Chap 2 - Electrostatics.pdf (2996k)

CHAPTER 2

ELECTROSTATICS

2.1 COULUMB’S LAW

2.2 ELECTRIC FIELD INTENSITY

2.3 LINE, SURFACE & VOLUME CHARGES 2.4 ELECTRIC FLUX DENSITY

2.5 GAUSS’S LAW

2.6 ELECTRIC POTENTIAL

INTRODUCTION

 Electromagnetics is the study of the effect of

charges at rest and charges in motion.

 Some special cases of electromagnetics:

 Electrostatics: charges at rest

 Magnetostatics: charges in steady motion (DC)  Electromagnetic waves: waves excited by

Maxwell’s equations Fundamental laws of

classical electromagnetics

Special cases

Electro-statics

Magneto-statics

Electro-magnetic

waves

Kirchoff’s

Statics:  ₀

 

t





d

Geometric Optics

Transmission Line Theory Circuit

Theory

Input from other



Electrical phenomena caused by friction

are part of our everyday lives, and can be

understood in terms of

electrical charge

.



The effects of electrical charge can be

observed in the attraction/repulsion of

various objects when “charged.”

 Charge comes in two varieties called “positive”

and “negative.”

 Objects carrying a net positive charge attract

those carrying a net negative charge and repel those carrying a net positive charge.

 Objects carrying a net negative charge attract

those carrying a net positive charge and repel those carrying a net negative charge.

 On an atomic scale, electrons are negatively

charged and nuclei are positively charged.



Electric charge is inherently quantized such

that the charge on any object is an integer

multiple of the smallest unit of charge which

is the magnitude of the electron charge

e = 1.602



10

_C

_.



On the macroscopic level, we can assume

that charge is “continuous.”

2.1 COULUMB’S LAW

In the late 18

_{century, Colonel Charles}

COULUMB’S LAW (Cont’d)

He found that the force is proportional to the product of two charges, inversely proportional to the square of the distance between the charges and acts in a line containing the two charges.

2

2

1

Q

Q

k

The proportional constant, k is:

m

F

m

F





36

10

10

85

.

8

9 12

0





_





Where the free space permittivity with a value given by:

0

4

1

_

_

_



k



,



Charge Q₁ exerts a vector force F₁₂ in Newton's (N) on charge Q₂,

2 1

Q

Q



COULUMB’S LAW (Cont’d)

If more than two charges, use the principle of superposition to determine the force on a particular charge.

If there are N charges, Q₁,Q₂...Q_N located

COULUMB’S LAW (Cont’d)













QQ

QQ

QQ

r

r

r

r

r

r

...

r

r

r

r

r

r

r

r

r

r

r

r

F

























4

4

4

















Q

Q

r

r

F

32 2 32 0 2 3 32 12 2 12 0 2 1 12 32 12

4

4

,

a

R

F

a

R

F

F

F

F





Q

Q

Q

Q









COULUMB’S LAW (Cont’d)

EXAMPLE 1

To employ Coulomb’s Law, first find vector

_R

 

 

a

a

a

a

r

r

R

4

4

4

4













Magnitude of

_R

4

4

12









R

And

a

a

a

a

R

R

a

2

1

2

1

2

4

4

4

12





















 

Q

Q

a

a

a

R

F

2

1

2

1

2

4

10

854

.

8

4

10

10

10

10

4













_















2.2 ELECTRIC FIELD INTENSITY

It becomes convenient to define electric field intensity E₁ or force per unit charge as:

2 12 1

Q

F

E



This field from charge Q₁ fixed at origin results

from the force vector F₁₂ for any arbitrarily

Coulomb’s law can be rewritten as

to find the electric field intensity at any point in space resulting from a fixed charge Q.

R

Q

a

R

E

2 0

4





Let a point charge Q

= 25nC be located

at P

(4,-2,7). If

ε

=

ε

, find electric field

intensity at P

(1,2,3).

