• Tidak ada hasil yang ditemukan

The Electric Stress Tensor

Dalam dokumen Modern Electrodynamics (Halaman 103-107)

Problems

Application 3.2 The Ionization Potential of a Metal Cluster

3.7 The Electric Stress Tensor

UsingR=RSN1/3, the predicted ionization potential has the form anticipated in (3.85):

IN =UEfinalUEinitial e2 4π 0RS

9 10 +1

2N−1/3

. (3.89)

3.7 The Electric Stress Tensor

LetE(r) be the electric field produced by a charge densityρ(r) which occupies a volume. The force exerted on the charge densityρ(rV) by the charge densityρ(rV) is

F=

V

d3r ρ(r)E(r). (3.90)

For many applications, it proves useful to eliminateρfrom (3.90) using0∇ ·E=ρ. To do this easily, we adopt the Einstein convention and sum over repeated Cartesian indices (see Section 1.2.4). In that case,∇ ·E=iEi, and

Fj =0

V

d3r EjiEi =0

V

d3r

i(EjEi)−EiiEj

. (3.91)

Now,∇ ×E=0 implies thatiEj =jEi. Therefore, Fj =0

V

d3r

i(EiEj)−EijEi

=0

V

d3r ∂i(EiEj12δijE·E). (3.92) The final member of (3.92) motivates us to define the Cartesian components of the Maxwellelectric stress tensoras

Tij(E)=0(EiEj12δijE2). (3.93) With this definition and the divergence theorem, (3.92) takes the form

Fj =

V

d3r ∂iTij(E)=

S

dSnˆiTij(E). (3.94)

Often, the stress tensor components in (3.93) are used to define the dyadic (Section 1.8)

T(E)=ˆeiTij(E)ˆej =xTˆ xx(Ex+xTˆ xy(Ey+xTˆ xz(Ez+yTˆ yx(Ex+ · · ·. (3.95) In this language, the net force on the charge inside the closed surfaceSis

F=

S

dSnˆ·T(E). (3.96)

Direct substitution from (3.93) confirms that an explicit vector form of (3.96) is

F=0

S

dS

(ˆn·E)E12(E·E)ˆn

. (3.97)

CUUK1954-03 CUUK1954/Zangwill 978 0 521 89697 9 August 9, 2012 14:57

82 ELECTROSTATICS: THE ELECTRIC FIELD PRODUCED BY STATIONARY CHARGE

nˆ E

ˆt

E θ

dS

Figure 3.16: The electric field line is coplanar with the surface unit normalˆnand the unit tangent vectorˆt. The angleθlies betweenˆnand the electric field line at the surface elementdS.

3.7.1 Applications of the Electric Stress Tensor

Let us apply (3.97) to a spherical balloon which carries a uniform charge per unit areaσ. The Gauss’

law electric field is zero everywhere inside the balloon and takes the valueE=rσ/ˆ 0 just outside the balloon’s surface. If we choose the balloon surface asS, the two terms in the integrand of (3.97) combine to give a force per unit areaf=rσˆ 2/20. This agrees with (3.52) where this situation yields f= 12σ[0+rσ/ˆ 0]. This force density tends to expand the size of the balloon (so each individual bit of charge gets farther apart from every other bit) while keeping the position of the center of mass fixed (F=0). More generally, Figure 3.16 shows that the electric fieldEat any surface elementdScan be decomposed into components along the surface normalˆnand a unit vector ˆtthat is tangent to the surface and coplanar withEand ˆn:

E=E( ˆncosθtsinθ). (3.98)

Figure 3.16 does not indicate the direction ofEbecause (3.98) transforms (3.97) into F=120

S

dS E2( ˆncos 2θtsin 2θ). (3.99) The great virtue of (3.97) and (3.99) is that they replace the volume integral (3.91) by a surface integral, which is often simpler to evaluate. We also gain the view that the net force on the charge inside V is transmitted through each surface element ˆndSby a vector force densityfwherefj =nˆiTij(E).

It is important to appreciate that the surface in question need not be coincident with the boundary of the charge distribution. Indeed, the fact that Tij(E) can be evaluated at any point in space leads very naturally to the idea that the vacuum acts as a sort of elastic medium capable of supporting stresses (measured byE) which communicate the Coulomb force. Faraday, Maxwell, and their con- temporaries believed strongly in the existence of this medium, which they called the “luminiferous aether”.

We can use (3.99) to extract mechanical information from the field line patterns shown in Figure 3.17 for two equal-magnitude charges. The key is to choose the perpendicular bisector plane to be part of a surfaceSthat completely encloses one charge when viewed from theothercharge (see Figure 3.18).

The charges are both positive in the left panel and the field lines bend toward tangency as they approach the bisector plane. This corresponds toθ=π/2 in Figure 3.16 and a force density−120E2nˆin (3.99).

Since ˆnis theoutwardnormal, this tells us that the net force on the bisector plane tends to push the bisector plane away from the viewing charge. This is consistent with Coulomb repulsion between the charges. By contrast, the field lines are normal to the bisector plane when the charges have opposite sign (right panel of Figure 3.17). This corresponds toθ =0 in Figure 3.16 and a force per unit area +120E2n. For this case, the net force tends to pull the bisector toward the viewing charge. This isˆ consistent with Coulomb attraction between the charges.

3.7 The Electric Stress Tensor 83

Figure 3.17: The electric field lines associated with two identical, positive point charges (left panel) and two equal but opposite point charges (right panel).

