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Coordinate Systems. Choice is based on symmetry of problem. 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. Examples:. - PowerPoint PPT Presentation
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Coordinate Systems
COORDINATE SYSTEMS• RECTANGULAR or Cartesian
• CYLINDRICAL
• SPHERICAL
Choice is based on symmetry of problem
Examples:Sheets - RECTANGULAR
Wires/Cables - CYLINDRICAL
Spheres - SPHERICAL
To understand the Electromagnetics, we must know basic vector algebra and coordinate systems. So let us start the coordinate systems.
Cylindrical Symmetry Spherical Symmetry
Visualization (Animation)
Orthogonal Coordinate Systems:
3. Spherical Coordinates
2. Cylindrical Coordinates
1. Cartesian Coordinates
P (x, y, z)
P (r, θ, Φ)
P (r, Φ, z)
x
y
zP(x,y,z)
Φ
z
rx y
z
P(r, Φ, z)
θ
Φ
r
z
yx
P(r, θ, Φ)
Rectangular CoordinatesOr
X=r cos Φ,Y=r sin Φ,Z=z
X=r sin θ cos Φ,Y=r sin θ sin Φ,Z=z cos θ
Cartesian CoordinatesP(x, y, z)
Spherical CoordinatesP(r, θ, Φ)
Cylindrical CoordinatesP(r, Φ, z)
x
y
zP(x,y,z)
Φ
z
rx y
z
P(r, Φ, z)
θ
Φ
r
z
yx
P(r, θ, Φ)
Cartesian coordinate system• 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 dzEx: Show that volume of a cube
of edge a is a3.
P(x,y,z)
X
Y
Z
3
000adzdydxdvV
aa
v
a
dxdy
dz
Cartesian Coordinates
Differential quantities:
Length:
Area:
Volume:
dzzdyydxxld ˆˆˆ
dxdyzsd
dxdzysddydzxsd
z
y
x
ˆ
ˆˆ
dxdydzdv
AREA INTEGRALS
• integration over 2 “delta” distances
dx
dy
Example:
x
y
2
6
3 7
AREA = 7
3
6
2
dxdy = 16
Note that: z = constant
Cylindrical coordinate system (r,φ,z)
X
Y
Z
rφ
Z
Spherical polar coordinate system
• 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 .
φ is azimuth angle
Cylindrical coordinate system (r,φ,z)
X
Y
Z
rφ
r dφ
dz
dr
r dφ
dr
dφ
Volume of a Cylinder of radius ‘R’ and Height ‘H’
HR
dzdrdr
dzddrrdvV
R H
v
2
0
2
0 0
Try yourself: 1) Surface Area of Cylinder = 2πRH . 2) Base Area of Cylinder (Disc)=πR2.
Differential quantities:
Length element:
Area element:
Volume element:
dzardadrald zr ˆˆˆ
rdrdasd
drdzasddzrdasd
zz
rr
ˆ
ˆˆ
dzddrrdv
Limits of integration of r, θ, φ are 0<r<∞ , 0<z <∞ , o<φ <2π
Cylindrical Coordinates: Visualization of Volume element
Spherically Symmetric problem (r,θ,φ)
X
Y
Z
r
φ
θ
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.
• Volume element is given by dv = dr r dθ r sinθ dφ
• Limits of integration of r, θ, φ are
0<r<∞ , 0<θ <π , o<φ <2πEx: Show that Volume of a
sphere of radius R is 4/3 π R3 .
P(r, θ, φ)
X
Y
Z
r
φ
θ
drP
r dθ
r sinθ dφ
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)
r cos θ
r sinθ
Volume of a sphere of radius ‘R’
33
0 0
2
0
2
2
342.2.
3
sin
sin
RR
dddrr
dddrrdvV
R
v
Try Yourself:1)Surface area of the sphere= 4πR2 .
Spherical Coordinates: Volume element in space
Points to rememberSystem Coordinates dl1 dl2 dl3
Cartesian x,y,z dx dy dzCylindrical r, φ,z dr rdφ dzSpherical r,θ, φ dr rdθ r sinθdφ
• Volume element : dv = dl1 dl2 dl3• If Volume charge density ‘ρ’ depends only on ‘r’:
Ex: For Circular plate: NOTEArea element da=r dr dφ in both the coordinate systems (because θ=900)
drrdvQv l 24
Quiz: Determine a) Areas S1, S2 and S3.b) Volume covered by these surfaces.
Radius is r,Height is h,
X
Y
Z
r
dφ
S1S2
S3
21
hrdzrddrVb
rrddrSiii
rhdzdrSii
rhdzrdSia
Solution
h r
r
r h
h
)(2
..)
)(2
.3)
2)
)(1))
:
12
2
0 0
12
2
0
0 0
120
2
1
2
1
2
1
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(Result of cross product is always perpendicular(normal) to the planeof A and B
A
B
n
Gradient, Divergence and Curl
• 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 Vector
xAA
.
The Del Operator
Fundamental theorem for divergence and curl
• Gauss divergence theorem:
• Stokes curl theorem
v s
daVdvV .).(
s l
dlVdaVx .).(
Conversion of volume integral to surface integral and vice verse.
Conversion of surface integral to line integral and vice verse.
Gradient:gradT: points the direction of maximum increase of the function T.
Divergence:
Curl:
Operator in Cartesian Coordinate System
kzTj
yTi
xTT ˆˆˆ
y zxV VV
Vx y z
kyV
xV
jxV
zVi
zV
yVV xyzxyz ˆˆˆ
kVjViVV zyxˆˆˆ
where
as
Operator in Cylindrical Coordinate System
Volume Element:
Gradient:
Divergence:
Curl:
dzrdrddv
zzTˆT
rr
rTT
1
1 1 zr
V VV rV
r r r z
zVrVrr
ˆrV
zVr
zVV
rV rzrz
11
zVVrVV zr ˆˆˆ
Operator In Spherical Coordinate System
Gradient :
Divergence:
Curl:
ˆTsinr
ˆTr
rrTT
11
2
2
sin1 1 1sin sin
rr V VVV
r r r r
ˆVrV
rr
ˆrVr
Vsinr
rV
Vsinsinr
V
r
r
1
111
ˆˆˆ VVrVV r
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.
Closed surface S, volume V, outward pointing normal
Basic Vector Calculus
2
( )
0, 0
( ) ( )
F G G F F G
F
F F F
Divergence or Gauss’ Theorem
SV
SdFdVF
dSnSd
Oriented boundary L
n
Stokes’ Theorem
S L
ldFSdF
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