By using the electric field intensity,

R

Q

a

R

E

2 0

4





This field will be:

12 12

0

9

4

10

25

a

R

E









z y

x

a

a

a

r

r

R









3



4



4

41



R







 



















a

a

a

R

R

a

R

E

Q

Q

4

4

3

41

10

854

.

8

4

10

25

4

4







ELECTRIC FIELD INTENSITY (Cont’d)

If there are N charges, Q₁,Q₂...Q_N located

respectively at point with position vectors r₁,r₂...r_N the electric field intensity at point r is:









Q

Q

r

r

r

r

r

r

r

r

r

r

r

r

E

















4

..

4



















Q

FIELD LINES

The behavior of the fields can be visualized using field lines:

Some of these field vectors can easily be joined by field lines that emanate from the positive point charge.

The direction of the arrow indicates the direction of electric fields

The magnitude is given by density of the lines

The field lines terminated at a negative point charge

The field lines for a pair of opposite charges

2.3 LINE,SURFACE &

VOLUME CHARGES

To determine the charge for each distributions:

Line charge:







L L L

dl

Q

dl

dQ





Surface charge:







S S S

dS

Q

dS

dQ





Volume charge:



dV

dQ



LINE,SURFACE &

Infinite Length of Line Charge:

To derive the electric field intensity at any

point in space resulting from an infinite

length line of charge placed conveniently

along the z-axis

Place an amount of charge in coulombs along the z axis.

The linear charge density is coulombs of charge per meter length,

Choose an arbitrary point P where we want to find the

electric field intensity.

 

C

_m

L



LINE CHARGE (Cont’d)

The electric field intensity is:

 

E

a

E

a

E

a

E









But, the field is only vary with the radial distance from the line.

There is no segment of charge

dQ anywhere on the z-axis that will give us . So,E_

LINE CHARGE (Cont’d)

Consider a dQ segment a distance z above radial axis, which will add the field

components for the second charge element dQ.

The components cancel each other (by symmetry) , and the adds, will give:

z

E

LINE CHARGE (Cont’d)

Recall for point charge,

R

Q

a

R

E

2 0

4





For continuous charge distribution, the summation of vector field for each charges becomes an integral,





dQ

a

R

E

LINE CHARGE (Cont’d)

The differential charge,

dz

dl

dQ

L L









The vector from source to test point P,

R

R

a

R







LINE CHARGE (Cont’d)

Which has magnitude, and a unit vector, 2 2

z







R

z

z











a

a

a

So, the equation for integral of continuous charge distribution becomes:







_























4

z

z

z

dz











a

a

LINE CHARGE (Cont’d)

Since there is no component,

Hence, the electric field intensity at any point ρ

away from an infinite length is:









a

E

0

2

L



For any finite length, use the limits on the integral.

EXAMPLE 3

Use Coulomb’s Law to find electric field

intensity at (0,0,h) for the ring of charge, of charge density,

centered at the origin in the x-y plane.

L

SOLUTION TO EXAMPLE 3

By inspection, the ring charges delivers only and contribution to the field.

component will be cancelled by symmetry.

z

dE



E

d



E





dQ

a

R

E

2 0

4



Each term need to be determined:

SOLUTION TO EXAMPLE 3 (Cont’d)

The differential charge,





The vector from source to test point, z R

h

a

R

a

a

a

R









Which has magnitude, and a unit vector, 2 2

h

a





R

h

a

h

a









a

a

a

SOLUTION TO EXAMPLE 3 (Cont’d)

The integral of continuous charge distribution becomes:







_





















4

a

h

h

a

h

a

ad

L

a

a

E

















_h

SOLUTION TO EXAMPLE 3 (Cont’d)

















4

_d

h

a

ah

a

E





EXAMPLE 4

An infinite length line of charge

exists at x = 2m and z = 4m. Find the

electric field intensity at the origin.

m

nC

SOLUTION TO EXAMPLE 4

SOLUTION TO EXAMPLE 4 (Cont’d)

The vector from line charge to the origin:

z

x

a

a

a

R





_





2



4

Which has magnitude, and a unit vector,

20



R

z x

R

a

a

a

a

20

4

20

2





Inserting into the infinite line charge equation:





 

m

V

a

a

a

a

a

E

4

.

