hkpho electricity

5/21/2018 HKPhO Electricity

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Hong Kong Physics Olympiad Lesson 16

16.1 Electric charge

16.2 Coulombs law

16.3 Shell theorems for electrostatics

16.4 Electric field

16.5 Electric field lines

16.6 Shielding and charging by induction

16.7 Electric flux

16.8 Gauss law

16.1 Electric charge

There are two kinds of charges, namely, positive (+) charge and negative () charge.

Like charges repel

Unlike charges attract

Objects with zero net change are said to be electrically neutral.

Electric charges are generated after rubbing between materials.

Example

After rubbing plastic rod (or amber rod) with fur, the plastic rod (or

amber rod) becomes negatively charged and the fur is positively

charged.

+ +

+


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After rubbing glass rod with silk, the glass rod becomes positively charged and the silk is

negatively charged.

A familiar example of an electrically neutral object is the atom.

Atoms have a small, dense nucleus with a positive charge surrounded

by a cloud of negatively charged electrons. All electrons have exactly

the same electric charge. This charge is very small, and is defined to

have a magnitude, e = 1.60 1019

C. S.I. unit of charge is coulomb,

C. Clearly, the charge on an electron, which is negative, is e. This is

one of the defining, or intrinsic, properties of the electron. Another

intrinsic property of the electron is its mass, me:

me= 9.11 1031

kg

In contrast, the charge on a proton one of the main constituents of nuclei is exactly +e. As

a result, since atoms have equal numbers of electrons and protons, their net charge is

precisely zero. The mass of the proton is

mp= 1.673 1027

kg.

Note that this is about 2000 times larger than the mass of the electron. The other main

constituent of the nucleus is the neutron, which, as its name implies, has zero charge. Its mass

is slightly larger than that of the proton:

mn= 1.675 10

27kg.

Example

How is it that rubbing a piece of amber with fur gives the amber a

charge?

Answer:

Rubbing the fur across the amber simply results in a transfer of

charge from the fur to the amber with the total amount of charge

remaining unchanged. Before charging, the fur and the amber are

both neutral. During the rubbing process some electrons are

transferred from the fur to the amber, giving the amber a net

negative charge, and leaving the fur with a net positive charge. At

no time during this process is charge ever created or destroyed. This,


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in fact, is an example of one of the fundamental conservation laws of physics: Conservation

of electric charge.

When charge is transferred from one object to another it is generally due to the movement of

electrons. In a typical solid, the nuclei of the atoms are fixed in position. The outer electrons

of these atoms, however, are often weakly bound and fairly easily separated. The atom that

loses an electron is a positive ion, and the atom that receives an extra electron becomes a

negative ions. This is charging by separation.

Example

Find the amount of positive electric charge in one mole of helium atoms.

Answer:

Note that the nucleus of a helium atom consists of two protons and two neutrons. The totalpositive charge in a mole is

CCeNA51923

1093.1)1060.1)(2)(1002.6()2( == .

16.11 Polarization

We know that charges of opposite sign attract, but it is

also possible for a charged rod to attract small objects

that have zero net charge. The mechanism responsible

for this attraction is called polarization. When a charged

rod is far from a neutral object the atoms in the object

are undistorted. As the rod is brought closer, however,

the atoms distort, producing an excess of one type of

charge on the surface of the object (in this case a negative charge). This induced charge is

referred to as a polarization charge. Since the sign of the polarization charge is the opposite

of the sign of the charge on the rod, there is an attractive force between the rod and the object.

16.12 Conductor and insulator

Conductors: materials that allow electric charges to move more or less freely, e.g. metals

Insulators: materials in which charges are not free to move, e.g. nonmetallic substances, say,

amber.


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On a microscopic level, the difference between conductors and

insulators is that the atoms in conductors allow one or more of their

outermost electrons to become detached. These detached electrons,

often referred to as conduction electrons, can move freely

throughout the conductor. The right figures show the charging of a

conductor by touching it with charged rod.

