ENE 325Electromagnetic Fields and Waves

Lecture 3 Gauss’s law and applications, Divergence, and Point Form of Gauss’s law

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Review (1)Review (1)

Coulomb’s law Coulomb’s force

electric field intensity

or V/m

1 212 122

0 124

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R

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Q

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QE a

R

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Review (2) Electric field intensity in different charge configurations

infinite line charge

ring charge

surface charge

02LE a

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2 2 3/ 2

02 ( )L

zah

E aa h

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02S

zE a

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Outline

Gauss ’s law and applications Divergence and point form of Gauss’s law

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Gauss’s law and applications

“ The net electric flux t hrough any closed surf

ace is equal to the tota l charge enclosed by t

hat surface”. If we completely enclo

se a charge, then the n et flux passing throug

h the enclosing surfac e must be equal to the

charge enclosed, Qenc.

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Gauss’s law and applications

The integral form of Gauss’s law:

Gauss’s law is useful in finding the fields for problems t hat have a high degree of symmetry by following these

steps: Determine what variables influence and what components

of are present. Select an enclosing surface, Gaussian surface , whose surface

vector is directed outward from the enclosed volume and i s everywhere either tangential to or normal to

encD dS Q ����������������������������

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Gauss’s law and applications

The enclosing surface must be selected in order for to be constant and to be able to pull it out of the integral.

D dS D dS ����������������������������

D��������������

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Ex1 Determine from a charge Q located at the origin by using

Gauss’s law. 1 .

2. Select a Gaussian surface

3. Dr at a fixed distance is constant and normal to a Gaussian surface,

can be pulled out from the integral .

D��������������

( ) rrD D r a��������������

2 sin rdS r d d a ��������������

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Ex2 Find at any point P ( , , z ) from an infinite length line of charg

e density L -on the z axis.1 . From symmetry,2. Select a Gaussian surface with ra

dius and length h.3 . D at a fixed distance is constant a

nd normal to a Gaussian surface, can be pulled out from the integra

l. ant and normal to a Gaussian su rface, can be pulled out from the i

ntegral.

D��������������

D D a��������������

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3Ex A parallel plate capacitor has surfac e charge +S located underneath a top pl

-ate and surface charge S located on a b ottom plate. Use Gauss’ s law to find a

nd between plates. D��������������

E��������������

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Ex4 Determine electric flux density for a coaxial cable.

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Ex5 A point charge of 0.25 C is located at r = 0 and uniform surface charge densities are located as follows: 2 mC/m2 - at r = 1 cm and 0.6 mC/m2 at r = 1.8 c

m. Calculate at D��������������

a) r = 0.5 cm

b) r = 1.5 cm

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c) r = 2.5 cm

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EE6 A uniform volume charge density of 80 C/m3 is present throughout region 8 mm < r < 10 mm. Let v = 0 for 0 < r < 8 mm,

a)Find the total charge inside the spherical surface r = 10 mm

b)Find Dr at r = 10 mm.

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c) If there is no charge for r > 10 mm, find Dr at r = 20 mm

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Divergence and Point form of Gauss’s law (1) Divergence of a vector field at a particular point i

n space is a spatial derivative of the field indicatin g to what degree the field emanates from the poin t. Divergence is a scalar quantity that implies whe

ther the point source contains a source or a sink o f field.

0limv

D dSdiv D

v

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where = volume differential element

v

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Divergence and Point form of Gauss’s law (2)

or we can write in derivative form as

.yx zDD D

Dx y z

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Del operator: x y z

It is apparent that0

lim encv

v

Qdiv D

v

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therefore we can write a differential or a point form of Gauss’s law as

vD ��������������

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Divergence and Point form of Gauss’s law (3)

For a cylindrical coordinate:

1 1 zD D

D Dz

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For a spherical coordinate:

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1 1 1sin

sin sinr

DD r D D

r r rr

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positive indicates a source of flux. (positive charge) negative indicates a sink of flux. (negative charge)

Physical example

The plunger moves up and down indicating net movement of molecules out and in,respectively.

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.D dS Ddv ������������������������������������������

An integral form of Gauss’s law can also be written as

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Ex7 Let . Determine

2 2 ��������������

x y zA x ya xyza z a

.A��������������

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EE8 Let in a cylindrical coordin ate system. Determine both terms of the di

vergence theorem for a volume enclosed by r = 1 m, r = 2 m, z = 0 m, and z = 10 m.

310 / 4 rD r a��������������

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