ch18 - physics.umanitoba.ca · Electric Field q q 0 Test charge, q 0 P!F The force on the test charge at P is: The electric ﬁeld at P is deﬁned by: F= kqq 0 r2 r E= F q 0 = kq

http://www.physics.umanitoba.ca/~birchall/PHYS1030/1Friday, January 11, 2008

2Friday, January 11, 2008

Schedule up to midterm break


PHYS 1030, General Physics II

J. Birchall: 205 Allen, 474–6205 [email protected]

Office hours: Mondays 9:30 – 10:30, Tuesdays 1:30 - 2:30 Other times by appointment

Textbook: Cutnell & Johnson, 7th edition. Editions 6 and 5 also OK, although problem numbers sometimes differ.

Physics 1030 Lab Manual (2008) from bookstore

Scientific Calculator: mostly OK, but storage of notes and formulae or remote communication not allowed in tests and exams.

Web Pagewww.physics.umanitoba.ca/~birchall/PHYS1030


Administration of course

• Coordinator: Dr. J. Birchall, keeper of the marks,

lab exemptions (if 80% average in PHYS 1030 lab in

last 2 years), room 205, Allen building, 474-6205

• Labs: Dr. H. Kunkel, in charge of labs.

Reschedule tests, labs (during week of test or lab),

room 402G, Allen building, 474-9214

• Tutorial tests: Dr. W. Ens/Dr. J. van Lierop

• Miscellaneous questions, problems: Physics main

office, room 301, Allen building

• Other – see the course handout or me


Evaluation: Labs 20%Tutorial tests 10%Mastering Physics 5%Term Test 20% (Thursday, March 6, 7 – 9 pm)Final Exam 45% (April exam period)

Term test and final exam: Multiple choice, closed book, formula sheet provided. The final exam must be written to receive credit for the course.

Tutorial tests (4): Multiple choice, following a tutorial, closed book, no formula sheet. Must take the tutorial tests, even if you have a lab exemption.

Labs (5): Begin next week. No mark for labs if more than one lab missed without valid reason!

PHYS 1030, General Physics II


“Mastering Physics”

Web-based homework assignments. Instant feedback, hints

point you to the solution.

About 5 assignments for credit for the whole course.

Worth 5% of final grade.

Experience from PHYS 1020: if you are using Windows

Vista, use Firefox, not Internet Explorer.

You can use your Mastering Physics username and password

from PHYS1020, but must use the new access code.


• These notes:

www.physics.umanitoba.ca/~birchall/PHYS1030/lectures.html

• Publisher’s web site for worked problems, interactive problems, simulations...

www.wiley.com/college/cutnell

• “Crisis centre” – room 105, Allen building

More Help


Overview of the Course

Electricity and Magnetism• Electric and magnetic fields and forces, electromagnetism• Electric circuits

Light and Optics• Electromagnetic waves, light – light rays: optics, optical instruments – light as a wave: interference, diffraction

Special Relativity• Special nature of light speed – special relativity

Modern Physics• Quantum physics – particles as waves, waves as particles• The atom, the nucleus, radioactive decay


Chapter 18:

Electric Fields &

Forces


Chapter 18: Electric Forces and Fields

• Electric charge, conservation of charge, electric force

• Conductors, insulators, flow of charge

• Coulomb’s law

• Electric field, electric field lines

• Electric field inside a conductor

• Omit 18.9, Gauss’ Law


Unlike charges attract

Two types of charge, ‘+’ and ‘–’

Like charges repel


Charge is Conserved

Charge is a property of many elementary particles (electrons, protons...), which they carry in multiples of a basic unit, “e”.

The charge of the electron is -e and +e for the proton.

– As charge is contained on the elementary particles in a fixed amount, charge is conserved.– Charges can be moved from one place to another, but the total charge is unchanged.

Charging by friction (eg, static build-up in winter, charging a balloon)

– charges flow from one surface to another – net charge is unchanged! – the force between charges is given by Coulomb’s law


Charging by contact

(electrical conductor)→ charge of same

sign as on the rod


Charging by induction

→ charge of opposite sign as on the rod


(electrical insulator)

Charges contained on molecules of

the plastic


Electrical Conductors

– charges (electrons) are free to move inside the conductor– flow of charge is an electric current

Electrical Insulators

– charges do not move freely

Similarities with flow of heat...


Conduction of charge

Rate of flow of heat:

Rate of flow of charge:

Conduction of heat

I =!A

L!V

Q =kA

L!T

k = thermal conductivity

σ = electrical conductivity


Coulomb’s Law

The force between two charges: F =kq1q2

r2

k = Coulomb’s constant = 9!109 N.m2/C2

Charges, q1, q

2 in Coulombs (C)

= 1/4!"0

(Coulomb, electrostatic force)


Gravitational force

F =Gm1m2

r2masses always positive, force always attractive

but anti-matter...?

