1

2

EELE 3331 – Electromagnetic I

Chapter 4

Electrostatic fields

Islamic University of Gaza

Electrical Engineering Department Dr. Talal Skaik

2012

Electrostatic Fields (time-invariant)

→ produced by static charge distribution.

Electrostatics Applications:

Electric power transmission, X-ray machines, lightning protection,

photocopiers, liquid crystal displays, spray painting…etc.

Two Fundamental Laws:

• Coulomb’s Law.

• Gauss Law

3

Electrostatic Fields

4

Electrostatics Application: Photocopiers and laser printers

e=1.602x10-19 C

Proton has a charge of e. Electron has a charge of –e. Atom (or group of atoms) that has lost one or more electrons gives

a net positive charge. If it gained one or more electrons, it gives net negative charge.

Deals with the force a point charge exerts on another point charge.

Two polarity of charges may be positive or negative.

→ Like charges repel, unlike charges attract.

Charges measured in Coulombs (C). 1C is approximately

equivalent to 6x1018 electrons.

5

Coulomb’s Law

Coulombs law states that the force F between two point charges Q1

and Q2 is:

1) Along the line joining them.

2) Directly Proportional to the product Q1Q2 of charges.

3) Inversely Proportional to the square of the distance R between

them.

6

Coulomb’s Law

912

0 0

9

0

10permitivity of free space, 8.854 10 F/m36

1 (constant) = 9 10 m/F4

k k

ε επ

πε

−−→ = ×

→ ×

1 2 1 22 2

04KQ Q Q QF

R Rπε= =

7

Coulomb’s Law

1 2 1 22 2

04KQ Q Q QF

R Rπε= =

( )12

12

12 2 1

1 2 2 11 212 32

0 0 2 1

12 2 112

The force F on due to is:r r

F = a4 4 r r

R r r a = , R

R

R

Q QQ QQ Q

R

where RR R

πε πε−

=−

−= =

Notes:

8

Coulomb’s Law

( )21 12

21 1 2

21 12 12

21 12

1 2

1 2

The force F on due to is:

F = F a F a

F F The distance R must be >> dimensions of charged

bodies and and must be static (at rest

R R

Q Q

or

Q QQ Q

→

= −

= −→

→

1 2

)Signs of and must be taken into account.Q Q→

( )12

1 2 2 11 212 32

0 0 2 1

r rF = a

4 4 r rR

Q QQ QRπε πε

−=

−

If there are N point charges Q1, Q2,…,QN located at points with

position vectors r1, r2,….,rN, the resultant force F on a charge

Q located at point with position vector r is given by using

principle of Superposition.

9

Coulomb’s Law

( ) ( ) ( )

( )

1 1 2 212 3 3 3

0 1 0 2 0

12 310

r rr r r rF =

4 r r 4 r r 4 r r

r r F =

4 r r

N N

N

NK K

k K

Q QQ Q Q Q

QQ

πε πε πε

πε =

−− −+ + +

− − −

−→

−∑

The Electric Field Intensity (or electric field strength) E is the

force per unit charge when placed in an electric field.

Note: If we increase Qt, force F increases by the same factor, and hence

E=F/Qt is the same at the location where E is to be found. 10

Electric Field Intensity (E)

( )

( )

30

3 200

FE

r r'F=

4 r r'

r r'FE a (Newton/Coulomb) or (V/m)44 r r'

E: Electric field intensity at point r due to a point charge (Q)

t

t

Rt

QQ Q

Q QQ R

πε

πεπε

=

−

−

−→ = = =

−

If there are N point charges Q1, Q2,…,QN located at r1, r2,….,rN,

the Electric field intensity at point r is:

11

Electric Field Intensity (E)

( ) ( ) ( )

( )

1 1 2 23 3 3

0 1 0 2 0

310

r rr r r rE=

4 r r 4 r r 4 r r

r r1or E=4 r r

N N

N

NK K

k K

QQ Q

Q

πε πε πε

πε =

−− −+ + +

− − −

−

−∑

12

Electric Field Intensity (E)

Point Charges 1 mC and -2 mC are located at (3,2,-1) and (-1,-1,4),

respectively. Calculate the electric force on a 10 nC charge

located at (0,3,1) and the electric field intensity at that point.

