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Transcript z - KL University
Coordinate Systems
To understand the Electromagnetics, we must know basic vector algebra and
coordinate systems. So let us start the coordinate systems.
COORDINATE SYSTEMS
• RECTANGULAR or Cartesian
• CYLINDRICAL
• SPHERICAL
Choice is based on
symmetry of problem
Examples:
Sheets - RECTANGULAR
Wires/Cables - CYLINDRICAL
Spheres - SPHERICAL
Cylindrical Symmetry
Spherical Symmetry
Visualization (Animation)
Orthogonal Coordinate Systems:
1. Cartesian Coordinates
z
P(x,y,z)
Or
Rectangular Coordinates
P (x, y, z)
y
x
z
z
P(r, Φ, z)
2. Cylindrical Coordinates
P (r, Φ, z)
X=r cos Φ,
Y=r sin Φ,
Z=z
3. Spherical Coordinates
x
y
z
θ r
P (r, θ, Φ)
X=r sin θ cos Φ,
Y=r sin θ sin Φ,
Z=z cos θ
r
Φ
x
Φ
P(r, θ, Φ)
y
z
θ r
x
Φ
z
P(r, θ, Φ)
Cartesian Coordinates
P(x, y, z)
P(x,y,z)
y
x
y
Spherical Coordinates
P(r, θ, Φ)
z
Cylindrical Coordinates
P(r, Φ, z)
z
P(r, Φ, z)
x
Φ
r
y
Cartesian coordinate system
Z
dz
dy
dx
P(x,y,z)
Y
X
• dx, dy, dz are infinitesimal
displacements along X,Y,Z.
• Volume element is given by
dv = dx dy dz
• Area element is
da = dx dy or dy dz or dxdz
• Line element is
dx or dy or dz
Ex: Show that volume of a cube
of edge a is a3.
a
V
a
a
dv
dx dy dz a
v
0
0
0
3
Cartesian Coordinates
Differential quantities:
Length:
d l xˆ dx yˆ dy zˆ dz
Area:
d s x xˆ dydz
d s y yˆ dxdz
d s z zˆ dxdy
Volume:
dv dxdydz
AREA INTEGRALS
• integration over 2 “delta” distances
dy
dx
Example:
y
7
AREA =
6
dy
3
2
6
dx
2
Note that: z = constant
3
7
x
= 16
Cylindrical coordinate system
(r,φ,z)
Z
Z
r
φ
X
Y
Cylindrical
Spherical
polarcoordinate
coordinatesystem
system
(r,φ,z)
Z
dz
r dφ
dr
φ
Y
dφ
r
dr
X
φ is azimuth angle
r dφ
• dr is infinitesimal displacement
along r, r dφ is along φ and
dz is along z direction.
• Volume element is given by
dv = dr r dφ dz
• Limits of integration of r, θ, φ
are
0<r<∞ , 0<z <∞ , o<φ <2π
Ex: Show that Volume of a
Cylinder of radius ‘R’ and
height ‘H’ is π R2H .
Volume of a Cylinder of radius ‘R’
and Height ‘H’
V
dv r dr d dz
v
2
R
H
rdr d dz
0
0
0
R H
2
Try yourself:
1) Surface Area of Cylinder = 2πRH .
2) Base Area of Cylinder (Disc)=πR2.
Cylindrical Coordinates: Visualization of Volume element
Differential quantities:
Length element:
d l aˆ r dr aˆ rd aˆ z dz
Area element:
d s r aˆ r rd dz
d s aˆ drdz
d s z aˆ z rdrd
Volume element:
dv r dr d dz
Limits of integration of r, θ, φ are 0<r<∞ , 0<z <∞ , o<φ <2π
Spherically Symmetric problem
(r,θ,φ)
Z
θ
r
Y
φ
X
Spherical polar coordinate system (r,θ,φ)
• dr is infinitesimal displacement
along r, r dθ is along θ and
r sinθ dφ is along φ direction.
P(r, θ, φ)
Z
• Volume element is given by
dr
dv = dr r dθ r sinθ dφ
P
r cos θ
r dθ
• Limits of integration of r, θ, φ
θ r
are
Y 0<r<∞ , 0<θ <π , o<φ <2π
Ex: Show that Volume of a
r sinθ dφ
φ
r sinθ
sphere of radius R is 4/3 π R3 .
