1

PHYS113 Electricityand Electromagnetism

Semester 2; 2002

Professor B. J. Fraser

This TRACE image was, I believe, taken during the windowincluding the large emissions of mid-July 2000. It does notillustrate a point about Satellite Anomalies, but is a lovely andfantastic image – like something for inclusion in Walt Disney’s“Fantasia”. It looks as if it could be a cosmic conductor risingfrom the surface of the Sun, ready to conduct some Wagnerianpiece, perhaps the “Ride of the Valkyries”.

This TRACE image was, I believe, taken during the windowincluding the large emissions of mid-July 2000. It does notillustrate a point about Satellite Anomalies, but is a lovely andfantastic image – like something for inclusion in Walt Disney’s“Fantasia”. It looks as if it could be a cosmic conductor risingfrom the surface of the Sun, ready to conduct some Wagnerianpiece, perhaps the “Ride of the Valkyries”.

2

1. Electric Charge

What is charge? 700 BC - Greeks write of effects of

rubbing amber (Electrum) 1600’s - Gilbert shows

electrification is a general phenomenon

1730 - C. Dufay concludes “there are 2 distinct Electricities”

1750 - Ben Franklin shows +ve & -ve charges

Electrostatics involves the forces between stationary charges.

Charge is a basic atomic property forces between electrons & nuclei unlike charges attract like charges repel

3

Transfer of Charge

Charge transfer touching charge sharing

(conduction) only the electrons move Can appear as though positive

charge has moved

Unit of charge: Coulomb, C 1 Coulomb = 1 Ampere

second Electronic charge, e = 1.602 x

10-19 C i.e. 1 C = 6.3 x 1018 electrons a small number!!

++

++ +++ + +

4

Conservation & Quantisation

Charge is always conserved it cannot be created or

destroyed Charge only comes in fixed

packets the packet size is ± e It cannot wear off The light from distant quasars

(billions of years old) shows evidence of exactly the same atomic charge.

5

Forces Between Charges

Coulomb’s Law 1785: Coulomb experimentally

determines force law between 2 charged point sources, q1 and q2.

Thus:

where k = 8.99 x 109 N m2 C-2

Electric force has direction (vector)

Hence Coulomb’s Law is:

F12 is the force on q1 due to q2

r12 is a unit vector from q2 to q1 along the line that joins them.

212 1

qqFr

F

221

rqq

kF

12221

12 r

rqq

kF

6

Hints for Problem Solving

Draw a clear diagram Forces are vectors

include coordinates i.e. Fx and Fy or i and j components add vectorially

Shortcuts due to symmetry?

Example: Electric Forces in a Plane

Calculate the forces on q1 and q3q3

q1 q2

-3 nC

+2 nC +2 nC2.0 m

2.0 m

7

Solution: Forces in a plane

Force on q1

This is due to q2 and q3q3

q1 q2

F13

F12

F1

jNiN

ji

jr

qi

r

qqk

rr

qqr

r

qqk

FFF

ˆ10x5.13ˆ 10x0.9

ˆ2

10x3ˆ2

10x210x210x9

ˆˆ

ˆˆ

99

2

9

2

999

213

32

12

21

13213

31122

12

21

13121

N10x16.2 9

221

yx FFF

j

i

8

Solution: Forces in a plane

Force on q3

This is due to q2 and q3

Find magnitude as before

q3

q1 q2

F31 F32

F3

jNiN

jNiNjN

ji

j

rr

qr

r

qqkF

ˆ 10x3.18ˆ10x8.4

ˆ10x8.4ˆ10x8.4ˆ 10x5.13

ˆ sinˆ cos22

10x2

ˆ2

10x2

10x310x9

ˆˆ

99

999

22

9

2

9

99

32232

2312

31

133

j

i

9

2.The Concept of the Electric Field

Why is there a force between charged particles?

How does each particle know that the other one is there?

What happens in space between charged particles?

This is an example of an action-at-a-distance force.

E.g. Gravitation, Magnetism These forces are described in

terms of a field in space surrounding the particle or object.

10

Electric Field Strength

Test an invisible force field? See if a test object experiences a force!