13

Example 4.1

( )

( ) ( ) ( )

( ) ( )( ) ( )

( ) ( )

31,20

1 2

1 2

3 3

30

r r1F=4 r r10 nC, 1 mC, 2 mC

r= 0,3,1 , r = 3,2, 1 , r = 1, 1,4

10 0,3,1 3,2, 1 2 10 0,3,1 1, 1,4F

4 0,3,1 3,2, 1 0,3,

K K

k K

QQ

Q Q Q

Q

πε

πε

=

− −

−

−

= = = −

− − −

− − × − − − = −− −

∑

( ) ( ) 31 1, 1,4

− − −

14

Example 4.1 - continued

( )( )

( )( )

( ) ( )

( )

9 3

9 3/2 3/2

2

3

9

3,1, 2 2 1,4, 3(10 10 )(10 )F10 9 1 4 1 16 9436

3,1,2 2, 8,6F=9 10

14 14 26 26F= 6.507 a 3.817 a 7.506 a , mN

10 6.507 a 3.817 a 7.506 aFAt that point, E=10 10

E

x y z

x y z

Q

ππ

− −

−

−

−

−

− −⋅= −

+ + + + ⋅

− − −× +

− − −

− − −→ =

⋅

→ = 650.7 a 381.7 a 750.6 a KV/mx y z− − −

Two point charges of equal mass m, and charge Q are suspended at a

common point by two threads of neglible mass and length l. Show

that at equilibrium the inclination angle α of each thread to the

vertical is given by

If α is very small, show that

15

Example 4.2

2 2 2016 sin tanQ mglπε α α=

2

32

016Q

mglα

πε=

At A or B, Since the point charges are at equilibrium, net horizontal and net vertical forces are zero.

Vertical → T cosα=mg

Horizontal→ T sinα=Fe

16

Example 4.2 - continued

2

20

2 2 2

2 2 30

2 2 20

2 230

sin 1cos 4

2 sin 4 sincos 16 sin

16 sin tan as required

For very small , tan sin /16

eF Qmg mg r

But r l r lQ mgl

Q mgl

Q mgl

αα πε

α α

α πε α

πε α α

α α α α α πε

= = ⋅

= → =

=

=

⇒ =

17

Electric Field due to continuous charge distributions

L2

3

Line Charge Density (in C/m)

Surface Charge Density (in C/m )

Volume Charge Density (in C/m )

To find the charge element :

( )

Line Ch ge

arL LL

S S

S

V

dQ

dQ dl Q dl

dQ dS Q dS

ρ ρ

ρ

ρ

ρ

ρ

ρ

= → =

= → =

∫ ( )

( )

Surface Charge

Volume ChargeS

V VV

dQ dV Q dVρ ρ= → =

∫

∫18

Electric Field due to continuous charge distributions

20

20

20

E of a point charge is:

E=4

Replace Q by charge elements and integrate:

E= ( )Line Charge

Surface Charg

4

E= ( )e4

R

LR

L

SR

S

Q aR

dl aR

dS aR

πε

ρπερπε

∫

∫

20

E= Volume Ch ( ) arge 4

VR

V

dV aR

ρπε∫

19

Electric Field due to continuous charge distributions

Consider a line charge with uniform charge density ρL extending from

A to B along the z-axis. Find E(x,y,z).

20

A. A line Charge

Charge element

Use ( , , ) for the field point.

Use ( ', ', ' ) for the source point.

