Lecture 5
Electric Flux and Flux Density,
Gauss Law in Integral Form
Sections: 3.1, 3.2, 3.3
Homework: See homework file
LECTURE 5
slide
1
Faraday’s Experiment (1837), Flux
• charge ‘transfer’ from inner to outer sphere
• electric induction: charge deposition
without contact
E
+Q
• observations on charge on outer electrode
it is of the same magnitude but opposite
sign (−Q) as that on inner electrode
it is the same regardless of the insulating
material used (εr)
it is the same regardless of electrode’s shape
b
a
εr
insulator
−Q
A
displacement flux ≡ electric flux Ψ =Q, C
flux depends on the charge and does not depend on the medium
LECTURE 5
slide
2
Flux Density and Flux – 1
• charge densities on inner and outer spheres
Q
−Q
ou
Q>0
=
ρ sin =
ρ
s
2
2
4π a
4π b
• flux has direction and density D corresponding to the enclosed
charge density (flux density points from +Q to −Q)
=
Q ∫∫ ρ s ds =
⇔ Ψ ∫∫ Dn ds
S
S
• in spherical coordinates (charged sphere +Q at origin)
 on the surface of the inner sphere
Dn = Dr
Ψ
Q
2 ⇒ D=
in
=
Ψ D
⋅
4
π
a
(
a
)
=
=
ρ
r
+Q
r
r
s

a
2
2
4π a
4π a
b
Dn
 on the surface of the outer sphere
S
Ψ
−
Q
ou
=
Ψ Dr ⋅ 4π b2 ⇒ D=
=
=
−
ρ
r ( b)
s
−Q
4π b2 4π b2
LECTURE 5
slide
3
Flux Density and Flux – 2
• flux density at any distance r (point charge at origin)
Q
1 Q
)
(
r
)
Ψ= Dr ⋅ 4π r 2 ⇒ Dr ( r=
⇒
D
=
⋅ 2 ar
2
4π r
4π r
• compare with E of a point charge at origin
1 Q
(r)
E=
⋅ 2 ar
4πε r
• E depends on the permittivity, D does not: D
describes the sources regardless of the medium
Dn = Dr
+Q
r
b
S
−Q
• in vacuum D = ε 0 E
• the principle of superposition applies to D as well
1
ρ v (r′)a R
dv′
D(r′) =
∫∫∫
2
4π v′
R
LECTURE 5
slide
4
Flux Density D of Fundamental Charge Configurations
• multiple point charges
1 N Qn (rn′ )
1 N Qn (rn′ )
E(r ) = ⋅ ∑
a ⇒ D( r ) = ⋅ ∑
a
2 n
2 n
4πε n 1 =
4π n 1 | r − rn′ |
| r − rn′ |
• line charge
ρl 1
⋅ ⋅ ( cosθ1 − cosθ 2 )
E ρ=
4πε ρ
ρl 1
⋅ ⋅ ( sin θ 2 − sin θ1 )
E z=
4πε ρ
ρl
Dρ = ⋅ ( cosθ1 − cosθ 2 )
4πρ
ρl
Dz = ⋅ ( sin θ 2 − sin θ1 )
4πρ
P2 θ 2
R2
dQ = ρl dz ′
dz ′ θ
α
′
z
L
z
0 y
x
θ1
az
R
ρ
aρ
P ( ρ , φ ,0)
R1
P1
ρs
• infinite sheet of uniform charge: =
E an
⇒=
D ?
2ε
5
Total Flux through Closed Surface
• total flux through closed surface depends neither on the surface
shape nor on the mutual positioning of the charge and its enclosing
surface
• it depends solely on the enclosed charge
• analogies: water flow, electrical current, etc.
 find the mathematical expression for flux so that the above is true
?
Ψ (D, S ) =
 procedure: consider a point charge at the origin and its electric
flux through: (a) sphere, (b) any surface
 if the expression is true for a point charge it will be true for any
collection of charges as per the superposition principle
LECTURE 5
slide
6
Background: Solid Angle and 1 Steradian
• 1 radian of a 2D angle (φ = 1 rad)
cuts out an arc from a circle of length
equal to the radius of the circle
if ϕ 1,=
larc r
larc = r ⋅ ϕ =
if ϕ 2=
=
π , larc 2π r
full circle
• 1 steradian of a solid angle (Ω = 1
sr) cuts out an area from a sphere
equal to (radius)2
ssph= r 2 ⋅ Ω
=
if Ω 1,=
ssph r 2
=
if Ω 4π=
, ssph 4π r 2
full sphere
LECTURE 5
slide
7
Differential Solid Angle and Differential Surface Area
• differential solid angle and differential surface area on a sphere
ds
= r 2d Ω
in SCS: ds = r 2 sin θ dθ dφ


