Elektromagnetika TeknikCHAPTER 2 COULOMB’S LAW AND ELECTRIC FIELD INTENSITY

Chapter 2Field Due to a Continuous Volume Charge

Distribution, Line Charge, and Sheet of Charge

Engineering Electromagnetics

Chapter 2 Coulomb’s Law and Electric Field Intensity

Field Due to a Continuous Volume Charge Distribution We denote the volume charge density by ρv, having the units of coulombs per cubic

meter (C/m3).

The small amount of charge ΔQ in a small volume Δv is

vQ v

We may define ρv mathematically by using a limit on the above equation:

0limvv

Q

v

The total charge within some finite volume is obtained by integrating throughout that volume:

vol

vQ dv

Chapter 2 Coulomb’s Law and Electric Field Intensity

ExampleFind the total charge inside the volume indicated by ρv = 4xyz2, 0 ≤ ρ ≤ 2, 0 ≤ Φ ≤ π/2, 0 ≤ z ≤ 3. All values are in SI units.

cosx

siny

24 sin cosv z

vol

vQ dv 23 2

2

0 0 0

(4 sin cos )( )z

z d d dz

23 2

3 2

0 0 0

4 sin cosz d d dz

23

2

0 0

16 sin cosz d dz

sin 2 2sin cos

3

2

0

8z dz 72 C

Field Due to a Continuous Volume Charge Distribution

Chapter 2 Coulomb’s Law and Electric Field Intensity

The incremental contribution to the electric field intensity at r produced by an incremental charge ΔQ at r’ is:

2

0

( )4

Q

r rE r

r rr r

The contributions of all the volume charge in a given region, let the volume element Δv approaches zero, is an integral in the form of:

2

0vol

( )1( )

4

v dv

r r rE r

r rr r

2

04

v v

r r

r rr r

Field Due to a Continuous Volume Charge Distribution

Field of a Line ChargeChapter 2 Coulomb’s Law and Electric Field Intensity

Now we consider a filamentlike distribution of volume charge density. It is convenient to treat the charge as a line charge of density ρL C/m.

Let us assume a straight-line charge extending along the z axis in a cylindrical coordinate system from –∞ to +∞.

We desire the electric field intensity E at any point resulting from a uniform line charge density ρL.

E z zd d dE E a a

Field of a Line ChargeChapter 2 Coulomb’s Law and Electric Field Intensity

The incremental field dE only has the components in aρ and az direction, and no aΦdirection.

• Why?

The component dEz is the result of symmetrical contributions of line segments above and below the observation point P.

Since the length is infinity, they are canceling each other ► dEz = 0.

The component dEρ exists, and from the Coulomb’s law we know that dEρ will be inversely proportional to the distance to the line charge, ρ.

E z zd d dE E a a

Field of a Line ChargeChapter 2 Coulomb’s Law and Electric Field Intensity

Take P(0,y,0),

E z zd d dE E a a

3

0

1 ( )

4

dQd

r rE

r r

2 2 3 2

0

( )1

4 ( )

L zdz z

z

a a

zz r a

yy r a a

2 2 3 2

0

1

4 ( )

L dz

z

a

2 2 3 2

0

1

4 ( )

L dzE

z

2 2 2 1 2

04 ( )

L z

z

02

LE

Field of a Line ChargeChapter 2 Coulomb’s Law and Electric Field Intensity

Now let us analyze the answer itself:

02

L

E a

The field falls off inversely with the distance to the charged line, as compared with the point charge, where the field decreased with the square of the distance.

Field of a Line ChargeChapter 2 Coulomb’s Law and Electric Field Intensity

Example D2.5.Infinite uniform line charges of 5 nC/m lie along the (positive and negative) x and yaxes in free space. Find E at: (a) PA(0,0,4); (b) PB(0,3,4).

PAPB

0 0

( )2 2

x

x y

L Ly

A

x y

P

E a a

9 9

0 0

5 10 5 10

2 (4) 2 (4)z z

a a

44.939 V mz a

0 0

( )2 2

x

x y

L Ly

B

x y

P

E a a

9 9

0 0

5 10 5 10(0.6 0.8 )

2 (5) 2 (4)y z z

a a a

10.785 36.850 V my z a a

• ρ is the shortest distance between an observation point and the

line charge

Field of a Sheet of ChargeChapter 2 Coulomb’s Law and Electric Field Intensity

Another basic charge configuration is the infinite sheet of charge having a uniform density of ρS C/m2.

The charge-distribution family is now complete: point (Q), line (ρL), surface (ρS), and volume (ρv).

Let us examine a sheet of charge above, which is placed in the yz plane.

The plane can be seen to be assembled from an infinite number of line charge, extending along the z axis, from –∞ to +∞.

Field of a Sheet of ChargeChapter 2 Coulomb’s Law and Electric Field Intensity

For a differential width strip dy’, the line charge density is given by ρL = ρSdy’.

The component dEz at P is zero, because the differential segments above and below the yaxis will cancel each other.

The component dEy at P is also zero, because the differential segments to the right and to theleft of z axis will cancel each other.

Only dEx is present, and this component is a function of x alone.

Field of a Sheet of ChargeChapter 2 Coulomb’s Law and Electric Field Intensity

The contribution of a strip to Ex at P is given by:

2 2

0

cos2

sx

dydE

x y

2 2

02

s xdy

x y

Adding the effects of all the strips,

2 2

02

sx

xdyE

x y

1

0

tan2

s y

x

02

s

Field of a Sheet of ChargeChapter 2 Coulomb’s Law and Electric Field Intensity

Fact: The electric field is always directed away from the positive charge, into the negative charge.

We now introduce a unit vector aN, which is normal to the sheet and directed away from it.

02

sN

E a

The field of a sheet of charge is constant in magnitude and direction. It is not a function of distance.

Chapter 2 Coulomb’s Law and Electric Field Intensity

Homework 2 D2.2.

D2.4.

D2.6. All homework problems from Hayt and Buck, 7th Edition.