Objectives
• T
To be
b able
bl to find
fi d the
h electric
l i flux
fl density
d i from
f
the
h
electric field intensity and vice versa.
• To
T be
b able
bl to
t use Gauss’
G
’ law
l to
t calculate
l l t the
th electric
l ti
field intensity of the charge distributed as the
following shapes: very long line (cylinder),
(cylinder) very
large plane, sphere. Also for the combination of those
p
shapes.
• To be able to find the divergence of the given function
in Cartesian, cylindrical and spherical coordinates.
• To be able to find the volume charge distribution from
the given electric field/flux intensity.
Gauss’ Law
• Faraday’s experiment
• Gauss
Gauss’ law and its application
• Divergence theorem
• Gauss’ law in the point form
EE3301
1
Faraday’ss Experiment
Faraday
2. Cover the sphere in 1. with a larger
conductor sphere.
sphere Divide each spere
with an insulator (dielectric)
3. Ground
3
G
d th
the larger
l
sphere.
h
Æ Charge
Ch
on outer surface = 0 C
4 Take ground out.
4.
out Then remove the
larger sphere. Measure the charge in the
g sphere.
p
larger
5. Charge in the larger sphere = −Q C
no matter which kind of the dielectric
between the two spheres.
EE3301
2
Faraday’ss Experiment: Description
Faraday
1 Add Q-C
1.
Q C charge to conductor sphere.
sphere
Q
EE3301
3
• Dielectric between sphere Æ charges cannot be
exchanged between sphere.
• Charge in the smaller sphere affect the outer sphere
such that there is –Q-C charge
g on the inner surface of
the larger sphere and +Q-C charge on the outer
surface (grounded out).
Faraday’s Law:
“The
The electric flux passing through any closed
surface is equal to the total charge enclosed by the
surface ”
surface.
EE3301
4
Gauss’ Law
Gauss
Gauss’ Law: Applications
Gauss
Gauss’ Law: explain Faraday’s law as maths eqn.
• Gauss’ law
l leads
l d to easy E, D calculation
l l i for
f
the following surface.
r r
∫ D ⋅ dS = Qenclosed
– Infinitely long line charge (constant charge
density)
– Charge distributed in an infinitely long cylinder as
a function of ρ.
– Infinitely large surface charge (constant charge
density)
– Charge distributed in a sphere as a function of r.
– The combination of the above distributions.
surface
where
h D : Electric
El i flux
fl density[C/m
d i [C/ 2]
r
r
D = ε 0E
EE3301
5
EE3301
Some Rules of Thumb
6
Gauss’ Law: Line Charge
Gauss
• Line charge on z-axis.
z axis
• Direction of D: cylindrical radius (aρ)
• Gaussian surface: cylinder r r
• NOTE: Used only for the shapes in Slide#6
• Gaussian surface follows the shape of charge
distribution. Æ Define coordinate according to
the shape.
shape
• Direction of D (E) is normal to the charge
distribution.
enclosed
surface
r r
D
∫ ⋅ dS +
top
h
D ρaρC/m2
dSside
ρl C/m
EE3301
∫ D ⋅ dS = Q
dStop
7
EE3301
dSbottom
r r
D
∫ ⋅ dS +
bottom
0 + 0 + Dρ
r r
D
∫ ⋅ dS = Qenclosed
side
∫ dS = ρ h
l
side
Dρ (2πρh ) = ρ l h
Dρ =
ρl
C/m 2
2πρ
8
Gauss’ Law: Cylinder
y
((Outside))
Gauss’ Law: Cylinder (Inside)
Gauss
• Charge
g on/in a cylinder
y
whose axis = z axis
• Charge density = function of ρ
• Similar
Si il to line
li charge
h
with
i h different
diff
Qenclosed
ρv(r) C/m3 or ρs C/m2
top
b
bottom
Ds
Dρ =
Dρ =
v
2πρ
Ds
9
r0
r
ra
r
surface
r
EE3301
∫ dS = Q
enclosed
surface
EE3301
(
)
Dr 4πr 2 = Qenclosed ⎯
⎯→ Dr =
10
• Similar to the effect of a point charge with the
ccharge
ge amount
ou equal
equ too charge
c ge enclosed
e c osed
located at the center of the sphere.
