Last Time…

•Electrostatic Phenomena•Coulomb’s Law

12212

2112 ˆ

4

1r

r

qqF

o

Text Reference: Chapter 22.1 through 22.3Examples: 22.1 – 22.5

q1•Superposition

qFtotal = F1 + F2 + ...q2

F1

F2

F

Today...

•Define Electric Field in terms of force on a test charge

•How to think about fields

•Electric Field Lines

•Example Calculation: Electric Dipoles

1Text Reference: Chapter 22.4 through 22.6

Examples: 22.5, 6, 7, 8, 9, 10, and 11

Lecture 2, Act 1Two balls of equal mass are suspended from the ceiling withnonconducting wire. One ball is given a charge +3q and the other is given a charge +q.

g+q+3q

Which of the following best represents the equilibrium positions?

(b)

+3q+q

(a)

+3q+q

(c)

+3q +q

Lecture 2, Act 1

+3q

(b)

+3q+q

(a)

+3q+q

(c)

+q

Which best represents the equilibrium position?

•Remember Newton’s Third Law!•The force on the +3q charge due to the +q charge must be equaland opposite to the force of the +3q charge on the +q charge

•Amount of charge on each ball determines the magnitude of theforce, but each ball experiences the same magnitude of force

•Symmetry, therefore, demands (c)

P.S. Knowing the form of Coulomb’s law you can write two equations with two unknowns (T and )

What is a Field?(from Lect. 1 notes)

A FIELD is something that can be defined anywhere in space

•It can be a scalar field (e.g., Temperature field)•It can be a vector field (e.g., Electric field)

•Fields represent physical quantities

7782

8368

5566

8375 80

9091

7571

80

72

84

73

82

8892

77

8888

7364

A Scalar Field (from Lect. 1 notes)

These isolated temperatures sample the scalar field(you only learn the temperature at the point you choose,

but T is defined everywhere (x, y) )

77

82

8368

5566

83

75 80

9091

7571

80

72

84

73

57

8892

77

5688

7364

A Vector Field (from Lect. 1 notes)

It may be more interesting to know which way the wind is blowing...

That would require a vector field(you learn both wind speed and direction)

The Electric Field

•A simple, yet profound observation- The net Coulomb force on a given charge is alwaysproportional to the strength of that charge.

q

q1

q2

F1

F

F2

F2F1F = +

22

222

1

11

0

ˆˆ

4 r

rq

r

rqqF

test charge

- We can now define a quantity, the electric field, whichis independent of the test charge, q, and depends only onposition in space:

q

FE

The qi are the sourcesof the electric field

The Electric Field

q

FE

With this concept, we can “map” the electric fieldanywhere in space produced by any arbitrary:

“Net” E at origin

+

F

These charges or this charge distribution“source” the electric field throughout space

ii

i rr

qE ˆ

4

12

0

++

+ +

+

+-

-

--

-

Bunch of Charges

rr

dqE ˆ

4

12

0

+ ++ ++ +++++ +

Charge Distribution

Example: Electric Field

The x and y components of the field at (0,0) are:

1) Notice that the fields from the top-rightand bottom left cancel at the origin?

2) The electric field, then, is just the fieldfrom the top -left charge. It points awayfrom the top-left charge as shown.3) Magnitude of E-field at the origin is:

E2a2

kq=

What is the electric field at the origin due to this set of charges?

x

y

+q+q

+q

a

a

a

a a

2a

Ex 2a2

kq= cos

2

12a2

kq=

Ey 2a2

kq= sin-

2

12a2

kq= -

Example: Electric Field

222 a

qQkQEF xx

222 a

qQkQEF yy

Now, a charge, Q, is placed at the origin. What is the net force on that charge?

+q

x

y

+q

+q

a

a

a

a a

2a

Q

Note: If the charge Q is positive, the force will be in the direction of the electric fieldIf the charge Q is negative, the force will be against the direction of the electric field

F is

F is

2

1

2 2a

qkEy

2

1

2 2a

qEx k

Let’s Try Some Numbers...

2

If q = 5C, a = 5cm, and Q = 15C.

