Problems
Application 3.1 Field Lines for a Point Charge in a Uniform Field
3.6 Electrostatic Total Energy
Thetotal electrostatic energyUEis defined as the total work required to assemble a charge distribution from an initial state where all the charge is dispersed at spatial infinity. We imagine that the work is performed quasistatically by an external agent in such a way that no dissipative effects occur. This
3.6 Electrostatic Total Energy 77
ensures that the process is reversible in the thermodynamic sense.UEis important for many reasons, not least because it achieves its minimum value for the ground state of the electrostatic system.
An example is a model charge densityρ(r) which depends on one or more variational parameters.
MinimizingUEwith respect to these parameters produces an approximation of the ground state charge density.
It is not difficult to computeUEfor a collection ofNstationary point chargesqklocated at positions rk. No work is required to bring the first charge into position from infinity becauseE=0 everywhere.
The workW12required to bring the second charge into position is thenegativeof the work calculated in (3.15) because the external agent does workagainstthe Coulomb force exerted by the first charge.
Therefore, ifϕ1(r) is the electrostatic potential (3.26) produced byq1, W12=q2[ϕ1(r2)−ϕ1(∞)]=q2ϕ1(r2)= 1
4π 0
q1q2
|r1−r2|. (3.73) The third charge interacts with bothq1 andq2 so additional workW13+W23 must be done. This generalizes to
UE=W = 1 4π 0
N j=1
N i>j
qiqj
|ri−rj|, (3.74)
or
UE = 1 4π 0
1 2
N i=1
N j=i
qiqj
|ri−rj| = 1 2
N i=1
qiϕ(ri). (3.75)
The factor of12in (3.75) corrects for overcounting; the double sum counts each distinct pair of charges twice rather than once as in (3.74). Recalling (3.10), the analog of (3.75) for a continuous distribution of charge is
UE= 1 8π 0
d3r
d3rρ(r)ρ(r)
|r−r| = 1 2
d3r ρ(r)ϕ(r). (3.76) The total energy cannot depend on exactly how a charge distribution is assembled. In practice, the symmetry of a problem often suggests a convenient method of assembly to findUE. An example is a ball of charge with uniform charge densityρ=Q/43π R3. Rather than evaluate (3.76), we can compute the work required to “build” the ball by successively adding spherical layers of uniform charge and thicknessdr. At an intermediate stage of assembly, the ball has radiusr and (3.73) identifies the work done against the Coulomb force to add an increment of chargedq=4π r2drρasdW =ϕSdq, whereϕSis the electrostatic potential at the surface of the ball. The method of Example 3.4 gives this potential asϕS =ρr2/30. Therefore, the energy we seek is
UE = 4π 30ρ2
R 0
drr4= 3 5
Q2
4π 0R. (3.77)
A variation of the method used to obtain (3.77) can be exploited to derive (3.76) without passing through the point charge result (3.75). Ifρ(r) andϕ(r) are the final charge density and electrostatic potential, the idea is to let the charge distribution at any intermediate stage of the assembly process beλρ(r), whereλis a real number which increases continuously from 0 to 1. By linearity, the scalar potential at the same intermediate stage isλϕ(r). Therefore, as in the preceding charged-ball problem, the work done against the Coulomb force when we add an infinitesimal bit of charge (whereupon λchanges toλ+δλ) to any volume element is [(δλ)ρ(r)][λϕ(r)]=[(δλ)λ][ρ(r)ϕ(r)]. In agreement
CUUK1954-03 CUUK1954/Zangwill 978 0 521 89697 9 August 9, 2012 14:57
78 ELECTROSTATICS: THE ELECTRIC FIELD PRODUCED BY STATIONARY CHARGE
with (3.76), the total work done is UE=
1 0
δλλ
d3r ρ(r)ϕ(r)= 1 2
d3r ρ(r)ϕ(r). (3.78) Finally, it is worth remarking that the electrostatic total energy UE is a quantity which clearly illustrates the fundamental limitations of the point charge concept in classical electrodynamics.UE
diverges for a single point charge, whether we try to evaluate (3.75) withN=1 or consider (3.77) in the limitR→0. This unphysical behavior is the price we pay for the computational advantages of writing charge densities as a sum of delta functions as in (2.6). There are circumstances where this divergence can be hidden (see Section 23.6.3), but its presence manifests itself in other ways.
