ene 325 electromagnetic fields and waves
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ENE 325 Electromagnetic Fields and Waves. Lecture 2 Static Electric Fields and Electric Flux density. Review (1). Vector quantity Magnitude Direction Coordinate systems Cartesian coordinates (x, y, z) Cylindrical coordinates (r, , z) Spherical coordinates (r, , ). Review (2). - PowerPoint PPT PresentationTRANSCRIPT
ENE 325Electromagnetic Fields and Waves
Lecture 2 Static Electric Fields and Electric Flux density
Review (1)
Vector quantityMagnitudeDirection
Coordinate systemsCartesian coordinates (x, y, z) Cylindrical coordinates (r, , z)Spherical coordinates (r, , )
Review (2) Coulomb’s law
Coulomb’s force
electric field intensity (V/m)
1 212 122
0 124
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R
121
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�������������� FE
Q
2
04
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QE a
R
Review (3) Key variables:
Coordinate system and its corresponding differential element
charge Q a unit vector
Outline Electric field intensity in different charge configurations
infinite line charge ring charge surface charge
Examples from previous lecture
Electric flux density
Infinite length line of charge The derivation of and electric field at any point in space
resulting from an infinite length line of charge. (good approximation)
Infinite length line of charge only varies with the radial distance select point P on - z axis for convenience. select a segment of charge dQ at distance –z, we then
have
��������������E
��������������
p zzE E a E a
Infinite length line of charge Consider another segment at distance z, z components
are cancelled out, we then have
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pE E a
Infinite length line of charge
From
We can write
Total field
2
04
��������������R
QE a
R
2
04
��������������R
dQdE a
R
2
04
��������������R
dQE a
R
Infinite length line of charge
Consider each segment
Ez components are cancelled due to symmetry.
2 2
��������������
��������������
L
z
zR
dQ dz
R a za
R a zaa
R z
Infinite length line of charge
Ring of charge
determine at (0,0,h)
cancels each other
��������������E
��������������dE
Ring of charge
Consider each segment:
2 2
��������������
��������������
L
z
zR
dQ dL
R aa ha
R aa haa
R a h
Surface charge
Surface charge density S (c/m2)
dQ = Sdxdy
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x y zx y zE E a E a E a
Since this is an infinite place, Ex and Ey components are cancelled due to symmetry.
Surface charge Consider each segment:
Devide the whole area into infinite length of line charges
02
��������������
L S
L
dy
dE a
Integrate over length y to get total electric field. Convert the radial component into cylindrical coordinates
y za ya ha
Ey components are cancelled out due to symmetry.
Surface charge
No dependence on a distance from the sheet
Concentrate ring (alternative approach)
Total field is integrated from = 0 to
2 2 3/ 2
0
( )
2 ( )
�������������� zSd hadE
h
for each ring
Then
2 2 3/ 20 0
0
2 ( )
.2
zS
Sz
ha dE
h
E a
��������������
��������������
h
z
Volume charge
Volume charge density V (c/m3) plasma doped semiconductor
Complicate derivation due to so many differential elements and vectors.
2
04
��������������V V
Rd
E aR
Ex1 Determine the distance between point P (5, 3/2, 0) and point Q (5, /2, 10) in cylindrical coordinates.
Ex2 Determine a unit vector directed from
(0, 0, h) to (r, , 0) in cylindrical coordinates.
Ex3 Determine a unit vector from any point on z = -5 plane to the origin.
Ex4 Find the area between on the surface of a sphere of a radius a. Given
= 0 and = .
Ex5 A charge Q1 = 0.35 C is located at (0, 4, 0). A charge Q2 = -0.55 C is located at (3, 0, 0). Determine at point (0, 0, 5).E
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Ex6 Determine at point (-2, -1, 4) given a line charge located at x = 2 and y = -4 with a charge density L = 20 nC/m.
E��������������
Ex7 Determine at the origin given a square sheet of charge located at z = -3 plane. The sheet is extended from -2 x 2 and -2 y 2 with a
surface charge density S = 2(x2+y2+9)3/2 nC/m2.
E��������������
Electric flux density
Negative charges are drawn to the outer sphere Electric flux lines are radially directed away from inner sphere to outer sphere or begin from positive charges +Q and
terminate on negative charges -Q.
Electric flux density
Electric flux density, (C/m2)
Note: (chi) is a flux in Coulomb unit and is equal to charge Q on the sphere
24
rD ar
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2
04r
QE a
r
��������������
So we have 0D E
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where 0 = 8.854x10-12 Farad/m
The amount of flux passing through a surface is
given by the product of and the amount of surface normal to. Same polarity charges repel one another
Note: = surface vector
Dot product:
cosD S ����������������������������
S��������������
cos ABA B A B ��������������������������������������������������������
x x y y z zA B A B A B A B ���������������������������� for Cartesian coordinates.
Dot product is a projection of A on B multiplies by B
Electric flux density
In case the flux is varied over the surface,
Electric flux density
The flux through a surface that is an angle to the direction of flux a) is less than the flux through an equivalent surface normal to the direction of flux b)
.D dS ����������������������������
Ex8 C/m2. Given the surface defined by = 6 m, 0 90 and -2 z
2, calculate the flux through the surface.
10 5D a a ��������������
Ex9 A charge Q = 30 nC is located at the origin, determine the electric flux density at point (1, 3, -4) m.
Ex10 Determine the flux through the area 1x1 mm2 on a surface of a cylinder at r = 10 m, z = 2 m, = 53.2 given 2 2(1 ) 4x y zD xa y a za ��������������
C/m2.