ENE 325 Electromagnetic Fields and Waves

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

�� QQF a

R

121

2��

�� FE

Q

2

04

��R

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

��

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


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


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

��

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

��

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��


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, (C/m2)

Note: (chi) is a flux in Coulomb unit and is equal to charge Q on the sphere

24

rD ar

��

2

04r

QE a

r

��

So we have 0D E

��

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


In case the flux is varied over the surface,


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.

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ENE 325 Electromagnetic Fields and Waves