16.2 Coulombs law

Electric force Coulombs law

2

21

r

qqkF

e= , where k(electrostatic constant > 0) is a constant.

The electrostatic constant229

0

/1099.84

1CmNk ==

, where 0 is called permittivity

constant of free space, and 012 2 2

885 10=

. /C N m .

Gravitational force Newtons gravitational law

2

21

r

mmGF

g = , where2211

/1067.6 kgmNG = .

The negative sign is inserted to represent an attractive force.

Remarks:

1. Fundamental laws cannot be derived!

etc.

law,Newtons

law,Coulombs

M

are concluded according to results in experiments and have survived in

every experimental test.

2. Objects are considered as point particle or point charge, if rdd


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4. The magnitude of the force of interaction between two point changes is directly

proportional to the product of the charges and inversely proportional to the square of

the distance between them.

F F1 2= Obey Newtons third law

5. Unit: International System of Units or Metric System (SI)

Charge q: measured in Coulomb or C.

Remark:

Electrostatic constant is related to the speed of light c:

00

1

=c ,

where smc /10998.2 8= , and 0 = 410-7

N/A2, the permeability of free space.

Example

Compare the electric and gravitational forces between a proton and an electron in a hydrogen

atom.

Answer:Taking the distance between the two particles to be the radius of hydrogen, mr

111029.5

= ,

we find that the electric force has a magnitude

Nm

CCCmN

r

qqkF

pe

e

8

211

1919229

21022.8

)1029.5(

)1060.1)(1060.1()/1099.8(

=

== .

Similarly, the magnitude of the gravitational force between the electron and the proton is

N

m

kgkgkgmN

r

mmGF

pe

g

47

211

27312211

2

1063.3

)1029.5(

)10673.1)(1011.9()/1067.6(

=

== .

Hence, we obtain the ratio of the two forces

000,000,000,000,000,000,000,000,000,000,000,000,260,21026.21063.3

1022.8 3947

8

==

=

N

N

F

F

g

e

+ +1q 2q

1Fv

2Fv


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Example

We study the classical model for Hydrogen atom. The electron undergoes a circular motion

with a radius a0 called Bohr radius. Find the speed of electron.

The constants are given as below.

a m A011

529 10 0529= =

. .o

me= 9.11 1031

kg

Answer:

The force acting on the electron is obtained in the last example, where ege FFFF += . But,

0

2

a

vmF e= (the centripetal force). Hence, v

Fa

me

2 0=

vFa

mm s

e

= =

=

0

8 11

31

68 22 10 529 10

911 10218 10

. .

.. / .

Example

An electron and a proton, initially separated by a distance d, are released from rest

simultaneously. The two particles are free to move. When they collide, are they (a) at the

midpoint of their initial separation, (b) closer to the initial position of the proton, or (c) closer

to the initial position of the electron?

Answer:

Because of Newtons third law, the forces exerted on the electron and proton are equal in

magnitude and opposite in direction. For this reason, it might seem that the particles meet at

the midpoint. The masses of the particles, however, are quite different. In fact, the mass of

the proton is about 2000 times greater than the mass of the electron; therefore, the protons

acceleration (a = F/m) is about 2000 times less than the electrons acceleration. As a result,

the particles collide near the initial position of the proton. That is, the answer is (b).

16.21 Superposition of Coulombs force

The force exerted on charge 1 by charge 2:

122

12

21

0

12

4

1r

r

qqF

=

r

Proton

electron

0a


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where 12r : unit vector along 12rr

.

q1 and q2 same sign repulsion

opposite sign attraction

Similarly, The force exerted on charge 2 by charge 1:

212

21

21

0

21

4

1r

r

qqF

=

r

where 21r : unit vector along 21rr

.

The total force acting on charge q due to coulombs forcesF1,F2andF3.

F F F F= + +1 2 3 (Principle of superposition)

The direction of forces shown in the right figure, representing that the

charge qis of opposite charge of q1, q2andq3.

Example

Three charges, each equal to +2.90 C, are placed at three corners of a square 0.500 m on a

side. Find the magnitude and direction of the net force on charge number 3.