LdQ dl

x y z

x y z

ρ∗

=

∗

∗

20

3 3/2 3/22 2 2 2 20

'

'

'E a4

R ( , , ) (0,0, ')

R a a ( ')a R a ( ') a

a ( ')a a ( ')aa R E '4( ') ( ')R

B

A

B

A

L LZ

LZ

ZL

RZ

x y z z

z zR L

dQ dl dz

Q dz

dzR

x y z zx y z z z z

z z z zdz

R z z z z

ρ

ρ ρ

ρ ρ

ρ

ρπε

ρ

ρ ρρπερ ρ

= =

=

=

= −

= + + − → = + −

+ − + −= = → =

+ − + −

∫

∫

∫

21

A line Charge

( )2

1

2 2

2

3/22 20

22

0

2

0

( ') sec ' 0T tan ' sec ' sin

( ')aE '

4 ( ')

cos sin aE sec

4

sec cos sin aE

4

zL

zL

zL

R z zzdz dz z R

a z zdz

z z

R a Rd

R R

a

ρ

αρ

α

ρ

ρ ρ αρ α

ρ α αα

ρρπε ρ

α αρ ρ α απε

ρ α α αρπε

= + − =

= −

= −− =

→

+ −=

+ − +

→ = −⋅

+→ = −

∫

∫2

1

2 2secd

α

α

αρ α

∫22

A line Charge

23

A line Charge

( ) ( )

2

1

2

1

2

2 20

0

1 10

sec cos sin aE

4 sec

E cos sin a4

, For finite Line Charge:

E sin sin cos cos a 4

zL

Lz

Lz

ad

a d

Hence

a

αρ

α

α

ρα

ρ

ρ α α αρ απε ρ α

ρ α α απε ρ

ρ α α α απε ρ 2 2

+ = −

= − +

→ = − − + −

∫

∫

24

A line Charge

( ) ( )1 10

1 2

0

For Finite Line Charge:

For Infinite Line Charge

E sin sin cos cos a 4

B at (0,0, ), A at (0,0, )

/ 2, / 2, z-componet vanishes,

E=

:

2

Lz

L

a

a

ρ

ρ

ρ α α α απε ρ

α π α π

ρπε ρ

2 2 → = − − + −

∞ −∞

→ = = −

⇒

If the line is not along the z-axis: ρ is the perpendicular distance from the line to the point of interest, and aρ is a unit vector along the distance directed from the line charge to the field point.

Infinite sheet of charge in the xy plane with uniform charge density.

dQ=ρS dS (charge associated with elemental area dS)

25

B. Surface Charge

( )2

0

2 2R

3/22 20

E field at P(0,0,h) due to dQ on elemental surface 1 is:

E= a4

R a a ,

RR , aR

a aE=

4

R

z

S S

S z

dQdR

h

h

dQ dS d d

d d hd

h

ρ

ρ

πε

ρ

ρ

ρ ρ ρ φ ρ

ρ ρ φ ρ ρ

πε ρ

= − +

= + =

= =

− + +

26

Surface Charge

3/22 20 0

Due to symmetry of charge distribution, for every element 1,there is element 2 whose contribution along a cancels thatof element 1.

E has only z-component.

E= E

E a4

ZS

Sz

d

h d dh

ρ

φ ρ

ρ ρ ρ φπε ρ

∞

= =

→

= +

∫

∫

( )

2

0

1/22 23/22 2 00 00

E= 2 a a4 2

S Sz z

h d h hh

π

ρ

ρ ρ ρ ρπ ρπε ερ

∞ ∞−

=

= − + +

∫

∫27

Surface Charge

2 2

3/2 1/23/2

: Let 2 Integratio 1 12 2

n u h du ddu u du uu

ρ ρ ρ

− −

= + → =

⇒ = = −∫ ∫

( )

1/22 2

00

0

0

0

In general, for infinite sheet of charge

E a2

E a (For h>0)2

For (h<0) E a2

E= a , a is unit vector normal to the sheet. 2

Sz

Sz

Sz

Sn n

h hρ ρερε

ρε

ρε

∞− = − +

→ =

→ = −

⇒

28

Surface Charge

0

E= a 2

Notes:E is independent of the distance between the sheet

and point P.