Test for an electric field by measuring force experienced by a positive test charge.

We know that E F and: E 1/q0

where q0 = charge of test charge E = electric field, N C-1

Hence: and since:

Then:

Electric field seen by q0 due to q.

0qF

E

++

+++

+ +++

+++

Charged Object

Q Electric Field

Region+

Test Charge

q0

0qF

E

F

rr

qqkF ˆ2

0

rrq

kE ˆ2

11

Electric Field Lines

Electric field strength is a vector quantity.

Much easier to represent using vectors pointing in field direction - electric field lines.

Concept due to M. Faraday “lines of force” Electric Field lines point away

from positive charges Field lines point in the

direction of the force or electric field

Density (spacing) of field lines depends upon magnitude of E.

Field lines never intersect.

12

Electric Fields in Nature

All charges (fixed & moving) produce an electric field that carries energy through space at the speed of light.

Field Description Strength(NC- 1=Vm- 1)

I nterplanetary space 10-3 – 10-2

At Earth’s surf ace in clearweather

100 - 200

I n a thunder storm 103

Electrical breakdown of dryair

3 x 106

Van der Graaff generator 106

Fermilab accelerator 1.2 x 107

Atom at electron orbit radius 109

13

Field Due to Point Charges

Electric fields add vectorially:E = E1 + E2 + E3 + …

Thus:

Worked Example Find the electric field at point A for the dipole shown

Field at A due to q1.

i

i

i

i rrq

kE ˆ2

j

i

q1 = +12 nC q2 = -12 nC

6 cm 4 cm

A

+ -

1-422

99

21

11

NC ˆ 10x0.3ˆ10x6

10x1210x9

ˆ

ii

rrq

kE A

14

Field Due to Point Charges

Field at A due to q2

Total electric field at A:

No component in the j direction Example of an electric dipole Often found in nature (e.g. molecules) For more: See Section 21.11

1-422

99

22

12

NC ˆ 10x8.6ˆ10x4

10x1210x9

ˆ

ii

rrq

kE A

1-4 NC ˆ 10x8.9 iE A

+q1 - q2

A

EA1

EA2

15

Field due to a line segment

Charge, Q, distributed uniformly along length, L, with charge density:

= Q/L

Worked Example What is the electric field at a

distance R from a rod of length 2L carrying a uniform charge density,?

Consider an infinite collection of charge elements, dQ.

L

-L

y

dydE

dE

PO

r

r

R

dQ

j

i

16

Field due to line segment (contd)

Centre rod at origin For every charge at + y, there is

another corresponding charge at -y Thus, fields in j component add to 0.

rrdy

k

rrdQ

kE

L

L

L

L

ˆ

ˆ

2

2

But = Q/L

and thus dQ = dy

i

i

i rR

i cos j sin-i cos

2322

3

2

22

L

L

L

L

L

L

L

L

L

L

dyyR

Rk

dyrR

k

rdy

k

rdy

krdy

kE

cos= R/r

r2 =(R2 + y2)

Can you do this integral?

17

Solution to Field Due to a Line Segment

The solution to the field due to a line segment is:

So, what is the big deal? Well, what happens if L >> R?

i 2

22 RLR

LkE

i 2Rk

E

Thus, the field from a line charge is proportional to 1/R and not 1/R2.

18

Field Due to a Surface

Consider a charge Q uniformly distributed across surface of area A

Surface charge density is: = Q/A

Worked Example Find the electric field at distance R

from an infinite plane sheet with surface charge density

P

R r

L

dx

x

j

i

kDivide the sheet into an infinite collection of line segments, L, long and, dx, wide

19

Field Due to a Surface

Charge on each strip: dQ = dA = L dx

Charge per unit length: = dQ/L = dx

From previous example, each strip sets up electric field:

E = 2k/r = 2 k dx/r

Summing for all the strips:

dxxR

Rk

xR

dx

xR

Rk

rdx

k

rdx

kE

22

21222122

2

k 2

k sin

2

i cos-k sin

But i components

sum to 0

Can you do this integral?