Example 3.6 Confirm that the stress tensor formalism reproduces the familiar Coulomb force law between two identical point charges separated by a distance 2d. Hint: Integrate (3.97) over a surface Sthat includes the perpendicular bisector plane between the charges.

x

r

q

d z

y

P q

Figure 3.18: The force between two identical charges may be calculated by integrating the electric stress tensor over the bowl-shaped surface (shaded) in the limit when the bowl radius goes to infinity.

Solution: Locate the charges at±d on thez-axis. To find the force on the charge atz= −d we chooseSas the surface of the solid bowl-shaped object shown in Figure 3.18. This surface encloses the half-spacez <0 in the limit when the bowl radius goes to infinity. In the same limit, the hemispherical portion of the bowl makes no contribution to the stress tensor surface integral (3.97) becausedSE2→0 asr→ ∞. Therefore, it is sufficient to integrate over the bisector plane z=0 where the outward normal ˆn=z

The charge distribution has mirror symmetry with respect toz=0. Therefore, the fieldEP at any pointPon the bisector must lie entirely in that plane. That is, ˆn·EP =0. The distance betweenP and either charge isd/cosθ whereθ is the angle between thez-axis and the line which connects either charge toP. Counting both charges, the component of the electric field in the plane of the bisector at the pointP is

E=2× q

4π 0d2cos2θsinθ.

Sincer=dtanθ, the differential element of area on the bisector is dS=rdrdφ=d2 sinθ

cos3θdθ dφ.

CUUK1954-03 CUUK1954/Zangwill 978 0 521 89697 9 August 9, 2012 14:57

84 ELECTROSTATICS: THE ELECTRIC FIELD PRODUCED BY STATIONARY CHARGE

Only the second term in (3.97) contributes, so the force on the charge enclosed bySis F= −0

2ˆz

z=0

dS E2= −0

2 ˆz q

2π 0d 22π

0

π/2

0

cosθsin3θ = −ˆz 1 4π 0

q2 (2d)2. This is indeed Coulomb’s law.

Sources, References, and Additional Reading

The quotation at the beginning of the chapter is taken from Letter XXIII of Volume II of

L. Euler,Letters on Different Subjects in Natural Philosophy Addressed to a German Princess(Arno, New York, 1975).

Section 3.1 The most complete treatment of electrostatics is the three-volume treatise E. Durand,Electrostatique(Masson, Paris, 1964).

Volume I is devoted to basic theory and the potential and field produced by specified distributions of charge.

Readers unfamiliar with the French language will still benefit from the beautiful drawings of electric field line patterns and equipotential surfaces for many electrostatic situations. Four textbooks which distinguish electrostatics from boundary value potential theory are

A. Sommerfeld,Electrodynamics(Academic, New York, 1952).

M.H. Nayfeh and M.K. Brussel,Electricity and Magnetism(Wiley, New York, 1985).

L. Eyges,The Classical Electromagnetic Field(Dover, New York, 1972).

W. Hauser,Introduction to the Principles of Electromagnetism(Addison-Wesley, Reading, MA, 1971).

Section 3.3 See Section 15.3.1 for a classical argument due to Eugene Wigner which relates the freedom to choose the zero of the electrostatic potential to the conservation of electric charge.

Example 3.2 is an electrostatic version of a scaling law derived by Newton for the gravitational force. See Section 86 of the thought-provoking monograph

S. Chandrasekhar,Newton’s Principia for the Common Reader(Clarendon, Oxford, 1995).

Figure 3.4 is taken from this reprint of the 3rd edition (1891) of Maxwell’sTreatise:

J.C. Maxwell,A Treatise on Electricity and Magnetism(Clarendon, Oxford, 1998).

The problems associated with projecting three-dimensional field line patterns onto two-dimensional diagrams are discussed in

A. Wolf, S.J. Van Hook, and E.R. Weeks, “Electric field lines don’t work”,American Journal of Physics64, 714 (1996).

Section 3.4 Application 3.1 is taken from

T.M. Kalotas, A.R. Lee, and J. Liesegang, “Analytical construction of electrostatic field lines with the aid of Gauss’ law”,American Journal of Physics64, 373 (1996).

Our discussion of symmetry and Gauss’ law is adapted from

R. Shaw, “Symmetry, uniqueness, and the Coulomb law of force”,American Journal of Physics33, 300 (1965).

Section 3.6 Entry points into the literature of Thomson’s problem and charge inversion are, respectively, E.L. Altschuler and A. P ´erez-Garrido, “Defect-free global minima in Thomson’s problem of charges on a sphere”,Physical Review E73, 036108 (2006).

A.Yu. Grosberg, T.T. Nguyen, and B.I. Shklovskii, “The physics of charge inversion in chemical and biological systems”,Reviews of Modern Physics,74, 329 (2002).

Problems 85

The empirical formula (3.85) in Application 3.2 comes from the paper

C. Br ´echignac, Ph. Cahuzak, F. Carlier, and J. Leygnier, "Photoionization of mass-selected ions: A test for the ionization scaling law",Physical Review Letters63, 1368 (1989).

Figure 3.17 and the discussion of the electric stress tensor in Section 3.7 are based on the treatment in E.J. Konopinski,Electromagnetic Fields and Relativistic Particles(McGraw-Hill, New York, 1981).

Problems

Dalam dokumen Modern Electrodynamics (Halaman 103-107)