Date post: | 22-Dec-2015 |
Category: |
Documents |
Upload: | blanche-crawford |
View: | 227 times |
Download: | 0 times |
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 Coordinates
Or
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 dz
Ex: Show that volume of a cube of edge a is a3.
P(x,y,z)
X
Y
Z
3
000
adzdydxdvVaa
v
a
dxdy
dz
Cartesian Coordinates
Differential quantities:
Length:
Area:
Volume:
dzzdyydxxld ˆˆˆ
dxdyzsd
dxdzysd
dydzxsd
z
y
x
ˆ
ˆ
ˆ
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
r
φ
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
drdzasd
dzrdasd
zz
rr
ˆ
ˆ
ˆ
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
3
42.2.
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 remember
System 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
hr
dzrddrVb
rrddrSiii
rhdzdrSii
rhdzrdSia
Solution
h r
r
r h
h
)(2
..)
)(2
.3)
2)
)(1))
:
12
2
0 0
12
2
0
0 0
12
0
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 plane
of 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
xA
A
.
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
kz
Tj
y
Tix
TT ˆˆˆ
y zxV VV
Vx y z
ky
V
x
Vj
x
V
z
Vi
z
V
y
VV xyzxyz ˆˆˆ
kVjViVV zyxˆˆˆ
where
as
Operator in Cylindrical Coordinate System
Volume Element:
Gradient:
Divergence:
Curl:
dzrdrddv
zz
TˆT
rr
r
TT
1
1 1 zr
V VV rV
r r r z
zV
rVrr
ˆr
V
z
Vr
z
VV
rV rzrz
11
zVVrVV zr ˆˆˆ
Operator In Spherical Coordinate System
Gradient :
Divergence:
Curl:
ˆT
sinrˆT
rr
r
TT
11
2
2
sin1 1 1
sin sinrr V VV
Vr r r r
ˆVrV
rr
ˆrVr
V
sinrr
VVsin
sinrV
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