3.6.1 U
Eis Positive-Definite
The electrostatic total energy is a non-negative quantity. To see this, it is simplest to eliminateρ(r) from the far right side of (3.76) using Gauss’ law and then integrate by parts. The result is
UE= −1 20
d3rE· ∇ϕ+1 20
d3r∇ ·(ϕE). (3.79)
The divergence theorem applied to the last term in (3.79) gives zero becauseE(r)→0 at infinity for any realistic source. Then, becauseE= −∇ϕin electrostatics,
UE =1 20
d3r|E|2≥0. (3.80)
UE ≥0 appears to contradict (3.75) for, say, two equal and opposite point charges. There is no contradiction because the latter explicitly excludes the positive but unphysically infinite self-energy of each point charge [computed by insertingE=qˆr/4π 0rinto (3.80)]. This characteristic pathology (remedied only by quantum electrodynamics) shows that the point charge concept must be used with care in classical calculations. By contrast, the self-energy part of (3.80) is integrable and makes a positive (but finite) contribution to the total energy for smooth and non-singular distributionsρ(r) of either sign.
Equation (3.80) invites us to ascribe a positive-definite electrostatic energy density to every point in space,
uE(r)= 120|E(r)|2. (3.81) This turns out to be correct (see Chapter 15), but a deduction based on (3.80) is unwarranted because (3.76) would lead us to make the same claim for 12ρ(r)ϕ(r), which is numerically a very different quantity.
Thomson’s Problem
The classical “plum pudding” model of the atom was devised by J.J. Thomson in 1904. A con- temporary variation of this model asks: what mechanically stable arrangement ofNnegative point chargesqk has the lowest energy when all the charges are constrained to lie on the surface of a spherical shell of uniform compensating positive charge? The presence of the positive charge vitiates Earnshaw’s theorem but, by symmetry, has no other effect on the spatial arrangement of the point charges. The problem thus reduces to minimizing the total energy (3.74) subject to the constraint that each vectorrkhas fixed length.
Surprisingly, the solutions donotalways correspond to configurations of maximal symmetry.
Thus, forN=8, the minimum energy does not occur when the charges are placed at the corners of
3.6 Electrostatic Total Energy 79
a cube. Instead,UEis minimized when the charges are placed at the corners of a twisted, rectangular parallelepiped. The search for the ground state is non-trivial because the number of configurations that correspond tolocalenergy minima increases exponentially withN. For largeN, the charges arrange themselves locally onto a triangular lattice with six nearest neighbors per charge. Many, but not all, of the ground states are found to possess exact or distorted icosahedral symmetry. If qk= |rk| =1, numerical studies show that the energy of the minimum energy configuration is very well approximated by
UE(N) 1 4π 0
1
2(N2−N3/2).
Roughly speaking, the first term is the average energy ofN charges distributed randomly over the surface of the sphere. The second term corrects for the fact that the charges are not randomly distributed but arrange themselves to maximize their distances from each other.
Thomson’s problem has a direct bearing on a remarkable phenomenon calledcharge inversion that occurs when large spherical ions with chargeZqare placed in solution with small, mobile particles with charge−q. It is reasonable to suppose thatZmobile particles would attach to the surface of each macro-ion to neutralize its charge. However, the form ofUE(N) above actually produces a lower energy whenN > Zparticles attach!
3.6.2 Interaction Total Energy is Potential Energy
Finally, it is instructive to computeUE for a charge distribution that is the sum of two parts so ρ(r)=ρ1(r)+ρ2(r).Using (3.76), the total energy is
UE[ρ1+ρ2]=UE[ρ1]+UE[ρ2]+ 1 4π 0
d3r
d3rρ1(r)ρ2(r)
|r−r| . (3.82) The first two terms on the right side of (3.82) are the total energies ofρ1 andρ2 in isolation. We identify the third term as theinteraction energybetweenρ1(r) andρ2(r):
VE= 1 4π 0
d3r
d3rρ1(r)ρ2(r)
|r−r| . (3.83)
We use the symbolVE in (3.83) because the interaction total energy is exactly the potential energy (3.70). We make this nearly self-evident point here because the same statement willnotbe true when we turn to magnetostatic energy in Chapter 12.
We note also a special case of (3.82) that arises whenρ2(r) is unspecified but serves to create an external potentialϕext(r). In that case, the termUE[ρ2] is absent from (3.82) and we deduce that the total electrostatic energy of a charge distributionρ1(r) in an external potential is
UE= 1 8π 0
d3r
d3rρ1(r)ρ1(r)
|r−r| +
d3rρ1(r)ϕext(r). (3.84)