Electrostatics
page 1.1
Capstones in Physics: Electromagnetism
1. ELECTROSTATICS
1.1. Electric Field
1.2. Electric Potential
1.3. Work and energy in Electrostatics
1.4. Conductors
(References to) "Introduction to Electrodynamics" by David J. Griffiths
3rd ed., Prentice-Hall, 1999. ISBN 0-13-805326-X
©2005 Oregon State University
Philip J. Siemens (revised by Yun-Shik Lee)
Electrostatics
page 1.2
1.1 Electric Field
Coulomb’s law
Force on a test charge Q due to a point charge q
r
r
Q
q
r
r'
r
F=
r
R
qQ )
r
2
4πε 0 r
r r r
r = R − r'
1
ε 0 = 8.85 ×10
−12
C2
N ⋅ m2
Permitivity of free space
q1
q2
qi
r
ri '
r
ri
Q
r
R
r r r
F = F1 + F2 + L
=

Q  q1 ) q2 )
 2 r1 + 2 r2 + L
4πε 0  r1
r2

=
qi )
∑ ri
4πε 0 i =1 ri 2
Q
n
Electric field
r
r
r
1 n qi )
F = QE ⇒ E ≡
∑ ri
4πε 0 i =1 ri 2
For continuous charge distribution :
©2005 Oregon State University
superposition principle
r
ρ (r )
1
E=
rˆdv
2
∫
4πε 0 r
Philip J. Siemens (revised by Yun-Shik Lee)
Electrostatics
page 1.3
Gauss’s law
The flux through any surface enclosing charge Qenc is Qenc/ε0.
For any closed surface,
r r 1
∫ E ⋅ da = Qenc
ε0
S
Divergence theorem:
(
)
r r
r
∫ E ⋅ da = ∫ ∇ ⋅ E dτ
S
V
Qenc in terms of the charge density ρ :
⇒
∫(
V
)
r
ρ
∇ ⋅ E dτ = ∫   dτ
ε
V 0 
Qenc = ∫ ρdτ
V
r 1
⇒ ∇⋅E = ρ
ε0
The Curl of E
Integral of E around a closed path is zero:
r r
r
E
⋅
d
l
=
0
⇒
∇
×
E
=0
∫
r r
r r
Stokes’ theorem, ∫ E ⋅ dl = ∫ ∇ × E ⋅ da
(
)
S
1.2 Electric potential
r r
E
Because ∫ ⋅ dl = 0 , the line integral of E from a to b is the same for all
paths.
(i)
a
©2005 Oregon State University
b
(ii)
Philip J. Siemens (revised by Yun-Shik Lee)
Electrostatics
page 1.4
Because the line integral is independent of path, we can define a function
r
r
r r
r
V (r ) ≡ − ∫ E ⋅ dl
(electric potential).
O
O is some standard reference point, thus V depends only on the point r .
The differential version:
r
E = −∇V
Poisson’s equations and Laplace’s equations
r
E = −∇V
r ρ
∇⋅E =
r ε0
∇× E = 0
⇒ ∇ 2V = −
ρ
ε 0 Poisson’s equation
In regions ρ=0, ∇ V = 0 : Laplace’s equation
2
Potential of a localized charge distribution
The potential of a point charge q :
r
V (r ) =
1
q
, where r is the distance from the charge to
4πε 0 r
r
For a volume charge, V (r ) =
4πε 0
∫
r
ρ (r ' )
r
dτ '
P
r
q
r
r'
1
r
r
©2005 Oregon State University
r
r.
r
r'
dτ '
r
P
r
r
Philip J. Siemens (revised by Yun-Shik Lee)
Electrostatics
page 1.5
1.3 Work and energy in Electrostatics
ENERGY OF A CHARGE DISTRIBUTION
potential energy of distribution
=
energy required to assemble it
construct from pair potential energies Uij =
qi qj
4πεorij
energy
step
1. bring charge 1
q
0
2. bring charge 2
U12
3. bring charge 3
U13 + U23
r 12
1
r 13
q
2
r 23
q3
______________
4, 5,....
total energy
=
Utotal =
U12 + U13+U23 + …
∑ Uij
pairs
compare
U total =
qj
qi q j
1
1
1
1
=
=
= ∑ qiVi
q
U
∑∑
∑∑
∑
ij
i∑
2 i j ≠i
2 i j ≠i 4πε 0 rij 2 i
2 i
j ≠ i 4πε 0 rij
1
⇒ generalization to continuous distribution has factor 2 :
Utotal
=
1
⌡d3r
2 ⌠
ρ(r) V(r)
1
= 2 ⌠
⌡d3r ⌠
⌡d3R
ρ(r) ρ(R)
4πεo|r–R|
note this form doesn't work for idealized distributions,
including point charges: the energy integral diverges!
©2005 Oregon State University
Philip J. Siemens (revised by Yun-Shik Lee)
Electrostatics
page 1.6
ENERGY OF AN ELECTRIC FIELD
potential energy of a charge distribution = energy required to assemble it
(note: can only compute if finite!)
see "energy of a charge distribution"
1
Utotal = 2 ⌠
⌡d3r ρ(r) V(r)
counting argument →
express ρ in terms of electric field E(r):
ρ(r) = εo∇•E(r)
εo
Utotal = 2 ⌠
⌡d3r V(r) ∇•E(r)
substitute:
∇•(aA) = (∇a) • A + a (∇ •A)