Answer:

The magnitude of force exerted on charge 3 by charge 1:

Nm

CCmNr

qkF 151.0)500.0(2

)1090.2()/1099.8()2(

2

26

229

2

2

31 ===

The magnitude of force exerted on charge 3 by charge 2:

Nm

CCmN

r

qkF 302.0

)500.0(

)1090.2()/1099.8(

2

26229

2

2

32 =

==

q

1q 3q

2q

1Fv

2Fv

3Fv

1rv

2rv

1q

2q

Point charge

2112 rrr vvv

=

O x

y

31F

r

32Fr

F

r

1 x

y

2

r2

3

r=0.500 m

r=0.500 m


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The components of 31Fr

and 32Fr

:

NNFF ox 107.0)707.0)(151.0(0.45cos31,31 ===

NNFF oy

107.0)707.0)(151.0(0.45sin31,31 ===

NNFF ox 151.0)1)(302.0(0cos32,32 ===

NNFF oy

0)0)(151.0(0sin32,32 ===

The components of the resultant force:

NNNFFF xxx 409.0302.0107.0,32,31 =+=+=

NNNFFFyyy 107.00107.0,32,31 =+=+=

The resultant force acting on charge 3:

NFFFyx 423.0

22=+=

The direction of the resultant force on charge 3:

o

x

y

F

F7.14tan

,3

,31=

=

.

Example

Two charges, q1 and q2, have equal magnitudes q and

are placed as shown in the following figure. The net

electric field at point P is vertically upward. We

conclude that (a) q1is positive, q2, is negative; (b) q1is

negative, q2 is positive; or (c) q1 and q2have the same

sign.

Answer:

If the net electric field at Pis vertically upward, the horizontal components ofE1andE2must

cancel, and theycomponents must both be in the positive ydirection. The only way for this

to happen is to have q1negative and q2positive.

O x

y

d

P

d

Enet

q1

q2


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16.3 Shell theorems for electrostatics

Theorem 1:

A uniform spherical shell of charge behaves, for external points, as if all its charge wereconcentrated at its center.

Fq q

r=

1

4 0

1

2

Theorem 2:

A uniform spherical shell of charge exerts no force on a charged particle placed inside the

shell.

F= 0

Remarks: The theorems are similar to the gravitational case.

16.4 Electric field

Gravitational field (a vector field)

rr

MG

m

rr

mMG

m

Fg e

e

2

2

=

==

rr

Gravitational field: Gravitational force per unit mass

Electric fieldEr

Total charge q

on spherical shell

1q

Earth

m

Test body

A test charge

with positivecharge

0qeF

v

Charged

particle q

O x

y

d

P

d Enet

q1

q2

E1 E2


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rr

q

q

FE e

4

12

00 =

rr

Electric field: Electrostatic force per unit charge

SI unit of electric field: Newton/Coulomb orN/C

Remark:

1. Why do we need to introduce the concept of Electric field?

Introducing the field as an intermediary between the charges, we can represent the interaction

as:

Our problem of determining the interaction between the charges is therefore reduced to two

separate problems: (1) determine, by measuring or calculation, the electric field established

by the first charge at every point in space, and (2) calculate the force that the field exerts on

the second charge placed at a particular point in space.

2. Principle of superposition in electric field:

The resultant electric field Eat a point is given by E E E E= + +1 2 3 , whereE1, E2, andE3

are the electric fields experienced at that point due to charge 1, 2 and 3 respectively.

Example (Optional)An electric dipole phas the configuration shown in figure, which has two charges of equal

magnitude and opposite sign, separated by a distance d. The electric dipole moment is

defined as dqdp .

Calculate the electric field at point Pfor the following cases.

(a) P is located along the direction of electric dipolep.

(b) Pis located along the perpendicular bisector of a dipole momentp.