20

Solution to Field Due to a Surface

The solution to the field due to a surface is:

So, what’s the big deal this time? How does the field vary with R?

k 2 kE

Thus, the field from a surface in independent of the distance R!

21

Particles in an Electric Field

A particle of charge, q, in an electric field, E, experiences a force:

E = F/q

F = qE = ma The particle accelerates at a +ve particle moves in direction of E I.e. from +ve to -ve charge regions. Thus an electron will be deflected

toward a +ve charged plate as its moving past it.

Examples: operation of CRT’s, TV tubes, etc.

+ + + + + + + + +

- - - - - - - - -

e-

screen

22

Particles in an Electric Field

Worked Example An electron in near-Earth space is

accelerated Earthward by an electric field of 0.01 NC-1. Find its speed when it strikes air molecules in the atmosphere after travelling 3 Earth radii (19 000 km).

The electron experiences a force:F = ma = qE

a = qE/m

For motion at constant acceleration:

v2 = u2 + 2as = 2as

v = 2.6 x 108 ms-1

i.e. 0.8 x speed of light

21

31

7219

10x1.910x9.11010x6.122

mqEs

v

23

Fishnets and Flux: The Gaussian Surface

Consider a fishnet with water flowing through it.

The rate of flow through net is the flux.

w = vA

v = velocity of flow A = area of net

If the net is angled at to the flow:w = vA cos

In vector form:w = v A

where the direction of A is normal to net

A

24

Defining Electric Flux

For an irregular shape, area A is sum of infinitesimal elements dA.

Thus, summing over 2-D surface S:

Now, replace water with electric Now, replace water with electric field, i.e. there is no physical field, i.e. there is no physical motionmotion..

The electric flux through a surface of area A is:

The electric flux through a surface The electric flux through a surface is proportional to the number of is proportional to the number of field lines passing through a field lines passing through a surfacesurface.

If the fishnet is formed into a closed shape (e.g. lobster pot) its called a Gaussian surface.

S

w Adv

S

E AdE

25

The Gaussian Surface

The total electric flux (number of field lines) passing through this surface is:

where A points perpendicularly away from each element dA.

If the flux in one side is the same as that out then the total flux is zero.

If there is no net charge inside a Gaussian surface the electric flux through it adds to zero.

Gaussian surfaces are imaginary Gaussian surfaces are imaginary constructions!constructions!

AdEE

EdA

26

3. Gauss’ Law

Consider a point charge surrounded by a Gaussian sphere.

The electric field is:

where 0 = permittivity of free space

= 8.85 x 10-12 C2 N-1 m-2 The electric flux through the

surface is then:

rrq

rrq

kE ˆ4

1ˆ 2

02

22

0

20

44

4

rr

q

dAr

q

dAE

AdE

sphere

S

S

E

Gaussian surface

Flux lines

Radial field lines are always normal to sphere

27

Gauss’ Law in General

Example: Coulomb’s Law from Gauss’ Law

What is the electric field due to a point charge?

Consider a Gaussian sphere of radius r centred on a charge q.

Only interested in radial field direction.

All fields in other directions cancel.

Gauss’ Law states that the electric flux through any closed surface enclosing a point charge Q is proportional to Q.

The surface need not be centred on Q and can be any shape.

iiencl

encl

S

E

qQ

QAdE

0

28

Coulomb from Gauss

Consider surface elements dA If E is along dA then:

E.dA = E dA cos(0º) = E dA Hence:

From Gauss’ Law:

Rearranging:

Which, since F = qE, gives Coulomb’s Law, where we put E radially outward from the charge q.

2 4

rE

dAE

dAEAdE

S

SS

0

2 4 qrE

20 4 rq

E

Gaussian surface

E

+q

dAr

29

Applications of Gauss’ Law

Use Gauss’ law to find electric flux or field in a symmetrical situation.

Shape of the Gaussian surface is dictated by the symmetry of the problem.

Worked Example Find the electric field due to an

infintely long rod, positively charged, of constant charge density, .

P

+ + + + + + + + + + + +

30

Electric Field of Long Rod

Consider motion of a test charge.

Only field lines radially away from the rod are important.

