Pablo Laguna Georgia Institute of Technology, USAlaguna.gatech.edu/EM1/Notes/Chapter_05.pdf ·...

ElectromagnetismChapter 5: Conducting Matter

Pablo LagunaGeorgia Institute of Technology, USA

Fall 2012

Notes based on textbook: Modern Electrodynamics by A. Zangwill

Laguna Electromagnetism

Perfect Conductor

A perfect conductor is a material in which both the macroscopic electric fieldand the macroscopic charge density vanish everywhere inside its volume V ;that is,

E(r) = ρ(r) = 0 r ∈ V

In a conductor, any excess of charge accumulates on its surface as a surfacecharge density σ(rS).

The simplest case is an isolated conducting sphere with net charge Q andradius R. By symmetry, the surface charge density is uniform soσ = Q/4πR2. The external electric field is identical to the electric field of apoint charge, and its electrostatic potential is

ϕ(r) =

Q

4πε0Rr ≤ R,

Q4πε0r

r ≥ R.


The electric field outside a conductor depends on its shape, net charge, andwhether or not an external field Eext(r) is present.

Since the field inside a conductor always vanishes,

E(r) = Eext(r) + Eself(r)

= Eext(r) +1

4πε0

∫S

dS′ σ(r′)r− r′

|r− r′|3 = 0

Thus, inside the conductor Eself(r) = −Eext(r)

Far outside the conductor, Eself(r) is a dipole electric field parameterized bythe macroscopic dipole moment

p =

∫S

rσ(r) dS

this is because in a perfect conductor in the presence of an external field,electrostatic induction creates charge polarization.


Thomson’s Theorem of Electrostatics

The electrostatic energy of a body (fixed shape and size) is minimized whenits charge Q distributes itself to make the electrostatic potential ϕ = constantthroughout the body and thus E = −∇ϕ = 0.

Proof: Let ρ(r) be the charge density of a body of volume V . If ϕext(r) is thepotential due to a external charge density ρext(r), then electrostatic energy isgiven by

UE [ρ] =1

8πε0

∫V

d3r∫V

d3r ′ρ(r)ρ(r′)|r− r′| +

∫V

d3r ρ(r)ϕext(r)

Next, we minimize UE subject to the constraint that Q =∫

V d3r ρ(r). Thus,

I [ρ] = UE [ρ]− λ∫

Vd3r ρ(r).

with λ a Lagrange multiplier.


Therefore

δI =1

8πε0

∫V

d3r∫V

d3r ′[ρ(r)δρ(r′)|r− r′| +

δρ(r)ρ(r′)|r− r′|

]

+

∫V

d3r δρ(r)ϕext(r)− λ∫V

d3r δρ(r)

Exchanging the variables r and r′ in the first term in square brackets yields

δI =

∫V

d3r δρ(r)

14πε0

∫V

d3r ′ρ(r′)|r− r′| + ϕext(r)− λ

= 0

Thus

14πε0

∫V

d3r ′ρ(r′)|r− r′| + ϕext(r) = λ

ϕint(r) + ϕext(r) = λ

Therefore, ϕtot(r) = λ.


ExampleFind the potential produced by a conducting sphere of radius R in a uniformelectric field E0. Show also that the sphere acquires a dipole momentp = αε0E0 with α = 4πR3 the polarizability.

Solution: By symmetry, a uniform external field induces an azimuthal chargedensity σ(θ) on the surface of the sphere. Since E0 = E0z, the chargedistribution σ(θ) must produce an electric field Eself = −E0 inside the sphere.Therefore ϕself(r < R, θ) = E0 z = E0r cos θ in order to haveEself = −∇ϕself(r < R, θ) = −E0.

Outside the sphere, σ(θ) produces a potential

ϕself(r > R, θ) =1

4π ε0

∞∑`=0

A`r `+1 P`(cos θ)

=Ar

+Br 2 cos θ +

Cr 3 P2(cos θ) + · · ·

The potential must be continuous at r = R. Therefore,

ϕself(r > R, θ) =E0R3

r 2 cos θ =1

4πε0

pr 2 cos θ

where p = αε0E0 with polarizability α = 4πR3.Laguna Electromagnetism

The Surface Charge Density

Recall the electric field matching conditions (with n the outward normal to theconductor surface),

n× [Eout − Ein] = 0

n · [Eout − Ein] = σ(rS)/ε0

Because Ein(r) = 0, then n× Eout|S = 0. Thus, Eout(rS) ∝ n, and from thesecond matching condition

Eout(rS) = nσ(rS)/ε0

Thereforeσ(rS) = ε0n · Eout(rS).