Answer:

(a) P is located along the direction of electric dipolep.

dqdp =r

q+ q

charge chargefield


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E E E= ++

Eq

zd

q

zd

=

+

+

42

42

0

2

0

2 ( ) ( )

22

0

22

0 )2

1(4)2

1(4z

dz

q

z

dz

q

+

=

Hence, ])21()21[(4

22

20

+= z

d

z

d

z

q

E

Remark:

If we are interested inE-field at large distance, i.e.,z >> d

Use Taylor expansion on ( )1+x for smallx,

( )( )

1 11

2

2+ = + +

+ +x x x

higher order terms

+1 x

That is, forz >> d, we have

( )12

12

+ d

z

d

z ( )1

21

2 +

d

z

d

z

z

d

z

q

z

d

z

d

z

qE

2

4)]1()1[(

42

0

2

0

=+=

Plug in the magnitude of electric dipole moment:p = qd

3

02

1

z

pE

= forz >> d

The negative sign indicates that the field is pointing to the electric dipole.

(b) Pis located along the perpendicular bisector of a dipole momentp.

The electric field is alongxdirection, since theE-fields alongydirection cancel to each other

cos||cos|| + += EEE (As ++= EEEvvv

)

-q+q

d

origin

z

xP

Ev

dqdp =r


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= 1

42

0

2

q

rcos

rd

z

d

r

2 2 2

2

2

= +

=

( )

cos

= =

+

1

4

1

42

03

0 2 2 3 2

qd

r

p

z d[ ( ) ]/

= + 1

41

203

2 3 2

p

z

d

z[ ( ) ]

/

Remark:

Forz >> d, that is the long distance behavior:

[ ( ) ] ( )/

12

13

2 2

2 3 2 2+

d

z

d

z

If we keep only the first order in dz

, then

[ ( ) ]/

12

12 3 2

+ d

z

Ep

z=

1

4 03

forz>> d

16.41 Discrete and continuous charge distribution

a) Discrete case

E E E E= + + + 1 2 3

b) Continuous case

i) When charge is uniformly distributed along a line.

Linear charge density

q

l

charge per unit length ( =dq

dl

)

ii) Charge on a surface (uniformly distributed)

Surface charge density

q

S charge per unit area (=

dq

dS)

l

qTotal

x

y

Ev

Ev

+Ev

origin

q+

z

P

r

q

dqdp =r

d

Surface area S

Total charge q


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iii) On a volume

Volume charge density

q

V charge per unit volume =

dq

dV

16.42 Electric field for a ring of charge

Due to the symmetry of the ring, the electric field along $z direction can be calculated as

follows.

= cos'EdE , where cos=

z

rand

22

0

2

0

'

4

1

4

1

Rz

ds

r

dqdE

+==

.

The linear charge density of the ring

= =q

R

dq

ds2, whereRis the radius of ring.

{

E zz R

ds

R

=+

14 0

2 2 3 2

2

( )

/

=+

1

4

2

0

2 2 3 2

z R

z R

( )

( )/

Plug in the expression, 2 R q= , hence, the electric field at any point P, a perpendicular

distancez from the plane and center of ring,2/322

0 )(4 Rz

zqE

+=

.

Remark:

Whenz >> R, when the distance is much larger than the dimension of the ring,

1 1

0

12 2 3 2 2 3 2 3( ) ( )/ /z R z z+

+

=

Eq

z

4 02

As if R 0

The ring seems to be a point charge!

VolumeV

Total charge q

'Edr

P

Er

'Edr

Total charge qds

z

'Edr

P

r

R


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16.43 A uniformly charged disk

The surface charge density

=q

R2

The differential area of the ring

rdrdA 2=

)2( rdrdAdq ==

The electric field experienced at a perpendicular distancezfrom the

center of disk due to the differential ring.

2/3220

2/322

0 )(

2

4)(4 rz

drrz

rz

zdqdE

+=

+=

+

==R

rz

drrzdEE

0 2/3220 )(

2

4

Now, we let 2ry= and we have rdrdy 2= . The integral +

R

rz

drr

0 2/322 )(

2becomes

+2

0 2/32 )(

R

yz

dy, which gives

2

0

2

1

2

0

))(2(4

R

yzz

E

+=

. Hence, we have

)1(2 220 Rz

zE

+=

.