Consider a Gaussian cylinder around part of the rod, radius r, height, h.

Total flux through cylinder is:

But, @ top & bottom E dA E.dA=0

For the side E is parallel to dAE.dA=E dA

+ +

+ +

+ +

+ +

+ +

dA

dA

h

rdA

sidebottomtop

total AdEAdEAdE

31

Field due to a Long Rod

sidesidetotal

E dAEdAEAdE

rhπ

dAside

2

h height, ofcylinder a of area side

00

2

hqrhE

Gauss’ Law

Rod Charge Density

rE

0 2

Compare this with our previous

result.E varies as 1/R

32

Charged Spherical Shell

Worked Example E-field inside & outside a charged

spherical shell (e.g. plane, car)Outside the shell Use a Gaussian sphere of radius r

centred on the shell. Then:E.dA = E dA (since E ||dA)

2

0

4 rE

dAE

AdEQ

r

R

E2

0 4 rQ

E

33

Inside a Charged Spherical Shell

Inside the shell r < R so consider a Gaussian

sphere inside the shell. no net charge enclosed by sphere Qencl = 0, so

Inside the shell the field is zero: a physically important result.

0 0

QE

E

No field inside the shellFaraday Cage!!

34

Solid Polarisable Sphere

Worked Example What is the electric field outside &

inside a solid nonconducting sphere of radius R containing uniformly distributed charge Q.

Outside the sphere: r > R consider spherical Gaussian

surface0

0

enclE

QAdE

20 4 r

QE encl

As before+

35

Inside the Solid Sphere

Inside the sphere r < R Charge enclosed by a Gaussian

sphere of radius r<R is:

3

3

3

3

34

34

volumedensity charge

Rr

Q

rR

Q

Q

r

R

36

Field Inside Solid Charged Sphere

Hence, from Gauss’ Law:

The same behaviour is found for other forces, e.g. gravity.

20

3

3

20 4

1

4 rRr

Qr

QE

rR

QE

4 30

20 4 RQ

E

rR

2

1~r

r~

37

Behaviour of Charges & Fields Near Conductors

The electric field is zero everywhere inside a conductor.

Electrons move to create an E field which opposes any external field.

Free charges move to the outside surfaces of conductors

A result of Gauss’ law.

The electric field near a conductor is perpendicular to its surface.

A parallel component would move charges and establish an electric field inside.

38

Why Doesn’t My Radio Work

The electric field outside a charged conductor is:

where:

Proof Consider a Gaussian cylinder

straddling the conductor’s surface.

Closed hollow conductors admit no electric field

EM shielding “Faraday Cages” Car Radios and biomagnetics

0E

areaQ

AEQ

dAE nnC 0

00

AQ

En

En

dA

E = 0

++

+ ++ +++

+++++

+++++++

39

Importance and Tests of Gauss’ Law

Coulomb’s law experimental evidence of Gauss’s law

1/r2 law is the key prediction Gauss’ law is so basic that its

essential to test its validity Tests of F 1/r2±Robinson 1769 = 06

Cavendish 1773 0.02

Coulomb 1785 0.10

Maxwell 1873 5 x 10-5

Plimpton & Lawton 1936 2 x 10-9

Williams, Faller &Hill

1971 3 x 10-16

40

4. Electric Potential: Technology Can’t Live Without It!

Technology relies on using energy associated with electrical interactions

Work is done when Coulomb forces move a charged particle in an electric field.

This work is expressed in terms of electric potential (energy)

Electric potential is measured in Volts.

Basic to the operation of all electric machines and circuits.

41

Mechanical Analogue

In mechanics

Work done in moving from point a b

results in a change in potential energy:

W a b= Ua - Ub

When W a b > 0

Ua > Ub

e.g. a mass falling under gravity

b

a

ba SdFW

42

What is Electric Potential?

In electricity

Consider a test charge q0 moving with respect to a charge, q, fixed at the origin.

The work done is:

When integrated along the path and thus:

This is the change in electric potential energy, for a charge q0 moving from a b.

b

a

ba SdEqW 0

b

a

SdEqU 0

43

Electric Potential Energy

Since:

By definition, a charge infinitely far away has zero potential energy.