• Charge
Ch
di
distributed
t ib t d iin a sphere
h bbehave
h
lik
like a
point charge.
r
⋅ dSa r = ∫∫∫ ρ v (r )dv
Dr
2πρ
ρ
r
r
Q
D(r ) = enclosed
ar
2
4πr
enclosed
r
0 0
• Investigate the value of D
surface
∫D a
ρ v ρdρdφdz
Comment on Spherical Charge
∫ D ⋅ dS = Q
S
0
ρ0
Assumption: center at the origin.
Charge density = function of r.
Direction of D: spherical radius (ar)
Gaussian surface: sphere
r
0
∫ ∫ ρ ρdρdϕ
S
Gauss’ Law: Spherical
p
Charge
g
•
•
•
•
ρ
v
v
EE3301
2π
∫ ∫
2π ρ
z1 − z0
S
∫ dS = ∫ ρ dv
Dρ (2πρh ) = (∫∫ ρ ρdρdϕ )h
∫∫ ρ ρdρdϕ
v
crosssection
h
0
side
dSbottom
Qenclosed = ∫
ρ
S
side
id
0 + 0 + Dρ
Dρaρ C/m2
dSside
Z
r r
D
∫ ⋅ dS = Qenclosed
r r surface r r
r r
D
⋅
d
S
+
D
⋅
d
S
+
D
∫
∫
∫ ⋅ dS = Qenclosed
dStop
h
• Similar to the previous case but with different
value for Qenclosed.
Qenclosed
4πr 2
11
EE3301
12
Gauss’ Law: Thin Plate
Gauss
Gauss’ Law: Thin Plate (2)
Gauss
• Assumption: surface charge = xy plane.
plane
• Direction of D: normal to the surface
• Gaussian surface: box
r r
• Direction of D:
∫ D ⋅ dS = Q
enclosed
Ds
Area = A
r
S
r
surface
r
r
+
–
Positive: outwards D
Negative: inwards D
r
r
∫ D ⋅ dS + ∫ D ⋅ dS + ∫ D ⋅ dS = Q
enclosed
S
S
top
4 sides
Dn ∫ dS + Dn
S
Ds
bottom
top
∫ dS + 0 = ρ A
s
bottom
Dn ( A + A) = ρ s A
S
Dn =
EE3301
ρs
13
2
EE3301
Adapted Case
14
Ex : Hollow Cylinder
Ex.:
z
• Combination
y
• A charge density is given as
ρ s = ρ v ( z )dz
z ρ ( z ) dz
v
Dn = ∫
⎧ ρ v C / m 3 ;0.25 < ρ < 0.5
ρv (ρ ) = ⎨
1
3
⎩ 0C / m ; otherwise
2
z0
dz
Find E everywhere.
everywhere
ρ1
• Hole
ρ0
Qenclosed =
Drawing:
2π ρ1
∫ ρ∫ ρ ρdρdϕ
0.25
v
0
0
2π ρ1
EE3301
0
0
2πρ
15
Theory: Gauss’ Law
r
r
D S ⋅ dS = Qenclosed
∫
surface
r
r D
E=
ε0
∫ ρ∫ ρ ρdρdϕ
v
Dr ,outside =
0.5
EE3301
r
r
Shape = cylinder so D = Dρ ( ρ )a ρ ,
Gaussian surface = cylinder. 16
Ex : Hollow Cylinder (2)
Ex.:
Ex : Hollow Cylinder (3)
Ex.:
Solution: Find D,, given
g
that Gaussian surface is a cylinder
y
whose
axis is the same as the charge distribution.
r
r
D
⋅
d
S
∫ S = Qenclosed
0
side
0
0
= ρv
r
∫D
surface
17
r
∫D
0.25
S
1
0
∫
0
0 0.25
1 2π
= ρv ∫
∫
0 0
= ρv
r
∫D
surface
EE3301
S
1
∫ ρv ρdρdϕdz = ρv ∫
(0.5
2
)
2π
∫
ρ
0
dφdz
0.25
0.1875 1 2π
− 0.252
dφdz = ρ v 2 ∫0 ∫0 1dφdz
2
0.1875
(2π ) = 0.1875πρv
2
r
⋅ dS = Qenclosed
2πρ
ρ0 Dρ ( ρ 0 ) = 0.1875πρ
ρv
r
ρ v
D( ρ 0 ) = 0.09375 v a ρ
ρ0
S
(ρ
2
0
dφdz
0.25
)
2
0
)
− 0.252
(2π ) = πρv ρ 02 − 0.0625
2
(
)
r
⋅ dS = Qenclosed
2πρ0 Dρ ( ρ 0 ) = πρv ( ρ 02 − 0.0625)
2) Find E.