Remark:

WhenR>>z (infinite sheet of charge)

E=

2 0

whenz>>R,z

z R R

z

R

z2 2 2

2

2

2

1

1

12+

=

+

E Rz

=

2

0

24,

= q

R2

2

04 z

qE

= (Result of point charge!)

16.44 Dipole in an electric field

dA

Rq

Er

dr

r

r

Disk of Radius R


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A dipole in a uniform electric field experiences a torque . Note that the total force acting on

the dipole is zero, e.g. F F F E E= + = + =+ q q( ) 0 .

Ifdis the vector pointed from qto q, then2

1

dd = ,

22

dd =

For simplicity, we take the moment at the center of dipole (In fact, you can take the moment

at any point). The torque is given by + += FdFd 21 .

Hence, we have Ep = , since

=

=

dp

EF

q

q.

In scalar form, we have = pEsin .

16.5 Electric field lines

Rules for drawing electric field lines

Electric field lines:

(ii) Point in the direction of electric field vectorEat every point;

(iii) Start at positive (+) charges or at infinity;

(iv) End at negative () charges or at infinity;(v) Are more dense whereEhas a greater magnitude. In particular, the number of lines

entering or leaving a charge is proportional to the magnitude of the charge.

Er

+Fr

Fr

2dr

1d

r

q

+q


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Example

Which of the following statements is correct: Electric field lines (a) can or (b) cannot

intersect?

Answer:

By definition, electric field lines are always tangent to the electric field. Since the electric

force, and hence the electric field, can point in only one direction at any given location, it

follows that field lines cannot intersect. If they did, the field at the intersection point would

have two conflicting directions.

Example

The electric field between the plates of a parallel-plate capacitor is horizontal, uniform, and

has a magnitudeE. A small object of mass 0.0250 kg and charge 3.10 C is suspended by a

thread between the plates. The thread makes an angle 10.5owith the vertical. Find (a) the

tension in the thread and (b) the magnitude of the electric field.


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Answer:

The net force in the x direction equal to zero:

0sin =+ TqE .

The net force in the y direction equal to zero:

0cos = mgT .

The tension Tis given by

Nmskgmg

To

249.0)5.10cos(

)81.9)(0250.0(

cos

2

===

.

The electric field E is given by

CNC

N

q

TE

o

/1046.11010.3

)5.10sin()249.0(sin 46

=

==

.

16.6 Shielding and charging by induction

In a perfect conductor there are enormous numbers f electrons

completely free to move about. This simple fact has some rather

interesting consequences. Consider, for example, a solid metal sphere

attached to an insulating base as shown in figure. Suppose a positive

charge Qis placed on the sphere. The question is: How does this charge

distribute itself on the sphere when it is in equilibrium? In particular,

does the charge spread itself uniformly throughout the volume of the

sphere, or does it concentrate on the surface?

The answer is that the charge concentrates on the surface . Why should

this be the case? First, assume the opposite that the charge is spread

uniformly throughout the spheres volume. If this were the case, a charge at locationAwould

experience an outward force due to the spherical distribution of charge between it and the

center of the sphere. Since charges are free to move, the charge at Awould respond to this

force by moving toward the surface. Clearly, then, a uniform distribution of charge within the

spheres volume is not in equilibrium. In fact, the argument that a charge at pointAwill move

toward the surface can be applied to any charge within the sphere. The preceding result holds

no matter what the shape of the conductor. In general,

excess charge placed on a conductor, whether positive or negative, moves to the exterior

surface of the conductor.


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16.61 Shielding

When electric charges are in equilibrium, the electric field within a conductor is zero; E= 0.

A straightforward extension of this idea explains the phenomenon of shielding, in which a

conductor shields its interior from external electric fields.

We also noted that the electric field lines contact conductor surfaces at right angles. If an

electric field contacted a conducting surface at an angle other than 90o, the result would be a

component of force parallel to the surface. This would result in a movement of electrons and

hence would not correspond to equilibrium.