The electric potential energy between 2 charges is then:

Since this is a scalar the total potential energy for a system of charges is:

b

a

SdFU

ba

b

a ab rrqq

rSdqq

U11

4 4 0

02

0

0

rqqk

rqq

rU 00

0 41

13

31

12

21

rqq

rqq

kU

44

Uranium Nucleus Example

Worked Example

Calculate the electrostatic potential energy between 2 protons in a Uranium nucleus separated by 2 x 10-15 m.

J10~

10 x 210 x 1.6

10 x 9.0

13

15-

219-9

21

rqq

krU

45

Electric Potential

Definition: Electric potential is potential

energy per unit charge:

where U(r) is the potential energy of test charge q0 due to a charge distribution.

V(r) is a property of the charges producing it, not q0.

Volt = unit of electric potential

1 V = 1 volt = 1 J/C Note also that 1 V/m = 1 N/C

0q

rUrV

46

Potential & Charge Distribution

For a single point charge; q, a distance r away, the electric potential is:

Potential is zero if r = For a collection of charges:

For a charge distribution:

r

qrqkrV

0 4

n

i i

i

rq

kV1

rdq

kV

47

Electric Potential Difference

Difference in electric potential for a charge q between points a and b.

i.e potential difference can be expressed as a path-independent integral over an electric field.

All charge distributions have an electric potential

The potential difference Va - Vb is the work/unit charge needed to move a test charge from a b without changing its kinetic energy.

abab rr

kqVVV11

b

a

ab sdFqq

UUV

00

1

EdsdEVb

a

For a uniform For a uniform field, d || field, d || EE

48

The electron volt

For the definition of volt, 1J of work is needed to move 1 C of charge through a potential difference of 1V

A more convenient unit at atomic scales is the electron-volt:

The energy gained by an electron (or proton) moving through a potential difference of 1 volt:

1 eV = (1.6 x 10-19 C)(1 V)= 1.6 x 10-19 J

Not an SI unit but a very useful one!

Worked Example In a hydrogen atom the e- revolves

around the p+ at a distance of 5.3 x 10-11 m. Find the electric potential at the e- due to the p+, and the electrostatic potential energy between them.

49

Worked Examples

Electric potential due to proton:

Electrostatic p.e. is given by:

p+

e -

r

V 27

10 x 5.310 x 1.610 x 9

11-

19-9

rq

krV

A very A very simplistic simplistic picturepicture

J10 x 4.3 2710 x 1.6 18-19-

12

2112

pVe

rqq

kU

50

Forces on Charged Particles

Worked Example In a CRT an electron moves

0.2 m in a straight line (from rest) driven by an electric field of 8 x 103 V/m. Find:

(a) The force on the electron.(b) The work done on it by the E-field.(c) Its potential difference from start to finish.(d) Its change in potential energy.(e) Its final speed.

51

Worked Examples

(a) Force is in opposite direction to the E-field, magnitude:

(b) Work done by force:

(c) Potential difference is defined as work/unit charge:

Alternatively (e- opposite to p+):

N 10 x 3.110 x 810 x 1.6 15-319- qEF

J 10 x 6.20.2 10 x 3.1 16-15- FsWork

V 10 x 1.6 10 x 1.610 x 6.2 3

19-

16-

qW

V

V 10 x 1.6 2.010 x 8

33

0

EddxEsdEVdb

a

52

Worked Examples

(d) Change in potential energy:

(e) Loss of PE = gain in KE = ½mv2

donework

J 10 x 2.6-

10 x 1.610 x 1.6-16-

319-

0

0

Vq

sdEqUb

a

1-7

31-

16-

ms 10 x 2.4

10 x 9.110 x 2.62

2

mKE

v

53

Worked Examples

Worked Example A proton is accelerated across a

potential difference of 600 V. Find its change in K.E. and its final velocity.

By definition, 1 eV = 1.6 x 10-19 J. Acceleration across 600 V Proton gains 600 eV.K.E. = 600(1.6 x 10-19) = 9.6 x 10-17 J

Final velocity is:

If it started from rest

1-5

27-

17-

ms 10 x 3.4

10 x 1.710 x 9.62

v

54

Equipotentials

Regions of equal electric potential may be joined by contour lines.