14

2

.

7

20

4

2

20

10

854

.

8

2

10

4

2







































SURFACE CHARGE

Infinite Sheet of Surface Charge:

To derive the electric field intensity at point

P at a height

h

above a charge sheet of

infinite area (x-y plane).

The charge distribution,



is in

SURFACE CHARGE (Cont’d)

Consider a differential charge,











d

d

dS

dQ

S S





The vector from surface charge to the origin:

 

a

ha

_z

Which has magnitude, and a unit vector,

2 2

h

h

R













a

a

a

2 2

h







R

Where, for continuous charge distribution:





dQ

a

R

E

2

4



The equation becomes:







_





















4

a

h

h

h

d

d

S

a

a

E























d

d

_h

_a

E













Since only z components exists,

SURFACE CHARGE (Cont’d)













h

h

d

h

h

h

d

d

h

a

E

a

a

a

E

0 0 2

SURFACE CHARGE (Cont’d)

A general expression for the field from a sheet charge is:

N S

_a

E

0

2







EXAMPLE 5

An infinite extent sheet of charge

exists at the plane y = -2m. Find the electric

field intensity at point P (0, 2m, 1m).

2

10

m

nC

SOLUTION TO EXAMPLE 5

SOLUTION TO EXAMPLE 5 (Cont’d)

The unit vector directed away from the sheet and toward the point P is

VOLUME CHARGE

A volume charge is distributed over a

volume and is characterized by its volume

charge density,



in

m C

The total charge in a volume containing a

charge

distribution,

is

found

by

integrating over the volume:

V







dV

EXAMPLE 6

Find the total charge

over the volume with

volume charge density,

3 105

5

m

C

e

V



SOLUTION TO EXAMPLE 6





dV

Q



The total charge, with volume:

dz

d

d

dV









VOLUME CHARGE (Cont’d)

To find the electric field intensity resulting from a volume charge, we use:









R

dV

dQ

a

R

a

R

E

2 0

2

0

4

4







Since the vector R and will vary over the

volume, this triple integral can be difficult. It can be much simpler to determine E using Gauss’s Law.

V

2.4 ELECTRIC FLUX DENSITY

Consider an amount of charge

+Q is applied to a metallic

sphere of radius a.

ELECTRIC FLUX DENSITY (Cont’d)

The outer shell is grounded. Remove the ground then we could find that –Q of charge has

accumulated on the outer sphere, meaning the

+Q charge of the inner sphere has induced the –Q

Electric flux, extends from the positive charge and casts about for a negative charge. It

begins at the +Q charge and terminates at the

–Q charge.

 

psi



The electric flux density, D in is:₂

m

C

a

D





_D





_E

This is the relation between D and E, where is the material permittivity. The advantage of using electric flux density rather than using electric

field intensity is that the number of flux lines

emanating from one set of charge and terminating on the other, independent from the media.



We can find the total flux over a surface as:

dS







_

D

We could also find the electric flux density, D

for:

 Infinite line of charge:

Where









a

E

0

2

L







a

D

2

L



Where

So,

_D

_a

2





N S

_a

E

0

2







 Infinite sheet of charge:

 Volume charge distribution:





dV

_a

R

E

0

4









dV

_a

R

D

2

4





EXAMPLE 7

Find the amount of electric flux through the

surface at z = 0 with

and

m

y

m

x

5

0

3

0





,





2

4

3

xy

a

_x

x

a

_z

C

m

D





SOLUTION TO EXAMPLE 7

The differential surface vector is

z

dxdy

d

S



a

We could have chosen but the

positive differential surface vector is pointing in

the same direction as the flux, which give us a

positive answer.





dxdy

EXAMPLE 8

Determine

D

at (4,0,3) if there is a point

charge at (4,0,0) and a line

charge along the y axis.

mC



5



m

C

m



SOLUTION TO EXAMPLE 8

Let total flux, _{  }





TOTAL

D

D

D

Where D_Q is flux densities due to point charge and D_L is flux densities due to line charge.