vector identity:
substitute a→V, A →E:
εo
Utotal = 2 ⌠
⌡d3r {∇•[V(r) E(r)]–E(r) •∇V(r)]}
express ∇V in terms of electric field E(r):
divergence theorem:
E(r) = –∇V(r)
∫V d r ∇•A(r) = ∫S dS •A(r)
3
substitute A →VE , apply to expression for Utotal:
εo
εo
Utotal = 2 ⌠
dS
•[V(r)
E(r)]
+
⌡d3r [E(r) •E(r)]
⌡
2 ⌠
first term → 0 when surface goes to ∞ :
area ~ R2, but localized charges ⇒ E ~ 1/R2, V ~ 1/R
εo
Utotal = d3r 2 E(r)2
∫
conclude
but surely
Utotal =
∫d d r
3
(energy density)
1
energy density = 2 εo E(r)2
©2005 Oregon State University
Philip J. Siemens (revised by Yun-Shik Lee)
Electrostatics
page 1.7
1.4 Conductors
STATIC CONDUCTORS
Conductors have charges free to move.
In steady state, constant current ⇒ motion of charges
In static state, charges aren't moving ⇒ no forces ⇒
E(r) = 0 everywhere inside static conductor
⇒
• charge density ρ(r) = 0
and
• potential V(r) = constant
⇒ charge density only on surface, σ
see "Surface charge"
Gauss' law ⇒ Eoutside – Einside = σ/εo normal to surface
but Einside = 0, so
σ
Eoutside =
n̂
εo
at surface of conductor
Pressure on a conductor due to electric field
consider surface charge on area dS
force due to other charges
F = Q Eexternal
1
Eexternal = 2 E
1
p = σ Eext = 2 σE
But Eexternal = Edue to surface charge ⇒
pressure p = Fnormal/dS ⇒
pressure
alternately, F = –
p =
δEpotential
δposition
ε
Epot = 2 E2 ×vol ,
©2005 Oregon State University
Q = σ dS
εo
σ2
= 2 E2
2εo
on surface of conductor
see "Electrostatic Forces"
move surface ⇒ δvol =S ×δposition
√
Philip J. Siemens (revised by Yun-Shik Lee)
Electrostatics
page 1.8
CAPACITANCE
recall a conductor is an equipotential,
claim: potential V on an isolated conductor ~ its charge Q
ρ
2
proof: ∇ V = −
ε 0 is a linear relation, determines V
C ≡ Q/V
definition: capacitance
for isolated conductor
2 conductors with equal, opposite charges are a capacitor .
definition: capacitance
Q
Q
C ≡ |V –V | =
∆V
2 1
Q
for pair of conductors
Q
⌠ dQ Q/C
energy stored by charging = ⌠
⌡ dQ V(Q) = ⌡
0
0
Q2
1
1
energy stored = 2C = 2 Q V = 2 C V2
Q
example: 2 large flat plates
area S, dielectric thickness d
assume charges ±Q on plates
σ
–σ
–Q
this is free charge!
S
⇒ σ = Q/S ⇒
E – Econd = σ/ε0
E
Econd = 0 ⇒ E = σ/ε0
∆V = – ∫ dL • E = d E =
©2005 Oregon State University
d
dσ
ε0
Q
⇒C =
∆V
σS
ε0 S
=
d σ / ε0 = d
Philip J. Siemens (revised by Yun-Shik Lee)
Electrostatics
page 1.9
MIRROR IMAGES
Uniqueness theorem: The solution to Laplace’s equation in some volume is
uniquely determined if V is specified on the boundary surface S.
At a plane interface, the potential can be described by a trick:
y
simplest example:
Q
point charge Q at (–D,0,0)
conducting plane x = 0
–Q
D
x
need to find induced charge
to make E ⊥ to plane
Potential for x<0 (V=0 for x>0)
V ( x, y , z ) =
Q 
1
1

−
2
2
2
2
4πε 0  ( x + D ) + y + z
( x − D) + y 2 + z 2

Field in midplane
Ex(x=0,y,z) = 2
Q
4πεo(D2+y2+z2)
surface charge density σ(y,z) = εoE•n̂ =



D
√D2+y2+z2
–QD
2
2π(D +y2+z2)3/2
Field on right = 0, field on left = EQ+Einduced
Eleft =
–Q(R–Dx̂)
Q(R+Dx̂)
+
4πεo[(x–D)2+y2+z2]3/2
4πεo[(x+D)2+y2+z2]3/2
Note total induced charge = –Q (proof: Gauss' law applied to surface
enclosing all charges, field ~ 1/r3)
generally, superposition ⇒ mirror-image charge distribution
More complicated: image in surface of dielectric insulator
©2005 Oregon State University
Philip J. Siemens (revised by Yun-Shik Lee)
Electrostatics
page 1.10
Other geometry: can also find images for surfaces:
• spherical surface
©2005 Oregon State University
• cylindrical surface
Philip J. Siemens (revised by Yun-Shik Lee)