Then Ex = 6.364 106 N/C

and Ey = -6.364 106 N/C

Fx=95.5 N Fy=-95.5 N

Fx = QEx and Fy = QEy

So... and

We also know that the magnitude ofE = 9.00 106 N/C

We can, therefore, calculate the magnitude of F

F = |Q| E = 135 N

x

y

+q+q

+q

a

a

a

a a

2a

Lecture 2, Act 2Two charges, Q1 and Q2, fixed along the x-axis asshown produce an electric field, E, at a point(x,y)=(0,d) which is directed along the negativey-axis.

Q2Q1 x

y

Ed- Which of the following is true?

(a) Both charges Q1 and Q2 are positive

(b) Both charges Q1 and Q2 are negative

(c) The charges Q1 and Q2 have opposite signs

Lecture 2, Act 2Two charges, Q1 and Q2, fixed along the x-axis asshown produce an electric field, E, at a point(x,y)=(0,d) which is directed along the negativey-axis.

- Which of the following is true?

E

Q2Q1

(a)

Q2Q1

(c)

E

Q2Q1

(b)

E

Q2Q1 x

y

Ed

(a) Both charges Q1 and Q2 are positive

(b) Both charges Q1 and Q2 are negative

(c) The charges Q1 and Q2 have opposite signs

Reality of Electric Fields

•The electric field has been introduced as a mathematicalconvenience, just as the gravitational field of Physics 111

•There is MUCH MORE to electric fields than this!

IMPORTANT FEATURE: E field propagates at speed of light

• NO instantaneous action at a distance (we will explain this whenwe discuss electromagnetic waves)

• i.e., as charge moves, resultant E-field at time t depends uponwhere charge was at time t - dt

• For now, we avoid these complications by restricting ourselves to situations in which the source of the E-field is at rest.

(electrostatics)

Mathematics of Fields

•Scalar Fields–Number associated with each point in space–May be time dependent (future)–Expressed as a function g(x, y, z, t) x

z

y

Pick a coordinatesystem…

This specifiesthe formula:

g(x, y, z, t)

G(x, y, z, t)

•Physical Fields–Obey a simple rule–Created by sources–Continuous and well behaved–What field looks like depends on rule and sources

–Vector associated with each point in space–May be time dependent (future)

–Expressed as a function G(x, y, z, t)

•Vector Fields

Ways to Visualize the E FieldConsider the E-field of a positive point charge at the origin

+

+ chg

field lines

+ chg

vector map

+

Rules for Vector Maps

•Direction of arrow indicates direction of field•Length of arrows local magnitude of E

+ chg

+

•Lines leave (+) charges and return to (-) charges•Number of lines leaving/entering charge amount

of charge•Tangent of line = direction of E•Local density of field lines local magnitude of E

• Field at two white dots differs by a factor of 4 since r differs by a factor of 2•Local density of field lines also differs by a factor of 4 (in 3D)

Rules for Field Lines

+ -

NOTE– from Lecture 3, ACT 1

Ex Solution: Draw some field lines according to our rules.

•Consider a dipole with the y-axis as shown.

+Q

x

a

a-Q

a

2a

y

-Which of the following statements about Ex(2a,a) is true?

(a) Ex(2a,a) < 0 (b) Ex(2a,a) = 0 (c) Ex(2a,a) > 0

Field Lines From Two Opposite ChargesDipole

Dipoles are central to our existence!

Molecular Force Model Basis of Attraction

Ion-dipole Ion and polar molecule

Dipole-dipole Partial charges of polar molecules

London dispersion

Induced dipoles of polarizable molecules

The Electric Dipolesee the appendix for further information

x

y

a

a

+Q

rE E

-Q

What is the E-field generated bythis arrangement of charges?

Calculate for a point along x-axis: (x, 0)

Ex = ??

Symmetry

Ex(x,0) = 0

Ey = ??

Electric Dipole Field Lines

• Lines leave positive chargeand return to negative charge

What can we observe about E?

• Ex(x,0) = 0 • Ex(0,y) = 0

• Field largest in space between two charges

• We derived:

... for r >> a,

Field Lines From Two Like Charges

• There is a zero halfwaybetween the two charges

• r >> a: looks like the fieldof point charge (+2q) at origin

3

Lecture 2, ACT 3• Consider a circular ring with total charge +Q.