Thus, R R Q

Q

Q

a

a

R

E

D

4





























SOLUTION TO EXAMPLE 8 (Cont’d)



 

 



a

R

3

3

,

0

,

0

0

,

0

,

4

3

,

0

,

4









Which has magnitude, and a unit vector,

z z

R

a

a

a



3



3



So,

 

138

.

0

9

4

10

5

4

m

C

m

Q

a

a

a

R

D



















And _





a

D

2



Where,



 





 



5

3

4

0

,

0

,

0

3

,

0

,

4

0

,

0

,

0

3

,

0

,

4

a

a

a













So,

 

5

3

4

5

2

3

L

a

a

D





















Therefore, total flux:



 



42

240

18

.

0

24

.

0

318

.

0

m

C



a

a

a

a

a

D

D

D

















2.5 GAUSS’S LAW

If a charge is enclosed, the net flux passing

through the enclosing surface must be equal to the charge enclosed, Q_enc.

Gauss’s Law states that:

The net electric flux through any closed surface is equal to the total charge enclosed by that surface





It can be rearranged so that we have relation between the Gauss’s Law and the electric flux.















V

V enc

enc

dV

Q

Q

d



S

D













V

V

S

dV

d

Q

D

S



GAUSS’S LAW (Cont’d)

Gauss’s Law is useful in finding the fields for problems that have high degree of symmetry.

• Determine variables influence D and what

components D present

GAUSS’S LAW APPLICATION (Cont’d)

Use Gauss’s Law to determine electric

field intensity for each cases below:



Point Charge



Infinite length of Line Charge

POINT CHARGE

• Point Charge:

It has spherical coordinate symmetry, where the field is everywhere directed

radially away from the origin. Thus,

D a

For a gaussian surface, we could find the differential surface vector is:

r

d

d

r

d

S



2

sin







a

So,













d

d

r

D

d

d

r

D

d

sin

sin















a

a

S

D

POINT CHARGE (Cont’d)

Since the gaussian surface has a fixed radius, D_r will be constant and can be taken from

integration to yield

POINT CHARGE (Cont’d)

By using Gauss’s Law, where:

Q

r

D

Q

d

r

enc









2

4



S

D

So, which leads to expected result:

2

4

r

Q

D





r

Q

a

E

2



INFINITE LENGTH LINE OF CHARGE

• Infinite length line of charge:

Find D and then E at any point

_P





_,



_,

_z



A Gaussian surface

LINE CHARGE (Cont’d)

An element of charge dQ along the line will give D_ρ and D_z. But second element of dQ will result in cancellation of D_z. Thus,





a

D



D

The flux through the closed surface is:























d

d

d

d

S

D

S

D

S

D

Where,





















a

S

,

a

S

,

a

S

dz

d

d

d

d

d

d

d

d









Then, we know that D_ρ is constant on the side of gaussian surface



 





d



dz

D

d







_



D

S

a

a

The charge enclosed by the gaussian surface:

h

dz

Q

h

L

enc











0

enc L

h

Q

D

h

d











D

S

2







We know that,

So,







2

L

D



_







a

E

0

2

L



INFINITE EXTENT SHEET OF CHARGE

• Infinite extent sheet of charge:

Determine the field everywhere resulting from an infinite extent sheet of charge ρ_S

placed on the x-y plane at z = 0.

Gaussian surface must contain this point and surround some portion of the charged sheet.