dΩ
example: the area and the solid angle of a full sphere (radius is r0)
Asph =
∫∫

2π π
ds =
Sphere
∫
2
2
2
2
π
r
sin
d
d
r
2
(
cos
)
4
r
,
m
θ
θ
φ
=
⋅
π
⋅
−
θ
=
π
0
0
0
0
∫
0 0
2π π
Ω
=
sph
d Ω ∫ ∫ sin θ dθ=
dφ
∫∫ =

Sphere
0 0
4π , sr
solid angle of full 3D space
LECTURE 5
slide
8
True or False
The solid angle defined by 0 ≤ θ ≤ π / 2 and 0 ≤ φ ≤ 2π
is equal to 2π.
The solid angle defined by a rectangular trihedral
corner is equal to π/2.
LECTURE 5
slide
9
Total and Differential Flux of Point Charge
flux through a sphere (charge at origin)
ds
2π π



2 sin=
D
ds
D
r
Ψ 
=
θ dθ d
φ Q
n
r
∫∫
∫ ∫ 
sphere
0 0
dΨ
due to spherical symmetry
Q
Dr ( r ) =
2
4π r 2
d Ψ =Dr r
dΩ
dS
⇒ dΨ =
Q
d Ω , d Ω =sin θ dθ dφ
4π
r1
r2
Q
4π
dΩ
dΨ
http://phys23p.sl.psu.edu/phys_anim/EM/sphere_to_sphere.avi
the differential flux dΨ is proportional to (QdΩ) and does not
depend on r
LECTURE 5
slide
10
Differential Flux of Point Charge
flux through arbitrary surface (charge is at origin)
• differential flux through surface element of an arbitrary surface
dssphere = dsother cos α
an
α ar
dΩ
0
http://phys23p.sl.psu.edu/phys_anim/EM/patch_to_sphere.avi
http://phys23p.sl.psu.edu/phys_anim/EM/fpbts3.avi
dssphere
dsother
• we want d Ψ sphere =Ψ
d other =Ψ
d  Qd Ω
d Ψ sphere =
Dr dssphere =
Dr cos α dsother =
Dn dsother =
D ⋅ ds

dssphere
• general expression for differential flux
d Ψ= D ⋅ ds
LECTURE 5
slide
11
Total Flux of Point Charge Through Arbitrary Surface
• regardless of the chosen surface
=
Ψ
ds
∫∫ D ⋅=

Q
S
dΩ
• as per superposition principle, result
holds for any collection of charges
P′
dΩ
Q
S2
Ssphere
P′
Q2
Q1
Q
= Q1 + Q2
Ψ = Ψ1 + Ψ 2
S1
P′′
Sother
single charge at origin
http://phys23p.sl.psu.edu/phys_anim/EM/blob_diced.avi
Sother
any two charges
12
Gauss’ Law
the electric flux over a closed surface is equal to the total
charge enclosed by the surface
Ψ=

∫∫ D ⋅ ds=
∫∫∫ ρv dv
Q=
S
flux equals enclosed
charge (no enclosed
charge means no flux)
Q
v
d Ψ ′′ =− d Ψ ′  Qd Ω
a n D( P′′)
a n dΩ
P′′
P′
D( P′)
d Ψ′ < 0
total flux Ψ =0
S
LECTURE 5
d Ψ ′′ > 0
slide
13
Gauss Law: Applications
Gauss’ law makes solutions to problems with planar, cylindrical
or spherical symmetry easy
procedure: choose integration surface so that
D is everywhere either normal or tangential to surface
if normal: D ⋅ ds =
Dds
if tangential: D ⋅ ds =
0
when normal to surface, D is also constant on surface
∫∫ D ⋅ ds = ∫∫ D ⋅ ds = D ⋅ S
S
S
LECTURE 5
slide
14
Gauss Law Applications: Field of Infinite Line Charge
z

