ENE 325 Electromagnetic Fields and Waves

ENE 325ENE 325Electromagnetic Electromagnetic Fields and WavesFields and Waves

Lecture 6 Lecture 6 Capacitance and Capacitance and MagnetostaticsMagnetostatics

Review (1)Review (1)

Conductor and boundary conditions tangential electric field, Et , 0= for equipotenti

al surface. normal electric flux density Dn = s .

Dielectric macroscopic electric dipoles and bound charges Polarization is dipole moment per unit volume,

electric flux density in dielectric medium,

0 1

1lim

n v

iv i

P pv

��

0D E P ��

Review (2)Review (2)

Dielectric and boundary conditions tangentialel ect r i c fi el d, Et1 = Et2 normal electric flux density .

1 221 ( ) Sa D D

��

OutlineOutline

Capacitance Static magnetic fields

Bio-Savart ‘s law Magnetic field in different current

configurations

CapacitanceCapacitance

Capacitancedependson t he shape of conduct or and t he permi ttivity of the medium. Capacitance has a u

nit of Farad or F.Q

CV

Q E dS ��

V E dl

��

.

��

��E dS

C FE dl

From

then

Capacitance Capacitance for parallel plate for parallel plate configurationconfiguration

At lower plate,

then

then

- - - - - - - - - - - - - - - - - - - - - - -

+ + + + + + + + + + + + + + + +

E

-s

s

z = 0

z = d zsD a

��

szE a

��

0lowers s

upper dV E dl dz d

��

.

A

C Faradd

The potential difference

Let A = the plate area then Q = sA

Total energy stored in the Total energy stored in the capacitancecapacitance

21

2Ev

W E dv

22 21 1 1

/2 2 2

Q

CV QV J mC

Ex1Ex1 Determine the relative Determine the relative permittivity of the dielectric permittivity of the dielectric material inserted between a material inserted between a parallel plate capacitor ifparallel plate capacitor if

a) C = 40 nF, d = 0.1 mm, and A = 0.15 m2

b) d = 0.2 mm, E = 500 kV/m, and s = 10 C/m2

Capacitance in various charge Capacitance in various charge distribution configurationsdistribution configurations (1)(1) Coaxial cable

ab

Use Gauss’s law, D dS Q ��

02

ln( )

ab

lQ QC Farad

bV Va

Capacitance in various charge Capacitance in various charge distribution configurationsdistribution configurations (2)(2) Sphere

Use Gauss’s law, D dS Q ��

a

b

041 1

C Farad

a b

Capacitance in various charge Capacitance in various charge distribution configurationsdistribution configurations (3)(3)

A parallel plate capacitor with horizontal A parallel plate capacitor with horizontal dielectric layersdielectric layers

d1

d2

1

2

A

+

-

V1

V2

+-

Capacitance in various charge Capacitance in various charge distribution configurationsdistribution configurations (4)(4)

A parallel plate capacitor with vertical dielectric layers

d 1 2

A

V

-

Ex5Ex5 From the parallel From the parallel capacitor shown, Find the total capacitor shown, Find the total capacitance.capacitance.

d3 = 0.4 mm

A=20 cm2

r1 1

r2 2

r3 3

d2 = 0.3 mm

d1 = 0.2 mm

Static magnetic fields Static magnetic fields (Magnetostatics)(Magnetostatics)

IntroductionIntroduction (1)(1) source of the steady magnetic field may be a per

manent magnet, and electric field changing linear ly with time or a direct current.

a schematic view of a bar magnet showing the m agnetic field. Magnetic flux lines begin and termi

nate at the same location, more like circulation.

IntroductionIntroduction (2)(2) Magnet i c nort h and sout h pol es are al w

ays together.

N

S

N

S

N

S

N

S

N

S

N

S

IntroductionIntroduction (3)(3) Oersted’s experiment shows th at current produ

ces magnetic fields that loop around the conducto r. The field grows weaker as one compass moves a

way from the source of the current.