A rectangular box is employed as the

Gaussian surface

surrounding a section of sheet charge with sides 2x, 2y and 2z

Only a D_Z component will be present, and the charge enclosed is simply:

xy

dy

dx

dS

Q

S

y

y x

x S S







4







_

_

_

 

No flux through the side of the box, so find













z z z top z z z bottom top

D

xy

dxdy

D

dxdy

D

d

d

d

4

2





































a

a

a

a

S

D

S

D

S

D

SHEET OF CHARGE (Cont’d)

Then we have:





2

4

4

2

S z

S z

D

xy

D

xy

Q

d

















D

S

z

S

_a

D

2





or

And electric field intensity, as expected:



GAUSS’S LAW (Cont’d)

Related to Gauss’s Law, where net flux is evaluated exiting a closed surface, is the concept of divergence.

Expression for divergence by applying Gauss’s Law might be too lengthy to derive, but it can be described as:

V







The expression is also called the point form of Gauss’s Law, since it occurs at some particular point in space. For instance,

Plunger stationary – no net movement of molecules

Plunger moves up – net

movement where air molecules diverging  air is expanding Plunger pushes in – net flux is

EXAMPLE 9

Suppose:

Find the flux through the surface of a cylinder with and by evaluating the left side and the right side of the divergence theorem.





a

D





h

z





SOLUTION TO EXAMPLE 9

Remember the divergence theorem?













V

dV

dS

D

D

We can first evaluate the left side of the divergence theorem by considering:

























A sketch of this cylinder is shown with differential vectors.

The integrals over the top and bottom surfaces are each zero, since:

0





a

a

Thus, 3 2 0 0 2

2

ha

dz

d

d

d































 





a

S

D

S

D

For evaluation of the right side of the divergence theorem, first find the divergence in cylindrical coordinate:

 

 

















3

1

1

3



















D

D

Performing a volume integration on this divergence,

 

2

0 0 0

2

3

3

dz

d

d

dz

d

d

dV

































  





D

2.6 ELECTRIC POTENTIAL

To develop the concept of electric potential and show its relationship to electric field intensity.

In moving the object from point a to b, the work can be expressed by:







a

d

W

F

L

ELECTRIC POTENTIAL (Cont’d)

The work done by the field in moving the charge from a to b is









b

a field

E

Q

d

W

E

L

If an external force moves the charge against the field, the work done is negative:









Q

d

We can defined the electric potential difference, V_ba as the work done by an external source to move a charge from point a to point b as:











a

ba

d

Q

W

V

E

L

Where,

a b

ba

V

V

V





ELECTRIC POTENTIAL (Cont’d)

Consider the potential difference between two points in space resulting from the field of a point charge located at origin, where the electric field intensity is radially directed, then move from point a to b to have:

















ba

dr

r

Q

d

V

E

L

a

a

2

Thus, a b b r a r ba

V

V

a

b

Q

r

Q

V















 





1

1

4

4





The absolute potential at some finite radius from a point charge fixed at the origin:

r

Q

V

4





If the collection of charges becomes a continuous distribution, we could find:





r

dQ

V

0

4



Where,





 

r dS V

r dL V

S L

0 0

4 4

 

 

Line charge

Surface charge

N N

Q

Q

Q

V

r

r

...

r

r

r

r













4

4

4









_



Q

V

4

1

r

r



The principle of superposition, where applied to electric field also applies to potential difference.

ELECTRIC POTENTIAL (Cont’d)

Based on figure, if a closed path is chosen, the integral will return zero potential:

Three different paths to calculate work moving from the origin to point P against an electric

EXAMPLE 10

SOLUTION TO EXAMPLE 10

Let and

_Q





₄



_C

_Q



₅



_C

So, 2 0 2 1 0 1

4

4

r



r



r



r







Q

Q

V



 

 





 

 



6

2

,

1

,

1

3

,

1

,

2

1

,

0

,

1

1















SOLUTION TO EXAMPLE 10 (Cont’d)





 

4

 

26

10

5

6

4

10

4

4

4

1

,

0

,

1









_

















r

r

r

r

Q

Q

V





kV

The electrostatic potential contours from a point charge form equipotential surfaces surrounding the point charge. The surfaces are always orthogonal to the field lines. The electric field can be determined by finding the max. rate and direction of spatial

Therefore,

V





E

The negative sign indicates that the field is

pointing in the direction of decreasing potential.