Let’s compute the dipole moment of the polarized sphere.

σ(θ) = ε0n · Eout(rS) = −ε0n · ∇[ϕext + ϕint]

= −ε0∂

∂r

[−Er cos θ +

ER3

r 2 cos θ]

r=R= 3ε0E cos θ.

Recall r = r sin θ cosφx + r sin θ sinφy + r cos θz, thus

p =

∫σ(θ) r dS =

∫σ(θ) r R2 sin θ dθ dφ

= 2πR3 zπ∫

0

σ(θ) sin θ cos θ dθ = 6π ε0 R3 E zπ∫

0

sin θ cos2 θ dθ

= 4πε0R3E z.


The Surface Charge Density

The surface charge density σ(rS) at the edge of conductors with sharp,knife-like edges diverges. For example, one can show that a conducting diskwith radius R, net charge Q, and a vanishingly small thickness is

σ(ρ) =Q

4πR√

R2 − ρ2


Screening & Shielding

Conductors are able to screen or shield the effects of an electric field.

For the conductor in the left above, E(r) = 0 not only at every point in or onthe conductor body but also inside the cavity, independently of the presenceof an external charge.

This can be seen from the following path integral in which inside the cavitythe electric field at A starts at a positive charge and ends at B in a negativecharge.

0 =

∮d` · E =

B∫A

d` |Ecavity|+A∫

B

d` · Econductor =

B∫A

d` |Ecavity|.


Screening & Shielding

On the other hand, for the conductor in the right above, the conductor is notable screen the space outside of itself from the charge placed inside a cavity.

This follows from applying Gauss’ law with a surface enclosing the conductor.


Capacitance

Capacitance measures the ability of a conductor to store charge,whether in isolation or in presence of other conductors.

Let V be the potential of an isolated conductor with volume Ω andsurface S.

From Thomson’s theorem, we have that ϕ(r) = V .

From Coulomb’s law, we also have that ϕ(|r| → ∞)→ 0.

Define ϕ(r) ≡ ϕ(r)/V

From Gauss’ law and E = −∇ϕ we have that

Q = −ε0V∫S

dS · ∇ϕ.


Define Self-capacitance as:

C =QV

= −ε0

∫S

dS · ∇ϕ

For a sphere of radius R, we have that ϕ(r) = 1/r , and thus C = 4πε0R

For a conducting disk, we have that

ϕ(0) =1

4πε0

∫S

dS′σ(r′S)

|r′S |= V .

Substitution ofσ(ρ) =

Q4πR

√R2 − ρ2

yields (careful, disk is not infinitely thin)

V =2

4πε0

Q4πR

2π∫0

dφ

R∫0

dρ√R2 − ρ2

=Q

8ε0R.

Thus, C = 8 ε0 R

Notice, C ∝ ε0 × length


The Capacitance Matrix

A capacitance matrix Cij relates the conductor charges Qi to theconductor potentials ϕj for a collection of N conductors; that is,Qi =

∑Nj=1 Cij ϕj

From the Green’s reciprocity relation,∫ρ(r) ϕ(r) d3r =

∫ρ(r)ϕ(r) d3r

applied to this caseN∑

i=1

ϕi Q′i =N∑

i=1

ϕ′i Qi

one can who that Cij = Cji .

Maxwell inequalities:Ckk > 0

Ckj < 0∑j

Ckj ≥ 0.


The Two-Conductor Capacitor

1Q Q

2Q Q

Recall the capacitor matrix Cij such that Qi =∑N

j=1 Cij ϕj .

If the matrix Cij is invertible, then ϕi =∑N

j=1 Pij Qj where P = C−1 is thematrix of coefficients of potential.

Since Q1 = −Q2, for the two-conductor capacitor above,

ϕ1 = P11Q1 + P12Q2 = (P11 − P12)Q

ϕ2 = P21Q1 + P22Q2 = (P12 − P22)Q.

which yields ϕ1 − ϕ2 = (P11 + P22 − 2P12)Q.