Electrostatic force

F =kq1q2

r2 charges positive or negative, force attractive or repulsive


Electric Field

q q0

Test charge, q0P

!F

The force on the test charge at P is:

The electric field at P is defined by:

F =kqq0

r2

r

E =F

q0=kq

r2

!E =!F

q0, or, !F = q0!E

E =kq

r2due to the charge q

= force per unit charge

Coulomb’s law


Electric field due

to many charges

Electric Field : !E =!F

q0

Each component of the force exerted on q0 is proportional to q0

! !E is the same for all values of q0


Electric Field: !E =!Fq0

, or !F = q0!E


The electric field at P is the vector sum of the electric fields due to the charge distributions at A and B


Electric field lines from a positive charge

The closer together the field lines, the stronger the field



Electric Dipole


Two positive charges

E = 0



Electric field lines ⇒ the direction of the

force on a positive test charge

• field lines point away from positive charges• point toward negative charges• the closer together the lines, the stronger the electric field


Field lines do not cross – they

represent the direction of the force

on a positive test charge. The force

cannot point in two directions at

the same place!

The number of field lines reaching

a charge should be proportional to

the charge – the larger the charge,

the greater the electric field and the

larger the number of field lines.

Drawing field linesIncorrect

Correct


Electric Field

!E =!F

q0, where !F is the force on a test charge q0

E =kq

r2at a distance r from a charge q

!0 = “permittivity of free space”

k =1

4!"0= 9!109 N.m2/C2

E =q

!0Abetween parallel charged plates

F =kqq0

r2, Coulomb’s law for force between charges q,q0

e= 1.6!10"19 C, magnitude of electron charge

(Newtons/Coulomb,

N/C)

= Coulomb constant

(from Coulomb’s law)


Prob. 18.27: A ball (mass = 0.012 kg) carries a charge of –18 !C. What electric field is needed to cause the ball to float above the ground?

What is the weight of the ball?

What magnitude and direction of electric field supplies an equal upward force?

mg

!F = q!E

q = –18 !C m = 0.012 kg


The story so far...

Charge is a property of certain elementary particles, such as electrons, protons. They each have a fixed amount of charge and so charge is conserved.

The effect of charge is the electrostatic (Coulomb) force. Like charges repel, unlike charges attract.

The force between charges is given by Coulomb’s law:

F = kqq0/r2,

charges q, q0 in Coulombs (C),

k = Coulomb constant = 9x109 N.m2/C2.


Coulomb’s Law, Electric Field

q q0

Test charge, q0P

!Fr

The force on the test charge at P is:

The electric field at P is defined by:

(Coulomb’s law)

!E =!F

q0

F =kqq0

r2

Electric field lines point away from

positive a charge, toward a negative

charge

E =kq

r2 (due to q)


Prob. 18C11: Three point charges are fixed to the corners of a square, one to a corner, in such as way that the net electric field at the empty corner is zero.

Do these charges all have

a) the same sign?

b) the same magnitude, but possibly different signs?


18.C9: On a thin, nonconducting rod, positive charges are spread evenly, so that there is the same amount of charge per unit length at every point.

On another identical rod, positive charges are spread evenly over only the left half, and the same amount of negative charge is spread evenly over the right half.

For each rod, deduce the direction of the electric field at a point that is located directly above the midpoint of the rod.

Pair off charges on equal sections of the rod equidistant from the midpoint.

Work out the direction of the field from the paired-off charges.


Where should q1 to placed so that the net force on it is zero?

Draw in the forces on q1 due to q2 and q3

Find where the forces are equal in magnitude and opposite in direction

When F = 0, the electric field due to q2 and q3 is also zero

q1

q2 q3L


Prob. 18.29/30: Where is the electric field equal to zero?

Let point P be a guess for where E = 0, a distance x from the charge at

the left.

q1 = –16 !C q2 = +4 !CL = 3 m

Px 3 – x

E = 0 ?


• Conductor contains charges that are free to move (electrons)

• At equilibrium, the charges are at rest

Electrical Conductor

E = 0

⇒ there can be no electric field inside the conductor, otherwise the

charges would be moving under the influence of the field and would not be in equilibrium.

The electric fields due to the rod and the charges on the surface cancel each other out inside the conductor


E = 0

The extra electrons are pushed to the surface of the conductor by the

repulsive Coulomb forces between them.

As the electrons end up at equilibrium (i.e. at rest) there can be no electric

field inside the conductor, otherwise the electrons would still be moving...

Put some extra electron charges inside the conductor

Electrical Conductor


E =q

!0A

Coulomb constant, k =1

4!"0

Uniform field inside a parallel plate capacitor

+q -q

Area A

!E

Uniform spacing of field

lines means constant

electric field


At equilibrium, electric field lines are made to hit the conductor at right angles – charges have moved around the surface until they reached an equilibrium in which there is no longer a field parallel to the surface to move them further.