- Bi o Savart l aw- Bi o Savart l aw(1)(1) - The law of Bio Savart states that at any point P the magn

itude of the magnetic field intensity produced by the differ ential element is proportional to the product of the current

, the magnitude of the differential length, and the sine of t he angle lying between the filament and a line connecting the filament to the point P at which the filed is desired.

The magnitude of the magnetic field intensity is inversely proportional to the square of the distance from the differe

ntial element to the point P.

- Bi o Savart l aw- Bi o Savart l aw(2)(2) The direction of the magnetic field intensity is n

ormal to the plane containing the differential fila ment and the line drawn from the filament to the point P.

- BioSavartlawi s a met hod t o det ermi ne t he magnet i c fi el d intensity. It is an analogy to Coulomb’s law of Elec trostatics.

- Bi o Savart l aw- Bi o Savart l aw(3)(3)

2 34 4

rId L a Id L RdH

r r

��

��

from this picture:

1 121

2 2124

I d L adH

R

��

��

Total field A/m 24

rId L aH

r

��

Magnetic field intensity resulting fr Magnetic field intensity resulting fr om om an infinite length line of current an infinite length line of current

(1)(1)

Pick an observation point P located on axis.

The current

The vector from the source to the test point is

zId L Idza

��

R zRa za a

a unit vector

2 2

zR

za aa

z

Magnetic field intensity resulting fr Magnetic field intensity resulting fr om om an infinite length line of current an infinite length line of current

(2)(2)

then

From a table of Integral,

3/ 22 24

z zIdza za aH

z

��

3/ 22 2.

4

�� I a dzH

z

3/ 2 2 2 22 2

dx x

a x ax a

then

2 2 24

I a zH

z

�� /

2

I aA m

Magnetic field intensity resulti Magnetic field intensity resulti ng from ng from a ring of current a ring of current (1)(1)

A ring is located on z = 0 plane with the radius a. The observation point is at z = h.


Id L Iad a��

R zRa ha aa

2 2

zR

ha aaa

h a

A unit vector


Consider a symmetry 2zId L R had a a d a

��

components are cancelled out due to symmetry of two segments on the opposite sides of the ring.

Therefore from 24

rId L aH

r

��

we have

2

3/ 22 20 4

zIad a ha aaH

h a

��


then

2 2

3/ 22 2 0.

4

zIa aH d

h a

��

We finally get

2

3/ 22 22

��z

IaH a

h a

Magnetic field intensity resulti Magnetic field intensity resulti ng from ng from a rectangular loop of c a rectangular loop of c

urrenturrent (1)(1)

Find the magnetic field intensity at the origin. By symmetry, there will be equal magnetic field intensity at each half width (w/2).


urrenturrent (2)(2)Consider 0 x w/2, y = -w/2

xId L Idxa

��

( / 2)R x yRa xa w a

A unit vector

2 2

( / 2)

( / 2)

x yR

xa w aa

x w

We have

3/ 22 2

( / 2)

4 ( / 2)

x x yIdxa xa w adH

x w

��


urrenturrent (3)(3)

3/ 22 2

( / 2)

4 ( / 2)

zIdx w a

x w

Then the total magnetic field at the origin is

/ 2

3 / 22 20

( / 2)8

4 ( / 2)

wzIdx w a

Hx w

��

Look up the table of integral, we find 3/ 2 2 2 22 2

dx x

a x ax a

then A/m. 2 2z

IH a

W

��

Bio-Savart law in different Bio-Savart law in different formsforms

We can express - Bio Savart law in terms of surface and volume current densities by re placing with and :

Id L��

KdS��

Jdv��

24

rKdS aH

r

��

where K = surface current density (A/m) I = K x width of the current sheet

and 24

rJdv aH

r

��

Ri ght handrul e Ri ght handrul e

Themethodtodet ermi ne t he resul t of cross product(x).

F qv B ��

ENE 325 Electromagnetic Fields and Waves

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