r
r D
E=
ε0
2 0.5
2
2
0
Ex Hollow Cylinder (5)
Ex.
r
⋅ dS = 2πρ0 Dρ ( ρ 0 )
2π 0.5
∫
ρ0
18
surface
Qenclose = ∫
)
ρ2
EE3301
Constant for the given standard shape
0.5
(ρ
0
2π
r
ρ
v
D( ρ 0 ) = v ( ρ 02 − 0.06
0625
5)a ρ
2ρ0
Ex: Hollow Cylinder (4)
0.5 < ρ0:
0 0.25
1 2π
(
0
r
r
2π 1
r
r
D S ⋅ dS = ∫ ∫ Dρ ( ρ 0 )a ρ ⋅ ρ 0 dzdφa ρ = 2πρ0 Dρ ( ρ 0 )
∫
1
∫ ρv ρdρdϕdz = ρv ∫
− 0.252
dφdz
0 ∫0
2
ρ 2 − 0.252 1 2π
= ρv 0
∫0 ∫0 1dφdz
2
Qenclosed = 0ÆD(ρ0 )= 0
r
r
− az
0.25 < ρ0 ≤ 0.5: + a z
r
r
r
r
r
r
r
r
D
⋅
d
S
=
D
⋅
d
S
+
D
⋅
d
S
+
D
⋅
d
S
S
S
S
S
∫f
∫top r ∫bottom r ∫side
surface
aρ
aρ
0
2π ρ 0
= ρv ∫
0.25
ρ0 ≤ 0.25:
0.25
∫
0.5
surface
0.5
EE3301
1
Qenclose = ∫
19
⎧
r
⎪
0 V/m; ρ ≤ 0.25
⎪
r
r
⎪ ρ
E( ρ ) = ⎨ v ρ 2 − 0.06252 a ρ V/m; 0.25 < ρ ≤ 0.5
⎪ 2ε 0 ρ
0.1875ρ v r
⎪
a ρ V/m; ρ > 0.5
⎪
2ε 0 ρ
⎩
(
EE3301
)
ANS
20
Ex : +Faraday
Ex.:
+Faraday’ss Experiment
Ex + Faraday’s
Ex.
Faraday s Experiment (2)
• Put Q-C charge to the metal sphere of the
p
with
radius r0 m. Cover the metal sphere
another sphere of the inner radius 2r0 m and
the outer radius of 3r0 m.
m The spheres have the
same center and are divided by air. Ground the
outer
t shell
h ll off the
th larger
l
sphere.
h
Find
Fi d D.
D
Drawing:
Air,
3r0
+Q C
r0
2r0
Theory:
Gauss’ Law:
Gauss
r
∫D
S
r
⋅ dS = Qenclosed
surface
r
r
sphere
Shape = sphere so D = Dr ((rr )a r , Gaussian surface = sphere.
Special
p
case:
EE3301
21
r
r
Q
D sphere (r ) = enclosed
ar
2
4πr
EE3301
Ex + Faraday’s
Ex.
Faraday s Experiment (3)
22
Ex + Faraday’s
Ex.
Faraday s Experiment (3)
Solution: (1) Find charge on the outer sphere.
(3) Summary.
Air,
3r0
+Q C
r0
-Q C
⎧ Q r
r
a r C/m 2 ; r0 < r < 2r0
⎪
2
D(r ) = ⎨ 4πr r
⎪⎩
0 C/m 2 ; otherwise
0C
20
2r
ANS
(2) Find D by Gauss’ law.
r
r
∴ D(r ) = 0
r
Q r
r0 < r < 2r0 ; Qenclosed = Q ∴ D(r ) =
a
2 r
r4πr r
r > 2r0 ; Qenclosed = Q + (−Q) = 0 ∴ D(r ) = 0
r < r0 ; Qenclosed = 0
EE3301
23
EE3301
24
Ex: Electric Flux (1)
Ex Electric Flux (2)
Ex.
• A circular disk of radius 3m with a uniform
charge density ρs C/m2 is enclosed by surface
S Find
S.
Fi d a total
t t l nett flux
fl crossing
i S.