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16.62 Charging by induction

One way to charge an object is to touch it with a charged rod; but

since electric forces can act at a distance, it is also possible to charge

an object without making direct physical contact. This type of

charging is referred to as charging by induction.

16.7 Electric flux

The electric flux is defined as , which is the product of the electric

field and the area of surface AE= .

Example

Consider a surface with area A and a uniform electric field

penetrating the surface perpendicularly. The electric flux is given by

EA= .


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Example

Consider a surface with area Aand a uniform electric field penetrating the surface with an

angle with the normal of surface. The electric flux is given by AE= . Or in scalar

form: cos EA .

Remarks:

1. For a plane surface, the area vectorAis defined as

A= Ae$ , where $e is the normal vector.

2. For a curved surface, we have d dA eA= $ .

When the electric fieldEis not uniform, i.e.,E=E(x,y,z), or if the surface is not a

plane, then =i

iAE or in the integral form AE d .

3. If the surface through which the flux is calculated is closed, the sign of the flux is as

follows:

The flux is positive for field lines that leave the enclosed volume of the surface.

The flux is negative for field lines that enter the enclosed volume of the surface.

16.8 Gauss lawIn order to understand Gausss law, we first look at the following. Consider a point charge q

and a spherical surface of radius r and centered on the charge. The electric field on the

surface of the sphere has the constant magnitude

2r

qkE= .

e

Surface areaA

Er

Normal of the surface

e

AreaA


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Since the electric field is everywhere perpendicular to the

spherical surface, it follows that the electric flux is simply E

times the areaA = 4r2of the sphere:

( ) kqrr

qkEA 44 2

2 =

==

Plug in k= 1/(40), we obtain the Gausss law

0

q= .

Thus we find the very simple result that the electric flux through a sphere that encloses a

charge qis the charge divided by the permittivity of free space, 0.This is a very nice result!!

As the electric field of many symmetrical system can be found readily with Gausss law.

Gausss law states that the flux of the electric field over the Gaussian surface equals to the

net charge enclosed by that surface.

Remarks:

1. Symmetrical situations arise in all fields ofphysicsand, when possible, it makes sense

to cast the laws ofphysicsin forms that take full advantage of this fact.

2. Gauss law is a new formulation of Coulombs law that can take advantage of

symmetry.

3. The integral form of Gausss law:

= qdAE

0 , where

0is the permittivity

constant.

Example

Consider the surface S shown in the figure. Is the electric flux through this surface (a)

negative, (b) positive, or (c) zero?

Answer:

Since the surface S encloses no charge, the net electric flux

through it must be zero, by Gausss law. That a charge +qis

nearby is irrelevant, because it is outside the volume enclosed

by the surface.

We can explain why the flux vanishes in another way. Notice

that the flux on portions of Snear the charge is negative, since

Er

+q

Gaussian surface with area 4r2


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22

field lines enter the enclosed volume there. On the other hand, the flux is positive on the outer

portions of S where field lines exit the volume. The combination of these positive and

negative contributions is a net flux of zero. That is, the answer is (c).

16.91 Gaussian surface

Gaussian surface is a closed surface, e.g. sphere, cube, cylinder, etc. And, Gauss law tells

how the fields at the Gaussian surface are related to the charges contained within that surface.

Example

Calculate the electric field at a external point due to a point charge.

Answer:

We form a spherical Gaussian surface, centered at the charge, to enclose the charge q. The

electric field,E,is uniform on the spherical Gaussian surface.

From Gauss law == qdAE00

As AE d// andEis uniform on the surface, we have

= qdAE0 or qrE =)4(

2

0

2

04 r

qE

= ,

which gives the Coulombs law.

Remarks: Gauss law equivalent

Coulombs law.

Remarks:

Suppose we have a spherical Gaussian surface, and if the electric field vectors are of uniform

magnitude and point radially outwards as shown. One can conclude that a net positive charge

must lie within the surface and that it must have spherical symmetry.