∫∫ D ⋅ ds = Q = ρl ⋅ l
S
due to symmetry D = Dρ a ρ
2π
⇒
l
∫ ∫
x φ ρ
Dρ ρ dφ dz =Dρ ⋅ 2πρ l =ρl l
y
=
φ 0=z 0
l
ρl
⇒D=
aρ
, C/m 2
2πρ
ρl
D
, V/m
⇒E=
= aρ
ε
2περ
ds = ρ dφ dza ρ
S
ρl
This result was already obtained in Lecture 4
by the superposition principle.
LECTURE 5
slide
15
Gauss Law Applications: Field of Coaxial Cable
problem has cylindrical symmetry
Gaussian surface chosen as cylinder of radius ρ
solution analogous to that of line charge
• for a ≤ ρ ≤ b
ρl
D = aρ
, C/m 2
2πρ
• for ρ > b, D =
0
ρ
Homework:
Prove that for a uniformly charged cylinder
 ρv 
=
of radius
a, Dρ   ρ , for 0 ≤ ρ ≤ a.
 2 
LECTURE 5
slide
16
Gauss’ Law Applications: Field of Sheet Charge
∫∫ D ⋅ ds = Q = ρ s ⋅ A

D top lx
an
ly
S
z y ρs
x
δ
A = lxl y
S
D
bottom
an
∫∫ Dz dxdy − ∫∫
Dz dxdy + ∫∫ D ⋅ ds =
ρs A
top
sides





 bottom
bottom
Dztop ⋅ A
due to
Dz
⋅A
symmetry: Dztop
ρs
flux through S is 0
=
− Dzbottom
ρs
=
D
⇒ 2 DA =
ρ s A, D =
2
ρs
D
⇒D=
± az ⇒ E = =
± az
2
2ε
ε
This result was already obtained in Lecture 4 by the superposition
principle.
LECTURE 5
slide 17
Gauss’ Law Applications: Field of Spherical Charge
A sphere of radius a has uniformly distributed charge of
volume density ρv, C/m3. Determine the electric flux density
in the sphere and out of it.
• spherical symmetry of the source implies spherical symmetry
of field
• choose integration surface as sphere
(1) inside the sphere

∫∫ D ⋅ ds=
Q ( r )=
S (r)
∫∫∫ ρ v dv=
ρ v ∫∫∫ dv
v(r)
v(r)
4 3
Dr ( r ) ⋅ 4π r =
ρv π r
3

2
 ρv 
⇒ Dr (r ) =
 r
 3 
v(r)
LECTURE 5
slide
18
Gauss’ Law Applications: Field of Spherical Charge
(2) outside the sphere
4 3

∫∫ D ⋅ ds = Q = ∫∫∫ ρ v dv = ρ v ⋅ 3 π a
S (r)
v
4 3
Dr ( r ) ⋅ 4π r ==
Q ρv π a
3
ρv a3
Q
⇒ Dr ( r ) = 2 = ⋅ 2
3 r
4π r
2
ρv a
Dr
3
 1/ R 2
0
a
R
• outside the sphere the field is the same as that of a point charge
LECTURE 5
slide
19
Gauss’ Law Applications: Field of Spherical Charge
E-field magnitude in 3-D space
LECTURE 5
slide
20
You have learned about:
the flux density vector D and how it relates to the charge Q and
the E vector
Gauss’ law of electrostatics in integral form
Ψ=

∫∫ D ⋅ ds=
S
Q=
∫∫∫ ρv dv
v
the application of Gauss’ law to the solution of symmetrical
problems
• infinite planar charge
• infinite line charge
• infinite cylinder (inside and outside)
• coaxial cable
• uniformly charged sphere (inside and outside)
LECTURE 5
slide
21