S
Drawing:
Φ
3m
ρsdS ρsρdρdφ
Theory: Gauss’
Gauss law
r
r
Φ Total = ∫ D S ⋅ dS = Qenclosed
surface
Φ Total
Solution:
= Qenclosed =
• A uniform line charge of ρl C/m lies along the
z axis. Find the net flux crossing
g the surface
parallel to xz plane and enclosed by y = 0.5m, 0 2 ≤ x ≤ 0.2,
0.2
0 2 -2
2 ≤ z ≤ 2.
2
∫ ρ dS
∫ dS = ρ π (3 ) = 9πρ C
s
ANS
25
x
EE3301
Example: Electric Flux (3)
ρm
0.4m
• A
Arrow showing
h i th
the magnitude
it d andd distance
di t
for
f
each point in vector field
• 2 ways for representing magnitude
(1) Find flux passing through cylinder of ρ-m
radius and 4-m height(-2 ≤ z ≤ 2)
Φ Total = Qenclosed
= ∫ ρ l dl = 4 ρ l C
x
– line density (the higher the number
number, the larger
larger.))
– line length (the longer, the larger)
−2
(2) Find the ratio of flux passing through red
plane located at x = 0.5.
y
• Direction from the arrow direction
• Type of arrow: going in and going out
α = tan −1 (0.2 0.5) = tan −1 (0.4) rad
α
0.5m
0.4m
Φ=
26
Flux Line (Streamlines)
2
4m
y
2
s
EE3301
z
Theory: Gauss’ law
r
r
Φ Total = ∫ D S ⋅ dS = Qenclosed
surface
4m
surface
Solution:
ρm
0.4m
s
surface
= ρs
z
Drawing:
2 ρ tan −1 (0.4)
2α
Φ total = l
C
2π
π
ANS
EE3301
27
EE3301
Going out = Source
Going in = Sink
28
Divergence
Divergence (2)
• Net flux
fl volume
l
flowing
i out from
f
a point
i in
i
vector field Net flux flowingg out of closed surface S
• Flux flowing in = negative
r r
• Maths definition:
r
r
∫ A ⋅ dS
div A = ∇ ⋅ A = lim S
Δv →0
Δv
Point’s volume = 0
Volume enclosed by
yS
r ∂Ax ∂Ay ∂Az
∇⋅A =
+
+
∂x
∂y
∂z
r
r
r
r
• Cylindrical coordinate A = Aρ a ρ + Aφ aφ + Az a z
r 1 ∂ (ρAρ ) 1 ∂Aφ ∂Az
∇⋅A =
+
+
ρ ∂ρ
ρ ∂φ ∂z
r
• Spherical coordinate A = Ar ar r + Aθ ar θ + Aφ ar φ
r 1 ∂ r 2 Ar
1 ∂ (sin θAθ )
1 ∂Aφ
∇⋅A = 2
+
+
r
∂r
r sin θ
∂θ
r sin θ ∂φ
(
• Output of divergence = Scalar value
EE3301
r
r
r
r
• Cartesian coordinate A = Ax a x + Ay a y + Az a z
29
EE3301
Ex Divergence (1)
Ex.
r cos 2 φ r
Find the divergence of D = 3 a r
r
r 1 ∂ (r 2 Dr )
i θDθ )
1 ∂ (sin
1 ∂Dφ
+
+
Sol.: ∇ ⋅ D = 2
r
∂r
∂θ
r sin θ
r sin θ ∂φ
⎛ ⎛ cos 2 φ ⎞ ⎞
∂⎜ r 2 ⎜ 3 ⎟ ⎟
⎜
⎟⎟
1 ⎜⎝ ⎝ r
1 ∂ (sin
i θ (0) )
1 ∂ (0)
⎠⎠
=
+
+
r sin θ
r sin θ ∂φ
∂r
∂θ
r2
2
⎛ ⎛ cos φ ⎞ ⎞
⎟⎟
∂⎜ ⎜
⎜ r ⎟⎟
2
⎜
2
−1
1 ⎝⎝
⎠ ⎠ cos φ ⎛⎜ ∂ (r ) ⎞⎟ = − cos φ ANS
=
=
EE3301
31
r 2 ⎜⎝ ∂r ⎟⎠
∂r
r4
r2
)
30
Divergence Theorem
• Sum the divergence of each point in a volume
Æ total flux out of the volume.
r
Total flux = ∫ ∇ ⋅ Ddv
v
• Total flux = Flux out of the closed surface
r r
enclosed
l d the
th volume
l
Total flux = ∫ D ⋅ dS
S
Note: positive dS going out.