By applying to the potential field:

r r

r

Q

r

Q

r

V

a

a

E

0

0

4

4



















IMPORTANT!!

Three ways to calculate E:

 If sufficient symmetry, employ Gauss’s Law.

 Use the Coulomb’s Law approach.

EXAMPLE 11

Consider a disk of charge ρ_S, find the

SOLUTION TO EXAMPLE 11

Find that,











d

d

dS

dQ

S S





and

_r



_h





With





r

dQ

V

0

4



then,

 



d

d

V

2







Let and leads to integral then,

2 2







h

u

How to calculate the integral?





d

du



2



u

du



h

a

h



h

V











2











To find E, need to know how V is changing with position. In this case E varies along the z-axis, so simply replace h with z in the answer for V, then proceed with the gradient equation.

S S z

z

z

z

V

V

a

a

a

E





































1

1

2

1









2.7 BOUNDARY CONDITIONS

So far we have considered the existence of electric field in a region consisting of two different media, the condition that the field must satisfy at the

interfacing separating the media called “boundary condition”. Thus, we could see how the fields

behave at the boundary between a pair of

First boundary condition can be determined by performing a line integral of E around a closed rectangular path,

Fields are shown in each medium along with normal and tangential components. For static fields,

0







E d

L

Integrate in the loop clockwise starting from a,

0

















_

_

_



d d

c c

b b

a

d

d

d

d

L

E

L

E

L

E

L

E

Evaluate each segment,





h

E

E

dL

E

dL

E

d

w

E

dL

E

d







































a

a

a

a

L

E

a

a

L

E





2

h

E

E

dL

E

dL

E

d

w

E

dL

E

d







































a

a

a

a

L

E

a

a

L

E

Summing for each segment, then we have the first

boundary condition:

_E

_

_E

Second boundary condition can be determined by applying Gauss’s Law over a small pillbox shaped Gaussian surface,

enc

Q

d







D

S

BOUNDARY CONDITIONS (Cont’d)

The Gauss’s Law,

Where,























side bottom

top

d

d

d

d

S

D

S

D

S

D

S

D

So, only top and bottom where:





D

S

dS

D

d

S

D

dS

D

d

































a

a

S

D

a

a

S

D

Which sums to:











And the right side of Gauss’s Law,

S

dS

Q













Thus, it leads to the second boundary condition:

S

N

N

D

D

₁



₂





This is when the normal direction from medium 2 to medium 1.

If the normal direction is from medium 1 to medium 2,

S

N

N

D

D

₂



₁





BOUNDARY CONDITIONS (Cont’d)

Generally,











_S



₁

₂

21

D

D

For a boundary conditions between a dielectric and a good conductor,

0



T

E

Because in a good conductor, E = 0. And since the electric flux density is zero inside the

conductor,

S

N





D

EXAMPLE 12

Consider that the field

E

is known as:

Find the field

E

in the other dielectrics.

m

V

z

y

x

a

a

a

BOUNDARY CONDITIONS (Cont’d)

We can employ Poisson’s and Laplace’s

equations to help find the potential function when conditions at the boundaries are specified.

From divergence theorem expression,

V









D

By considering ,

_D





_E

From the gradient expression,

V





E

Which gives us Poisson’s equation,





_V

V







2

In charge free medium in which , it becomes Laplace’s equation:

0





0

2





V

EXAMPLE 13

Determine the electric

potential in the dielectric region between a pair of concentric spheres that have a potential

SOLUTION TO EXAMPLE 13

The charge distribution is: 0 ₃

m

C

r







This employs Poisson’s equation and the potential is only a function of r,

Multiply with r2 _{and integrate to obtain,}

A

r

r

r

V

r











2

)

(







Dividing both sides with r2 _{and integrate again,}

B

r

A

r

r

V









2

)

(





