REVIEW:

• ELECTRIC FORCE, ELECTRIC FIELD, • ELECTRIC FIELD LINES, ELECTRIC FLUX, GAUSS’S LAW, • ELECTRIC POTENTIAL, • CONTINUOUS CHARGE DISTRIBUTIONS (ELECTRIC FIELD, ELECTRIC

POTENTIAL, GAUSS’S LAW), • ELECTRIC CURRENT, • MAGNETIC FORCE ON MOVING CHARGES AND WIRES, • BIO-SAVART-LAW, • FORCE BETWEEN PARALLEL CURRENT CARRYING WIRES

ELECTRIC FORCE: Coulomb’s law: • force on charge 1 due to charge 2 is

12221

e12 rrqqkF =

r

Net force on a charge due to several other charges: • VECTOR SUM of all forces on that charge due to other charges • Called Principle of SUPERPOSITON

• Each charge exerts a force on charge 1

Resultant force is 4131211 FFFF

rrrr++=

• says net force on charge 1 equals sum of force on 1

from 2, force on 1 from 3, and force on 1 from 4

ELECTRIC FIELD

• If the force on 0q at a point is Fr

, then electric field at that point is 0q

FEr

r=

• If the electric field at a point is Er

, then the force on 0q at point is EqFrr

0=

• Electric field at P due to a point charge is rrqkE ˆ

2eP =r

o Unit vector r points from q → P

• Electric field points away from positive charge • Electric field points toward negative charge

Superposition: Total E

r at point P due to an arrangement of point charges is the VECTOR SUM

of the electric field contributions from all charges around P • Total electric field at P is:

43212

ˆEEEE

rrqkE

i i

iieT

rrrrr+++== ∑

o iq is the charge at i

o ir is the distance from iq → P

o ir is the unit vector from iq → P

o the sum is a VECTOR SUM

Did example with electric dipole

ELECTRIC FIELD LINES: • E

r vector at a point in space is tangent to the EFL through that point

• “Density” of EFL is proportional to E (magnitude) in that region

o Larger E→ closer packing of lines

• EFL start on positive charges and end on negative charges

• Number of EFL starting/ending on charge is proportional to its magnitude

• Electric field lines do not cross

Looked at motion of a particle in a uniform electric field:

ELECTRIC FLUX

General result for Electric Flux through element of area iA∆

iiiiii AEAErr

∆⋅=∆=∆Φ θcosE

Total flux through a closed surface:

∫∫ =⋅=Φ dAEAdE n

surfaceclosedover

E

rr

GAUSS’S LAW (general statement):

0

enclosed

surfaceclosed

e εqAdE∫ =⋅=Φ

rr

• Powerful way to calculate electric field if we can factor nE out of integral

o Trick is to choose surface so that nE is uniform over all or part of surface No charge inside: • net number of lines leaving = 0 • all lines go through

Positive charge inside: • non-zero net number of lines leaving • lines start on charge inside sphere

Used Gauss’s Law to calculate electric field around a point charge

ELECTRIC POTENTIAL Difference in electric potential between points A and B is:

0qUVVV AB

∆=−=∆

SO: ∫ ⋅−=−=∆B

AAB sdEVVV rr

o Potential difference between two points depends on electric field o Note sign and order of integration limits

EQUIPOTENTIAL o All points on plane perpendicular to uniform

Er

field have same electric potential

SIGN OF ∆V o change in potential energy of 0q when moved

from A → B is

EdqVqU 00 −=∆=∆ o Says that if A → B is in same direction as E

r, then 0<−=∆ AB VVV

o If path is in the same direction as E

r, then potential difference V∆ is

negative

ELECTRIC POTENTIAL DUE TO A POINT CHARGE Electric potential at distance r from a point charge q :

rqkV e=

• assumes that electric potential at infinity is 0 IMPORTANT: • Electric potential V is a scalar.

o can just add contributions from different charges

• ALL points at distance r from a point charge q have the same potential o Spherical surface around point charge is an equipotential surface

Electric Potential due to multiple point charges: SUPERPOSITION • Electric potential at P is

∑=i i

i

rqkV e

o Not a vector sum. Contributions to V add as scalars

Can get components of E

r from derivatives of V

Implies: dxdVEx −= dy

dVEy −= dzdVEz −=

• Says: E

r is always perpendicular to equipotential surfaces

EQUIPOTENTIAL SURFACES: • like contour maps

o “valleys” centred on negative charge o “hills” centred on positive charge

• Example: Electric dipole (contours marked in volts)

o Electric field lines

start on positive charge, end on negative charge and are perpendicular to equipotentials at crossing

Cange in potential energy of charge 0q+ moving from A → B

• ( ) ( )( )V16V2400 −−=−=∆ qVVqU AB • So: V 400 ×=∆ qU • External force does V 400ext ×= qW