The charge is spread uniformly around the ring, as shown, so there is λ = Q/2R charge per unit length.

• The electric field at the origin is

R

x

y++++

+ +

+++

++

+ + + + ++++++

+

(a) zeroR

2

4

1

0

(b) (c)

• The key thing to remember here is that the total field at the origin is given by the VECTOR SUM of the contributions from all bits of charge.

• If the total field were given by the ALGEBRAIC SUM, then (b) would be correct. (exercise for the student).

• Note that the electric field at the origin produced by one bit of charge is exactly cancelled by that produced by the bit of charge diametrically opposite!!

• Therefore, the VECTOR SUM of all these contributions is ZERO!!

Electric Field inside a Conductor

• A two electron atom, e.g., Ca– heavy ion core– two valence electrons

• An array of these atoms– microscopically crystalline

– ions are immobile

– electrons can move easily

• Viewed macroscopically:– neutral

There is never a net electric field inside a conductor – the free charges always

move to exactly cancel it out.

2+

2+ 2+ 2+ 2+

2+ 2+ 2+ 2+

2+ 2+ 2+ 2+

2+ 2+ 2+ 2+

Summary

• Define E-Field in terms of forceon “test charge”

• How to think about fields

• Electric Field Lines

• Example Calculation: Electric Dipole

Reading Assignment: Chapter 23.1 through 23.2Examples: 23.1 and 23.4

Appendix A:Other ways to Visualize the E Field

Consider a point charge at the origin

+

+ chg

Field LinesEx, Ey, Ez as a function of (x, y, z)Er, E, E as a function of (r, , )

Graphs

x

Ex(x,0,0)

Appendix A

(a) (c)(b)

0 2

Er3A

0 2

Er

0 2

Er

Fixedr > 0

3B

Fixedr > 0

0 2

Ex

0 2

Ex

0 2

Ex

Consider a point charge fixed at the origin ofa coordinate system as shown.

–Which of the following graphs bestrepresent the functional dependence ofthe Electric Field for fixed radius r?

xQ

y

r

Appendix “ACT”Consider a point charge fixed at the origin ofa coordinate system as shown.

– Which of the following graphs bestrepresent the functional dependence ofthe Electric Field for fixed radius r?

xQ

y

r

(a) (c)(b)

0 2

Er3A

0 2

Er

0 2

Er

Fixedr > 0

• At fixed r, the radial component of the field is a constant, independent of !!

• For r>0, this constant is > 0. (note: the azimuthal component E is, however, zero)

Appendix “ACT” Consider a point charge fixed at the origin ofa coordinate system as shown.

–Which of the following graphs bestrepresent the functional dependence ofthe Electric Field for fixed radius r?

xQ

y

r

(a) (c)(b)

3B

Fixedr > 0

0 2

Ex

0 2

Ex

0 2

Ex

• At fixed r, the horizontal component of the field Ex is given by:

Appendix B: Electric Dipole

What is the E-field generated bythis arrangement of charges?

Calculate for a point along x-axis: (x, 0)

Ex = ??

Symmetry

Ex(x,0) = 0

Ey = ??

x

y

a

a

+Q

rE E

-Q

Electric Dipole

x

y

a

a

+Q

-Q

Coulomb ForceRadial

E

Now calculate for a pt along y-axis: (0,y)

Ex = ?? Ey = ??

What is the Electric Field generated by this charge arrangement?

Electric Dipole

x

y

a

+Q

-Q

Case of special interest:(antennas, molecules)

r > > a

r

For pts along x-axis: For pts along y-axis:

For r >>a, For r >>a

a

Field Lines• Since Coulomb’s Law is an inverse square law, we can use the density of the field lines to represent the magnitude of the field!

–Consider the E-field produced by a single point charge

+ chg

+

Note: density of field lines is only proportional to the magnitude of the field as a result of E being inversely proportional to R2

This concept will become more useful (and physical) whenwe discuss Gauss’ Law in terms of electric flux!

–The field is spherically symmetric

1) magnitude of E-field depends only on R2) at radius R, magnitude = E = kQ/R2

Suppose N field lines at surface of sphere of radius RThen this is the density ofthe field lines at the surface: