhere. The integral equation is

@

@n

S

=1

40

ZV(x0)

@

@n

1

Rd3x0

+1

4

IS

" @

@n

1

R

!@

@n0 (x0) @

2

@n@n01

R

#da0: (2.20)

These integral equations show that it is not possible to assign both and @=@narbitrarily on the boundary. Physically, S could be the surface of conductors whichare held at xed potentials. The charges in these conductors shift about and determinethe surface charge density , which is given by = 0@=@n. Note that n is directedinto the conductor (contrary to the usual practice).

We can view 0@=@n as a surface charge density and 0 as a dipole layer thatcause, respectively, a discontinuity in the eld and in the potential, in such a way that jumps to zero in crossing S.

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We can solve the boundary value problem anew for every (x), but there is a betterway. Consider for instance the Dirichlet case. We can construct the general solution ifwe can solve the particular case when S is at zero potential and (x) is a point charge ofstrength q=40 = 1, so that (x) =

(3)(x x0). This special solution G(x;x0) is the Greensfunction. It obeys the equation

r2G(x;x0) = 4(3)(x x0); (2.21)G(x;x0) = 0 when x is on S: (2.22)

Once G has been found (this is easier said than done), we can use Greens theorem with = G and = and proceed as in the derivation of Eq. (2.18). In other words, simplywrite G in place of 1=R in (2.18). What we nd is an explicit expression for (x):

(x) =1

40

ZVG(x;x0)(x0) d3x0 1

4

IS

(x0)@G

@n0da0: (2.23)

In particular, if = 0 on S, we get

(x) =1

40

ZVG(x;x0)(x0) d3x0; (2.24)

which is very much like Coulombs law. In fact, 1= jx x0j is nothing but the Greens functionfor all space. Similarly, one can dene and use a Greens function GN(x;x

0) for Neumannboundary conditions. Jackson discusses this case in detail.

If we put G = 1=R + F and we use r2G = 4(3)(x x0) and r2(1=R) =4(3)(x x0), we see that F satises Laplaces equation r2F = 0 everywhere in V . Wecan regard F as an electric potential due to (ctitious) charges located outside V . The ideais to continue F and G across S as if they were not discontinuous at S. In some cases thisapproach leads to a simple solution of the boundary value problem (method of images).

There are many applications of Greens theorems. A useful one is the mean valuetheorem, Jacksons Problem 1.10: In charge-free space, (x) is equal to the average of on any sphere centered at x. A consequence of this is that (x) cannot have maxima orminima in charge-free space. One could say that this is obvious, because in charge-free space satises Laplaces equation, r2 = 0, which in cartesian coordinates is

@2

@x2+@2

@y2+@2

@z2= 0 (2.25)

so that it is impossible for all the second derivatives to have the same sign. However, themean value theorem shows that it is also impossible to have a point where all the secondand third derivatives vanish but all the fourth derivatives are positive, for instance.

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