In parallel plate capacitor, the electric field existing between the two

Sn

ρε

→

→

( )0 0 0

plates having equal and opposite charges is given by:

E= a a a2 2

S S Sn n n

ρ ρ ρε ε ε

−+ − =

29

Surface Charge

A sphere with radius a centred at the origin, with volume charge

density ρV (in C/m3). Find E(0,0,z).

30

C. Volume Charge

20

20

E= a , = R4

R cos a sin a

Ra cos a sin a

components add up to zero.cosE

4

v

vR

z

R z

Vz

dQ dvdvd RR

R R

RE

dvR

ρ

ρ

ρ

ρρπε

α α

α α

ρ απε

=

= −

= = −

→ = ∫

[ ]

[ ]

20

2

20

3

0

cosE4

' sin ' ' ' 'See integration page 119

E a at point (0,0,z)4 is the total charge on sphere

4=3

Generally, the electric field at ( , , ) is:

E=4

vz

z

v v vv v

dvR

dv r dr d d

Qz

QaQ dv dv

P r

Q

ρ απε

θ θ φ

πε

πρ ρ ρ

θ φ

πε

=

=

→

=

→

= =

→

∫

∫ ∫

[ ]2 a Identical to point charge (see Gauss's law) rr31

Volume Charge

Circular ring of radius a, carries

a uniform ρL C/m.

(a) Show that

(b) What values of h gives Emax?

(c) If the total charge on the ring

is Q, find E as a→0.

32

Example 4.4

3/22 20

E(0,0, ) a2

Lz

ahhh aρ

ε=

+

20

2 2R

3/22 20

2

3/2 3/2 2 2 200 0

( ) E= a4

, R= ( a ) a

RR , a =

( a ) aE=

4

By symmetry, the contributionsalong a add up to zero.

aE= E=4 2

LR

L

z

zL

L z L

dlaR

dl ad a h

R a hR

a ha d

a h

ah ahda h a h

ρ

ρ

ρ

π

ρπε

φ

ρ φπε

ρ ρφπε ε

= − +

= = +

− +

+

→ + +

∫

∫

∫

2 a z

33

Example 4.4 - continued

3/2 1/22 2 2 2

32 2

1/22 2 2 2 2

2 2

(1) (3 / 2)(2 ) ( )E( )

EFor maximum E, 0

3 0

2 0 or 2

a h h a h hdb

dh a h

ddh

a h a h h

aa h h

+ − + = +

=

→ + + − =

− = = ±

34

Example 4.4 - continued

3/22 20

3/22 20

20

20

( ) Since the charge is unifromly distributed,

, 2

E= a2

E= a4

As 0, E= a4

or in general E= a (same as that of point charge)4

L

Lz

z

z

R

cQ

aah

a h

Qha h

QahQ

r

ρπρ

ε

πε

πε

πε

=

+

→ +

→

35

Example 4.4 - continued

( )

( )

( ) ( ) ( )

1 13/22 2

0 01 1

3/22 2

0 0

15/221 15/2 5/22 2

0 00

(a) Q= 25 nC

Let , 2 , 2 5 ( / 2) 2

251 1 26 25 2 5 / 2 5

SS

y u

dS xy x y dxdy

dux u du x dx xdx y u y du dy

u yy dy y y y dy

ρ

= =

= + +

= = = → + +

+ + = = + − +

∫ ∫ ∫

∫ ∫

∫ ∫36

Example 4.5

The finite sheet 0≤x ≤1, 0 ≤y ≤1 on the z=0 plane has a charge density

ρS = xy(x2+y2+25)3/2 C/m2. Find:

(a) The total charge on the sheet

(b) The electric field at (0,0,5)

(c) The force experienced by -1 mC charge located at (0,0,5).