Therefore capacitance C of the capacitor is

C =Q

ϕ1 − ϕ2=

1P11 + P22 − 2P12

=C11C22 − C2

12

C11 + C22 + 2C12

where we used

P = C−1 =

[C11 C12

C12 C22

]−1

=1

C11C22 − C212

[C22 −C12

−C12 C11

].

Notice that in the limit C12 C11,C22

1C

=1

C11+

1C22

.

which is the case of capacitors connected in series.


Because Q1 = Q2 = Q and |∆V | = |∆V1|+ |∆V2|

1Ceq

=1

C1+

1C2

Because Q = Q1 + Q2 and |∆V | = |∆V1| = |∆V2|

Ceq = C1 + C2


1Q Q

2Q Q

More generally, if S is any surface that encloses the positive conductor,

C =Q

ϕ1 − ϕ2=

ε0∫S

dS · E

2∫1

d` · E.

The line integral follows any path from the positive conductor to thenegative conductor.


The most familiar example of a two-conductor capacitor is theparallel-plate capacitor.

It consists of two flat conducting plates with uniform surface chargedensities ±σ = ±Q/A separated a distance d . Above the plate at thebottom is the one with positive charge.

If d √

A, we use that E = nσ/ε0. Thus, and E ≈ const.


For the lower plate

Elower = +σ

2 ε0above

Elower = − σ

2 ε0below

For the upper plate

Eupper = − σ

2 ε0above

Eupper = +σ

2 ε0below

Thus Ebetween = σ/ε0 and Eoutside = 0

ThereforeC =

QV

=Q

E d=

Q ε0

σ d=

Q ε0 AQ d

=ε0 Ad


In reality, E in a parallel plate capacitor d is not very small compared to√A.

There are fringe electric field lines which begin and end on the outersurfaces of the plates. That is, some of the charge of each plate resideson those outer surfaces.

Therefore, |E| between the plates is slightly less than the infinite platevalue Q/A ε0.

Thus, Cexact > C = ε0 A/d


The Energy of a System of Conductors

Consider a collection of N conductors.

The charge of the k th conductor Qk is distributed with density σk (rS)over its surface Sk where the potential ϕk is constant.

Thus

UE = 12

∫ρϕ d3r = 1

2

N∑k=1

ϕk∫Sk

σk dS = 12

N∑k=1

Qk ϕk .

Using Qi =∑N

j=1 Cij ϕj we get

UE = 12

N∑i=1

N∑j=1

ϕi Cij ϕj

Using ϕi =∑N

j=1 Pij Qj we also get

UE = 12

N∑i=1

N∑j=1

Qi Pij Qj .


Recall that for a two-conductor capacitor Q1 = −Q2 = Q and

P = C−1 =

[C11 C12

C12 C22

]−1

=1

C11C22 − C212

[C22 −C12

−C12 C11

].

thus

UE = 12 Q2(P11 + P22 − 2P12) =

Q2

2C=

12

C(ϕ1 − ϕ2)2.

Since UE ≥ 0, then C > 0

Also

UE =12

C(ϕ1 − ϕ2)2 =1

2 CC2(ϕ1 − ϕ2)2

=1

2 CQ2

(ϕ1 − ϕ2)2 (ϕ1 − ϕ2)2 =Q2

2C


Forces on Conductors

Recall p = 12σ(Ein + Eout) is the electrostatic force per unit area

(pressure).

Since Ein = 0 inside a conductor and Eout = σn/ε0, we have that theexternal field excerpts a force/area or pressure

p =1

2 ε0σ2 n

orp =

ε0

2E2

out

That is, the electrostatic pressure is equivalent to the energy density ofthe electric field.


Consider a conductor expanding a distance dx due to the electrostaticpressure.

The electric field will be excluded from this region dV = A dx .

Thus, the energy of the electric field decreases by an amountdE = U dV = (ε0/2) E2 dV , where U is the energy density of the field.

On the other hand, the work which the field does on the conductor inorder to make it expand is dW = p A dx

Therefore, from conservation of energy, dE = dW and

p =ε0

2E2

out

Notice that this pressure will make a closed conductor expand in thepresence of an external electric field.


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