These are equipotentials. In 3-D these can form

equipotential surfaces where the potential is the same at each point on the surface.

Field lines and equipotentials are always perpendicular.

No work is done in moving a charge along an equipotential surface because there is no change in potential.

The surface of a conductor is an equipotential since charge is uniformly distributed across the surface of conductors.

55

Obtaining E from the Electric Potential

Recall:

If the direction s is parallel to E for infinitesimal elements ds from a to b

Electric field is the rate of change of potential V in the direction ds.

In 3-D space we use x, y & z components to express in terms of partial derivatives.

b

a

b

a

ab sdEdVVV

b

a

b

a

sdEdV

dsEdV

dsdV

E

, ,dzdV

EdydV

EdxdV

E zyx

56

Vector Notation

In vector notation:

Where is the gradient operator.

V

Vz

ky

jx

i

kzV

jyV

ixV

E

ˆˆˆ

ˆˆˆ

57

Vector Notation Example

Worked Example A potential distribution in space is

described by:V = Axy2 - Byz

where A and B are constants. Find the electric field.

-By

Bz-2Axy

Ay 2

dzdV

dydVdxdV

VE

kByjBzAxyiAyE ˆ ˆ 2ˆ 2

58

Potential Due to Charge Distributions

If E is known, use:

If E is not known, use:

for continuous charge distributions:

b

a

ab sdEVVV

n

i i

i

rq

kV1

rdq

kV

59

Parallel Plates

Worked Example Two parallel metal plates have an

area A = 225 cm2 and are l=0.5 cm apart, with a p.d. of 0.25 V between them. Calculate the electric field.

This is obvious from the definition of units of electric field = V/m.

0V 0.25V

0.1V 0.2V

x =0 x =0.5m

ds

El

dxE

dxE

sdE

VVV

l

l

b

a

rightleft

0

0

1-Vm 505.025.0

lV

E

60

Worked Example Find the electric potential and

electric field along the axis of a uniformly charged disc of radius R and total charge Q.

Consider the disc divided into rings of radius,r, width, dr.

Uniformly Charged Disc

dy

(y2 + x2)½

P

yR

dq

xy

krdq

kV22

x

61

Uniformly Charged Disc (contd)

For the ring shown:

For the total potential we integrate over all rings:

By definition of charge density:

For the ring: dq = 2 y dy

22

xy

dqkdV

R

xy

dqkV

022

discfor area 2RQQ

xxR

xy

xy

dyyV

R

R

22

0

022

0

022

0

2

2

4

2

62

Uniformly Charged Disc

The field is only in the x direction.

dxdV

EE x

12

2

21

222

0

21

222

0

xR

xR

Q

xxRR

Qx

E

63

Why Sparks Occur at Pointed Tips

Recall: Conducting objects contain zero electric field.

Charge resides on outer surface This surface is an equipotential. Equipotential surfaces outside the

conductor are parallel to its surface.

For curved conductors, surface charge density:

Hence: (radius of curvature)

Small radius implies and E are large

E.g. at points and tips

r1

rE

1

64

St. Elmo’s Fire

Regions of strong E-field Ionisation of air

Corona discharge greenish glow (St. Elmo’s Fire) E > 3 x 106 V/m

Ionisation Current flow Carry away excess charge

Lightning conductors

Do not attract lightning

Introduce a lower potential difference region close to clouds.

+++++

+

+

+

+

+

+

++++++++

++++

65

Uses in Technology

Accelerators (1929)(Giancoli Section 44.2, p1115) Van der Graaf HV Accelerator Works because E-field inside

Gaussian sphere is zero 1m sphere 3 x 106 V Up to 20 MV producedPrecipitators (See Figure shown) Remove dust and particles from

coal combustion -ve wire @ 40 - 100 kV E-field particles to wall > 99% effective.Photocopiers (1940) (Giancoli Example 21.5, p555) Image on +ve photoconductive

drum Charge pattern -ve toner pattern Heat fixing +ve paper.