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, , )

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Review (2)

Coulomb’s law

Coulomb’s force
F 12

Q1Q2

a
2 12
4 0 R12
electric field intensity (V/m)
F 12
E1 
Q2
E
Q
4 0 R
2
aR
3
Review (3)

Key variables:
Coordinate system and its corresponding
differential element
 charge Q


a unit vector
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Outline

Electric field intensity in different charge configurations
 infinite line charge
 ring charge
 surface charge

Examples from previous lecture

Electric flux density
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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)
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Infinite length line of charge

E 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  E a p  E z a z
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Infinite length line of charge

Consider another segment at distance z, z components are
cancelled out, we then have
E  E a p
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Infinite length line of charge

From
E
Q
4 0 R
2
aR
We can write
dE 
dQ
a
2 R
4 0 R
Total field
E
dQ
a
2 R
4 0 R
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Infinite length line of charge

Consider each segment
dQ   L dz
R   a   za z
R  a   za z
aR  
R
 2  z2

Ez components are cancelled due to symmetry.
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Infinite length line of charge
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Ring of charge
determine E at (0,0,h)
d E  cancels each other
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Ring of charge
Consider each segment:
dQ   L dL
R  aa   ha z
aR 
R aa   ha z

R
a 2  h2
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Surface charge

Surface charge density S (c/m2)
dQ = Sdxdy
E  Ex a x  E y a y  Ez a z
Since this is an infinite place, Ex
and Ey components are
cancelled due to symmetry.
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Surface charge
Consider each segment:
Devide the whole area into infinite length of line charges

 L   S dy
L
 dE 
a
2 0 
Integrate over length y to get total electric field.
Convert the radial component into cylindrical coordinates
 a    ya y  ha z
Ey components are cancelled out due to symmetry.
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Surface charge
No dependence on a distance from the sheet
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Concentrate ring (alternative approach)
z
h

( S d  )  ha z
dE 
2 0 (  2  h2 )3/ 2
for each ring
Total field is integrated from  = 0 to 
Then
 S ha z 
d 
E
2 0 0 (  2  h 2 )3 / 2
S
E
az.
2 0
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Volume charge

Volume charge density V (c/m3)


plasma
doped semiconductor
V dV
E
a
2 R
4 0 R

Complicate derivation due to so many differential elements and
vectors.
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Ex1 Determine the distance between point P
(5, /2, 10) and point Q (1, /3, 5) in
cylindrical coordinates.
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Ex2 Determine a unit vector directed from (0,
0, h) to (r, , 0) in cylindrical coordinates.
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Ex3 Determine a unit vector from any point on
z = -5 plane to the origin.
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Ex4 Find the area between 90    135 on
the surface of a sphere of a radius 1 m.
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Ex5 A charge Q1 = 0.3 C is located at (1,4,0).
A charge Q2 = 0.2 C is located at (3,0,0).
Determine E at point (0,0,5).
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Ex6 Determine E at point (-2, -1, 4) given a
line charge located at x = 2 and y = -4 with a
charge density L = 20 nC/m.
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Ex7 Determine E at the origin given a square sheet
of charge located at z = -1 plane. The sheet is
extended from -1  x  1 and -1  y  1with a surface
charge density S = 2(x2+y2+1)3/2 nC/m2.
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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.
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Electric flux density

Electric flux density, (C/m2)

D
a
2 r
4r
Note:  (chi) is a flux in Coulomb unit and is equal to charge Q on
the sphere
Q
E 
a
2 r
40 r
So we have
D  0 E
where 0 = 8.854x10-12 Farad/m
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Electric flux density

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
  D S cos 
Note: S = surface vector

Dot product:
A B  A B cos  AB
A B  Ax Bx  Ay By  Az Bz for Cartesian coordinates.
Dot product is a projection of A on B multiplies by B28
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)

In case the flux is varied over the surface,
   D d S.
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Ex8 D  2a  1a C/m2. Given the surface
defined by  = 1 m, 0    90 and -1  z  1,
calculate the flux through the surface.
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Ex9 A charge Q = 30 nC is located at the origin,
determine the electric flux density at point (1, 2, 4) m.
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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 D  xa x  (1  y)a y  za z C/m2.
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