EE3301
32
Divergence Theorem (2)
∫
v
Ex Divergence (2)
Ex.
r
⎛ cos 2 φ r ⎞
cos 2 φ r
Evaluate ∫ ∇ ⋅ ⎜⎜ 3 a r ⎟⎟dv, ∫
a
⋅
d
S
r
v
S
r3
⎝ r
⎠
r
r r
∇ ⋅ Ddv = ∫ D ⋅ dS
S
where v is the volume of the hollow sphere
b
bounded
d d bby r = 1 and
d r = 2 and
d S is
i the
h surface
f
enveloped v.
S: Closed surface enclosing v
D: Volume function in vector field
(
)
r 1 ∂ r 2 Dr
1 ∂ (sin θDθ )
1 ∂Dφ
+
+
Sol.: 1) ∇ ⋅ D = 2
r
∂r
r sin θ
∂θ
r sin θ ∂φ
2
cos φ
=−
r4
EE3301
33
EE3301
Ex. Divergence
g
(2)
( ) ((2))
⎛ cos 2 φ r ⎞
⋅
∇
2) Evaluate ∫v ⎜⎜ 3 a r ⎟⎟dv
⎝ r
⎠
⎛ cos 2 φ r ⎞
⎛ cos 2 φ ⎞
⎟
⎜
∇
⋅
a
dv
=
∫v ⎜⎝ r 3 r ⎟⎠ ∫v ⎜⎜⎝ − r 4 ⎟⎟⎠dv
2π π 2 ⎛ cos 2 φ ⎞ 2
⎟r sin θdrdθdφ
= ∫ ∫ ∫ ⎜−
4 ⎟
0 0 1 ⎜
r
⎝
⎠
2
2π
φ⎞
π ⎛ 1 ⎞ ⎛ sin 2φ
= −(− cos θ ) 0 ⎜ − ⎟ ⎜
+ ⎟
2⎠0
⎝ r ⎠1 ⎝ 4
⎛ 1 ⎞⎛ π ⎞
= −(2)⎜ ⎟⎜ ⎟ = −π
ANS
EE3301
35
⎝ 2 ⎠⎝ 2 ⎠
34
Ex Divergence (2) (3)
Ex.
⎛ cos 2 φ r ⎞
3) E
Evaluate
l t ∫ ⎜⎜ 3 a r ⎟⎟ ⋅ ddSS
r
⎠
S⎝
r
r
⎛ cos 2 φ r ⎞ r
cos 2 φ r
cos 2 φ r
⎟
⎜
a
⋅
d
S
=
a
⋅
d
S
+
a
⋅
d
S
∫S ⎜⎝ r 3 r ⎟⎠
∫inner r 3 r
∫outer r 3 r
r
cos 2 φ r
a
⋅
d
S
∫inner r 3 r
2
2π π cos φ r
r
=∫ ∫
a r ⋅ r 2 sin θdθdφ (− a r )
3
0
0
r
1 2π π
= − ∫ ∫ cos 2 φ sin θdθdφ = −2π
1 0 0
EE3301
36
Ex Divergence (2) (4)
Ex.
Gauss’ Law: Point form
Gauss
r
cos 2 φ r
a
⋅
dS
d
S
∫outer r 3 r
2
2π π cos φ r
r
=∫ ∫
a r ⋅ r 2 sin θdθdφ (a r )
3
0
0
r
1 2π π
= ∫ ∫ cos 2 φ sin θdθdφ = π
2 0 0
r
r
⎛ cos 2 φ r ⎞ r
cos 2 φ r
cos 2 φ r
⎟
⎜
a
⋅
d
S
=
a
⋅
d
S
+
a
⋅
d
S
∫S ⎜⎝ r 3 r ⎟⎠
∫inner r 3 r
∫outer r 3 r
= −2π + π = −π
ANS
r
r
D
⋅
dS
∫ S = Qenclosed
surface
Divergence theorem:
r r
r
D
⋅
d
S
=
∇
⋅
D
∫S
∫v dv = Qenclosed
r
∇
⋅
D
dv = ∫ ρ v dv
∫
v
v
r
∇ ⋅ D = ρ v [C/
[C/m 3 ]
Di
Divergence
off electric
l i flux
fl density:
d i flux
fl source = volume
l
charge
h
density.
d i
EE3301
37
EE3301
38