( ) ( )

( ) ( )

( )

15/2 5/22 2

02

15/2 5/2

0

20

2 2

3/29 2 2

1 26 25 5

, Let , 2 , / 2

1Q= 26 25 33.15 nC10

(b) E= , 4

R=(0,0,5) - ( , ,0) ( , ,5), R 25

(10 ) 25E

SR

S

Q y y y dy

Similarly v y dv y dy y dy dv

v v dv

dS aR

x y x y x y

xy x y x

ρπε

−

= + − +

= = =

+ − + =

= − − = + +

+ + −=

∫

∫

∫

( )( )

1 1

3/22 20 0 0

a a 5a

4 25x y zy

dxdyx yπε

− +

+ +∫ ∫

37

Example 4.5 - continued

( ) ( )( )

3/29 2 21 1

3/22 20 0 0

1 1 1 1 1 12 2

0 0 0 0 0 0

(10 ) 25 a a 5aE

4 25

E 9 a a 5 a

1 1 1 1 1 1E=9 a a 5 a3 2 2 3 2 2

E= 1.5 a 1

x y z

x y z

x y z

x

xy x y x y dxdy

x y

x y dxdy xy dxdy xy dxdy

πε

− + + − − +=

+ +

= − − +

− − +

→ − −

∫ ∫

∫ ∫ ∫ ∫ ∫ ∫

.5 a 11.25 a

(c) F=qE=(1.5,1.5, 11.25) mN

y z+

−

38

Example 4.5

( )1

2

1 2 3

1 2 39

1 90

19

2 90

2

Let E=E +E +EE : due to sheet 1, E : due to sheet 2, E : due to Line

10 10E a2 (2)(10 / 36 )

E 180 a

15 10E a2 (2)(10 / 36 )

E 270 a

sx

x

sy

y

ρε π

πρε ππ

−

−

−

−

×= − =

= −

×= =

=39

Example 4.6

Planes x=2 and y=-3, respectively, carry charges 10 nC/m2 and

15nC/m2. If the line x=0, z=2 carries charge 10π nC/m, Calculate

E at (1,1,-1) due to the three charge distributions.

( )

30

9

3 9

1 2 3

E a2

R=(1,1,-1)-(0,1,2)=(1,0,-3)

R=a 3 a , R 10R a 3 a 1 3a a aR 10 10 10

10 10 a 3 aE 18 a 3 a 2 (10 / 36 )( 10) 10

Total field is E=E E EE 162 a 270 a 54 a

L

x z

x zx z

x zx z

x y

ρ

ρ

ρπε ρ

ρ

π ππ π

π π π

−

−

=

− = =

−= = = −

× − = = −

+ += − + − V/mz

40

Example 4.6

2

0

0

2

= D S

D (C/m )

D= E (for free space)

For an infinite sheet of charge, E= a D= a2 2

For a volume charge dist

Electric Flux

Elec

ribution: D= a

tr

ic Flux

4

Density

S

S Sn n

vR

v

d

dvR

ψ

ε

ρ ρε

ρπ

⋅

→

→

∫

∫

41

Electric Flux Density (D) [measure of the number of field lines passing through an area]. (in Coulombs)

D is a function of charge and position only, it is independent of the medium.

( )Q L

Q 0 32

32

z3

L

L

Let D=D +D

'D = E= a , ' (4,0,3) (4,0,0) (0,0,3)

4 R 4 '

5 10 (0,0,3)D 0.138 a mC/m4 (0,0,3)

(4,0,3) (0,0,0) (4,0,3)D a , a , (4,0,3) (0,0,0) 52 (4,0,3) (0,0,0) 5

3 10D

R

Q

L

Q r rQ r rr r

ρ ρ

επ π

ππ

ρ ρπρ

π

−

−

−= − = − =

−

− ×→ = = −

−= = = = − =

−

×=

32

2x z

(4a 3a ) 0.24 a 0.18 a mC/m(2 )(5) 5

D D D 240 a 42 a C/m

x zx z

Q L

πµ

+= +

⇒ = + = +42

Example 4.7

Determine D at (4,0,3) if there is a point charge -5π mC at (4,0,0) and

a line charge 3π mC/m along the y-axis.