Charges

Gaussian surface

Er

+q

Gaussian surface with area 4r2


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23

Example

Use Gausss law to investigate the space under a uniform electric field.

We form a cylindrical Gaussian surface, which is immersed in a uniform field.

The electric flux:

++==cba

dddd AEAEAEAE , where E A = d EdAcos.

(i) Surface a = 180 EAda

= AE

(ii) Surface b = 90 0=b

dAE

(iii) Surface c = 0 EAdc

= AE

00 =++= EAEA

No charge is enclosed in the Gaussian surface.

?

Charge inside?

Spherical Gaussian surface

E

dAr

b

Er

dAr

a

dAr

c Er

Er


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24

Example

Find the electric field at a point very near to the surface of a charged conductor.

Answer:

Suppose charges q are on the right surface, the surface charge density is given by = qA

, or

in the language of calculus, ( )r =dq

dA.

Now, we form the Gaussian surface as shown in figure. Near the conductor, the surface is flat,

so theE-field is //to $e of the surface.

Apply the Gauss law, we have

qd = AE0

++=cba

dAE

i) For surface a, E= 0 (inside the conductor)

ii) For surface c, eE , hence we haveE A = d 0 .

iii) For surface b, AEAE =b

d .

So,

0

0

EA q Eq

A= = or E=

0.

Example

Infinite long charged plastic rod with linear charge density ,

where q

h( q: total charge enclosed by Gaussian surface)

From cylindrical symmetry,Eis along radial direction.

By Gauss law: qd = AE0 .

Surfaces a and bdo not contribute to the integral 0 2E rh q h = =

So Er

=

2 0

E

ra

c

Gaussian

surfacee

b

Charges q on surface

Conductor

Gaussian surface

+

+

+

+

+

+

++

+

+

rEr

h

a

c

b

e

e

Er

Gaussian surface at top


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25

Example

Infinite plane sheet of charge, with surface charge density q

A

Gauss law qd = AE0

EAdba

2=+= AE

So qEA =02 00 22

==

A

qE

Example

Find the electric field inside and outside a spherical shell of

charge q.

Answer:

Due to spherical symmetry, the direction of E is along

radial direction andEis uniform on the sphere.

For Gaussian surface S2 : qd = AE0

or2

0

2

204

14

r

qEqEr

== ,

which gives the first shell theorem.

For Gaussian surface S1 : qd = AE0 , where q= 0 .

0042

10 == EEr ,

which gives the second shell theorem.

Example

Find the electric field E(r), inside and outside a uniformly

charged insulating sphere, with total charge Qand radiusR.

Answer:

In order to findEwith different r, we apply the Gauss law.

The volume charge density is given by

= =Q

V

Q

R4

3

3

.

r1

r2

S1

S2

Spherical shell of

Charge q

Spherical Gaussian surfaces

rR

S1S2

Er

Radial direction

Insulating

sphere with

charge inside


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26

The charge enclosed by the Gaussian surface S1 is

Q VQ

R

r Qr

Rencl= = =

'

4

3

4

33

33

3.

From the spherical symmetryEhas the same value on the surface S1.

enclQd = AE0 or 33

2

0 )4(R

QrrE =

That is, rR

QE

3

04

1

= , for r < R (E r).

At the surface of sphere, r = R, hence, we obtain EQ

R=

1

4 02

.

When r > R, the Gaussian surface S2 encloses all the charges.

Q Qencl =

02

4E r Q =

Hence, we obtain2

04

1

r

QE

= , which is the result of coulomb s law.

Example

Calculate the electric field between the metal plates of a

parallel-plate capacitor. Each plate has a charge per unit

area of magnitude .

Answer:

The electric field0

=E .

E

r

rR

QE

3

04

1

=

2

0

4

1

R

QE

=

2

04

1

r

QE

=

R


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27

Example

Calculate the electric field at a point distanced rfrom the charge Q, which is enclosed by a

metallic spherical shell.

Answer:

For 0 < r

hkpho electricity

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