1

EELE 3331 – Electromagnetic I

Chapter 4

Electrostatic fields

Islamic University of Gaza

Electrical Engineering Department

Dr. Talal Skaik

2012

encD S, Q

D S

Applying divergence theorem: D S D

D

Gauss Law - Integ

[

ral Form

Gauss Law - Differential For

Fir

m

st Max

enc

v

S v

v

S v

S v

v

Q

d dv

Q d dv

d dv

well's Equation] 2

Gauss Law – Maxwell’s Equation Gauss law states that the total electric flux Ψ through any closed

surface is equal to the total charge enclosed by that surface.

encQ

3

Gauss Law

Illustration of Gauss’s law: flux leaving v1 is 5 nC and that leaving v2 is 0 C.

4

Gauss Law

Gauss law provides an easy means of finding E or D for symmetrical

charge distribution such as point charge, an infinite line charge, a

spherical distribution of a charge.

Applications of Gauss Law

If there is symmetric charge distribution, we construct a mathematical

closed surface (known as Gaussian surface).

→D should be normal or tangential to the Gaussian surface.

Normal → D.dS=Dds (cos 0=1)

Tangential→ D.dS=0 (cos 90=0)

Gaussian surface is chosen so that D is constant in the parts of the

surface for which D is normal to the surface. Therefore we can bring D

outside the integration sign→ easier solution

5

Gauss Law

encQ

= Apply on Gaussian surface.

For point charge enclosed inside Gaussian surface.

Total charge enclosed inside Gaussian surface

Li for cn he

S

enc

enc L

D dS

Q Q Q

Q dl

2

arge distribution with denisty C/m.

Total charge enclosed inside Gaussian surface

for charge distribution with denisty C/m .

Total cha

Sur

rge enc

face

L

enc S

S

enc v

Q dS

Q dv

3

losed inside Gaussian surface

for charge distribution with denisty C/m .Volume v

2

2

0 0

2

2 of

Sphere

D S S

sin

[ not function of or ]

4 D= a4

r

S S

r

r

r r

Area

Q d D d

D r d d

D

QQ D r

r

6

A. Point Charge

Q is located at the origin. To determine D:

→choose Gaussian surface to be a spherical surface containing P (to

satisfy symmetry conditions).

→D is everywhere normal to the surface →D=Drar.

00

D S, D= a

D S D S D S

[D is tangential to the top and

bottom surfaces] D dS=0

S S= 2

2

enc

S

sides top bottom

L

Q d D

Q d d d

Q D d D d D l

l D

D a2

L

l

7

B. Infinite Line Charge

Infinite line of uniform charge ρL C/m along the z-axis.

ρLl is the total charge enclosed inside Gaussian surface.

0

D is normal to the sheet

D= a

D S

D S

D S D S

z z

S

sides

top bottom

D

Q d

Q d

d d

8

C. Infinite Sheet of Charge

Consider an infinite sheet of uniform charge ρS C/m2 lying on the z=0

plane.

Gaussian surface: choose rectangular box.

0

0 0

D S D S D S

=

Assume

[ ]

D a 2

DE = a

2

sides top bottom

S z z

S top bottom

S z

Sz

Sz

Q d d d

dS D dS D dS

dS A

A D A A

9

Infinite Sheet of Charge

10

D. Uniformly Charged Sphere

Consider a sphere of radius a with a uniform charge ρ0 C/m3.

Gaussian surface: choose a spherical surface.

Gaussian surface for a uniformly charged sphere when (a) r a and (b) r a.

2

2 3

0 0 0

0 0 0volume of sphere of radius r

2

2

0 0

2

area of sphere

2 3

0

0

For r a

4= sin

3

D S S sin

4

4 4

3

D a 03

r

enc V

v v r

r r

S S

r

enc r

r

Q dv dv r dr d d r

d D d D r d d

D r

Q D r r

r

<r a11

Uniformly Charged Sphere

Total charge enclosed inside Gaussian surface.

Apply on

Gaussian

Surface

2

2 3

0 0 0

0 0 0volume of sphere of radius a

2

2

0 0

2

area of sphere

2 3

0

3

02

For r a

4= sin

3

D S S sin

4

4 4

3

D a 3

a

enc v

v v r

r r

S S

r

enc r

r

Q dv dv r dr d d a

d D d D r d d

D r

Q D r a

a

r

r a12

Uniformly Charged Sphere

0

3

02

a 0<r a3

D=

a r a3

r

r

r

a

r

13

Uniformly Charged Sphere

2

3

2

2 2 1

2 2

2 0 0

D= cos

At 1, / 4,3 (1)cos( / 4) 0.5 C/m

To find the total charge enclosed by the cylinder:-

Method 1: Q= cos

cos 4( )(1 / 3) 4 / 3 C

zV

V

V

v

D

dz

dv d d dz

Q dz d d

2

1 cos2 1 sin 2cos

2 2 2d d

14

Example 4.8

D=zρcos2φ az C/m2, calculate the charge density at (1,π/4,3) and the

total charge enclosed by the cylinder of radius 1m with -2≤z ≤2 m.

1 2

2

z

0 0

1 2

2 2

0 0

Method 2: Use Gauss law

Q= = D S D S=

Since D has not component along a , 0

For : S= a , z=2 cos

2 cos 2(1 / 3) 2 / 3

F

S t b

S s t b

S

t t

t

d d

d d d z d d

d d

1 2

2

z

0 0

1 2

2 2

0 0

or : S= a , z= 2 cos

z= 2, 2 cos 2 / 3

0 2 / 3 2 / 3 4 / 3 C

b b

t

d d d z d d

d d

Q

15

Example 4.8 - continued

2

2 200

0 0 0

30

0

0

4 22 0 0

0

0

D S

for r<R

4 sin

(2 )( cos )

4 (2 )(2) E a4 4

enc v

S v

r

enc r

r

r r

Q d dv

rQ E r r d d dr

R

r drR

r rE r

R R

16

Example 4.9 A charge distribution with spherical symmetry has density:

, Determine E everywhere.

0 0

0 v

rr R

R

r R

0

2

2 2

0

0 0 0

2 2

2 20

0 0 0 0 0

2

0 0

0

0

for r>R

4 sin

sin 0 sin

4

v

r

enc r v

R r

R

r

rr R

R

r R

Q E r r d d dr

rr d d dr r d d dr

R

E r

3

3

0

2

0

E= a4

r

R

R

r

17

Example 4.9 - continued