X
θ is zenith angle( starts from +Z reaches up to –Z) ,
φ is azimuth angle (starts from +X direction and lies in x-y plane only)
Volume of a sphere of radius ‘R’
V
dv r
2
dr sin d d
v
R
r
0
2
dr
2
sin d d
0
R
0
3
3
. 2 . 2
4
R
3
3
Try Yourself:
1)Surface area of the sphere= 4πR2 .
Spherical Coordinates: Volume element in space
Points to remember
System
Coordinates
dl1
dl2
dl3
Cartesian
Cylindrical
Spherical
x,y,z
r, φ,z
r,θ, φ
dx
dr
dr
dy
rdφ
rdθ
dz
dz
r sinθdφ
• Volume element : dv = dl1 dl2 dl3
• If Volume charge density ‘ρ’ depends only on ‘r’:
Q
dv 4 r dr
2
v
l
Ex: For Circular plate: NOTE
Area element da=r dr dφ in both the
coordinate systems (because θ=900)
Quiz: Determine
a) Areas S1, S2 and S3.
b) Volume covered by these surfaces.
S3
Z
Solution :
2
h
1
r
ii ) S 2
0
h
dr dz
0
rh
S2
S1
0
2 r
iii ) S 3
dr .rd
1 0
r
2
2
h 2 r
b)V
Radius is r,
Height is h,
1 2
r
a ) i ) S 1 rd dz rh ( 2 1 )
dr .rd .dz
0 1 0
( 2 1 )
r
Y
dφ
2
2
( 2 1 ) h
X
Vector Analysis
• What about A.B=?, AxB=? and AB=?
• Scalar and Vector product:
A.B=ABcosθ
Scalar
or
(Axi+Ayj+Azk).(Bxi+Byj+Bzk)=AxBx+AyBy+AzBz
AxB=ABSinθ n
Vector
n
(Result of cross product is always
perpendicular(normal) to the plane
of A and B
B
A
Gradient, Divergence and Curl
The Del Operator
• Gradient of a scalar function is a
vector quantity.
• Divergence of a vector is a scalar
quantity.
• Curl of a vector is a vector
quantity.
f
. A
xA
Vector
Fundamental theorem for
divergence and curl
• Gauss divergence
theorem:
( .V ) dv V .da
v
s
Conversion of volume integral to surface integral and vice verse.
• Stokes curl theorem
( x V ). da V .dl
s
l
Conversion of surface integral to line integral and vice verse.
Operator in Cartesian Coordinate System
T ˆ T ˆ T ˆ
i
j
k
x
y
z
as
Gradient:
gradT: points the direction of maximum increase of the
function T.
T
Divergence:
Curl:
V
V x
x
V y
y
Vz
z
where
V V x iˆ V y ˆj V z kˆ
V V y V
V z ˆ V y V x ˆ
x
z
ˆ
i
k
V
j
z z
x x
y
y
Operator in Cylindrical Coordinate System
Volume Element:
Gradient:
dv rdrd dz
T
T
r
1
ˆ
r
rV r
1 T ˆ
T
ˆz
r
z
1 V
V
Curl:
1 V V V V
z
rˆ r z
V
z z
r
r
r r
r
Vz
Divergence:
z
V V r rˆ V ˆ V z zˆ
Vr
ˆ 1
rV
r r
ˆz
Operator In Spherical Coordinate System
Gradient :
T
T
r
rˆ
1 T ˆ
1
T ˆ
r
r sin
1 r Vr
2
Divergence:
V
r
Curl:
V
2
r
1
sin V
r sin
sin V V
r sin
1
V r
1
rV
r r
ˆ
1
V
r sin
1 1 V r
rV
rˆ
r sin
r
V V r rˆ V ˆ V ˆ
ˆ
Basic Vector Calculus
(F G ) G F F G
0,
F 0
( F ) ( F ) F
2
Divergence or Gauss’ Theorem
The divergence theorem states that the total outward flux of a
vector field F through the closed surface S is the same as the
volume integral of the divergence of F.
F dV F d S
V
S
Closed surface S, volume V,
outward pointing normal
Stokes’ Theorem
Stokes’s theorem states that the circulation of a vector field F around a
closed path L is equal to the surface integral of the curl of F over the
open surface S bounded by L
F d S F d l
S
d S n dS
Oriented boundary L
L
n