Uniform spacing of field lines means constant electric field

E = 0

Distortion of field lines due to charges on surface of conductor

–e E=0


Equipotential SurfaceAs the electric field lines are at right angles to the conducting surface, no work is done in moving an electron charge around the surface

→ the electron has constant

potential energy on the surface

→ the surface is an “equipotential

surface” (more in chapter 19)

– equipotential surfaces are at right angles to electric field lines

Electric potentials are measured in volts (V).

–eE = 0


Equipotentials and electric field

V1 volts

V2 volts

90º

90º 90º

90º


Experiment 1 : Equipotential Lines

Sketch out equipotential lines by using a voltmeter to find places that lie at the same potential (voltage).

– one probe of the voltmeter is fixed in position, the other is moved around to locate points that give the same voltage reading on the voltmeter.

– all of the points giving the same reading on the voltmeter lie on an equipotential line.

Electric field lines are sketched in so that they are always at right angles to the equipotential lines.


Interior of the cavity is shielded from outside fields – basis of “Faraday cage”

– example, shielding of electronic circuits (in a closed metal box) from stray

electric fields

Net charge = 0

Scooped out cylindrical conductor


Mastering Physics

The first assignment is now available

It is due Friday, January 25 at 11 pm

Based on chapter 18

4 questions, each worth 10 marks

Have avoided questions where formulae have to be entered

or objects have to be dragged into boxes

Can use your PHYS 1020 (or Chemistry...) username and password

but must enter new Mastering Physics access code


Click here, then “Yes, look me up” to enter old Mastering Physics username and password. Course ID is PHYS1030UM

www.masteringphysics.com


Click for access code


Inside an Electrical Conductor

Charges (electrons) are free to move, so that:

• forces between charges push charges to the surface of the conductor

• charges move around the surface until they come to equilibrium

• as the charges are at rest, there must be zero force acting on them

→ the electric field inside the conductor must be zero

• as the charges are at equilibrium, electric field lines

must hit the conductor at right angles, otherwise

charges would be moved around the surface

→ no work is done in moving charges around the surface

→ charges on the surface must have constant potential energy

→ the surface of the conductor is an “equipotential”

E = 0

e-


Inside an Electrical Insulator

Electric charges cannot move, so:

→ charges do not move around to reduce the electric field to zero

→ an electric field can exist inside an insulator


The number of field lines leaving +q is

equal to the number hitting the inner

surface of the conductor...

→ there must be an induced charge

of –q on the inner surface of the

conductor (the field lines point

toward the inner surface, so the

induced charge must be negative).

As the outer surface has a charge +q, it must have the same number of lines

leaving the surface

→ from the outside, the conductor has no effect on the electric field.

–q

+q

As the conductor is electrically neutral (has zero

net charge), there must also be an induced charge

+q on the outer surface of the conductor.

E = 0

Charge +q placed inside an

uncharged hollow conductor


18.65/37: A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and thereby increases its linear momentum to 5x10-23 kg.m/s from 1.5x10-23 kg.m/s in a time of 6.3 !s. What is the magnitude of the electric field?


Prob. 18.68/22: Two small, identical objects are 0.2 m apart. Each carries a charge and they attract each other with a force of 1.2 N.

The objects are brought into contact so that the net charge is shared equally, and then they are returned to their initial positions. There is now a repulsive force of 1.2 N between the objects.

What is the initial charge on each object?

Write down Coulomb’s law for charges q1 and q2

Apply conservation of charge to find the charge q on each object after the charges have equalized

Write down Coulomb’s law again


Prob. 18.24: There are four charges, each of magnitude 2 !C. Two are positive and two are negative.

The charges are fixed to the corners of a 0.3 m square, one to a corner, in such a way that the net force on any charge is directed to the centre of the square.

Find the magnitude of the electrostatic force experienced by any charge.

Work out how to arrange the four charges so that the net force on each is toward the centre of the square –"implies there must be some symmetry in the arrangement of charges.


18.45: Two point charges of the same magnitude but opposite signs are fixed to either end of the base of an isosceles triangle. The electric field at the midpoint M between the charges has a magnitude EM. The field at point P has magnitude EP.

The ratio of these two field magnitudes is EM/EP = 9.0. Find the angle αin the diagram.

LL

dd

cos ! =d

L


Prob. 18.19/67: Two small spheres are mounted on identical horizontal springs and rest on a frictionless table.

When the spheres are uncharged, the spacing between them is 0.05 m, and the springs are unstrained.

When each sphere has a charge of +1.6 !C, the spacing doubles.

Determine the spring constant, ks, of the springs.


18.42: An electron enters the lower left side of a parallel plate capacitor

and exits at the upper right, as shown. The initial speed of the electron is

7#106 m/s. The capacitor is 2 cm long, the plates are 0.15 cm apart.

Find the magnitude of the electric field between the plates, assuming that

it is uniform everywhere.

F = –eE

vx = constantx

y


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ch18 - physics.umanitoba.ca · Electric Field q q 0 Test charge, q 0 P!F The force on the test charge at P is: The electric ﬁeld at P is deﬁned by: F= kqq 0 r2 r E= F q 0 = kq