CONTINUOUS CHARGE DISTRIBUTIONS: CHARGED OBJECTS WITH FINITE SIZE

Strategy: • Break distribution into small charge elements • Find contribution to electric potential or electric field from a given element • Sum/integrate to find total V or E

r

ELECTRICAL POTENTIAL FROM A CONTINUOUS CHARGE DISTRIBUTION

• total V at P is: ∫∑ =∆

=→∆

chargeallover

ee0lim

rdqk

rqkV

i i

i

qi

• Identify rdqkdV e= as the contribution to V from the charge element dq

CHARGE DENSITY • to convert sum/integral over charge elements into a sum/integral over spatial

variables (i.e. x, y, z)

o Linear charge density: LQ

=λ

o Surface charge density: AQ

=σ

o Volume charge density: VQ

=ρ

ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS Approach: • Break charge distribution into small elements (treat each as a point charge) • Write vector sum of contributions from elements • Take limit as elements become infinitesimally small → INTEGRAL

o ir is unit vector pointing from iq∆ toward P

o iEr

∆ is the contribution to Er

due to iq∆

∫

∑

=

∆=

→∆

rrdqk

rrqkE

ii

i

iqi

ˆ

ˆlim

2e

2e0

r

Two Ways to use GAUSS’S LAW:

• 0

insidee ε

q=Φ

o relates flux to charge inside for surface of ANY shape

• 0

inside

surface closed over ε

qAdE =⋅∫rr

o gives a way to calculate E

r for SPECIAL cases

mostly useful if we normal component of E

r is constant over part of

the surface and can be factored out of the integral

Using Gauss’s law to calculate Electric Field Er

• Must know direction of electric field from symmetry of problem

o radial (spherical symmetry) for point charge

o radial (cylindrical symmetry) for a long line of

charge

o uniform for a large flat sheet of charge

• Must choose Gaussian surface that allows us to calculate ∫ ⋅=Φ

surface closed over

e AdErr

o Must be able to factor Er

out of flux integral in region of space where we want to find electric field

Two cases for which we can evaluate ∫ ⋅surface

AdErr

for all or part of surface

• E

r uniform and perpendicular to part or all of Gaussian surface

o then flux is ∫ =⋅ AEAdE n

rr for that part of the surface

• Er

parallel (tangent) to part of the Gaussian surface

CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM (19.11) PROPERTIES: ISOLATED CONDUCTOR IN ELECTROSTATIC EQUILIBRIUM

1st: 0=E

r everywhere inside a conductor

• Must be true or else charges would move until 0=E

r

2nd: Any NET CHARGE on conductor must be on surface • can prove with Gauss’s Law

ANY NET CHARGE ON THE CONDUCTOR MUST BE ON THE SURFACE

3rd: Electric field JUST OUTSIDE a CHARGED CONDUCTOR: • MUST be perpendicular to the surface. If not, charge flows

along surface

• MUST have magnitude 0

n εσ

=E

4th: Surface charge density is HIGHEST where radius of surface curvature is smallest • Means highest E

r at most “pointed” regions of surface

ELECTRIC POTENTIAL AT THE SURFACE OF AND INSIDE CHARGED CONDUCTORS (text section 20.6)

• ⊥E

rsurface means that the surface is an equipotential

• 0=E

r inside says potential V inside conductor must be

same as potential V at surface

o entire conductor is an equipotential

ELECTRIC CURRENT – FLOWING CHARGE (Quick Review from Chap 21)

CURRENT: rate at which charge crosses a specified surface

Average current is tQI∆∆

=ave

• Q∆ is amount of charge across surface in time t∆ DIRECTION OF CURRENT: • SAME as direction of POSITIVE charges crossing

surface • OPPOSITE direction of NEGATIVE charges

crossing surface CONDUCTIVITY: current density EvqnJ d σ== • dv is drift velocity (proportional to electric field) • n is density of charge carriers with charge q

+

+

-

-

Current (I)

RESISTIVITY: σρ 1=

IMPORTANT: resistivity and conductivity are properties of the material RESISTANCE: relates current through particular object to pot. diff. across it • Look at wire with cross-sectional area A

o Potential difference between point a and point b is VVV ∆=− ab

o lVAI

ρ∆

= Define resistance AlR ρ

= so that IRV =∆

Ohm’s Law: ρσ EEJ == Resistance unit is Ohm (Ω): 1Ω = 1 V/A

Resistivity unit is ohm-metre (Ω·m) Conductivity unit is (Ω·m)-1

A

I (current) a b

l

MAGNETIC FORCE & MAGNETIC FIELD (Chapter 22) • Magnetic Field Lines – RULES

o Lines start and stop on poles of a magnet OR lines form closed loops around current-carrying wires • No such thing as a magnetic monopole • For magnet – lines start at North pole and end at South pole

o MAGNITUDE of magnetic field || Br

indicated by DENSITY of lines o DIRECTION of magnetic field at a point is tangent to field lines at point

• Direction of magnetic field at some location is the direction that a compass needle would point at that location.

N S

MAGNETIC FORCE BFr

on a moving charge BvqFrrr

×=B Cyclotron motion: Charge with velocity perpendicular to magnetic field moves on circular path (cyclotron orbit) in plane perpendicular to magnetic field

Radius of orbit is Bqmvr =

• angular frequency for motion (cyclotron frequency) mqB

rv==ω

• Period of circular motion Bqm

vr

vT ππ 22ncecircumfere

===

• Period independent of speed:

F

+v

r

Devices that use magnetic force on a moving charge (22.4)

In space with Er

and Br

, force on moving charge is Force) (Lorentz BvqEqFrrrr

×+= VELOCITY SELECTOR • Contains region of space with uniform electric and magnetic fields

o perpendicular to each other o perpendicular to path of charged particle

• In selector region:

o BvqFrrr

×=B is “up” by RHR o EqF

rr=E is “down”

• Net force on charge is zero IF EqBvq =

• So particles with speed BEv= are undeflected

+ + + + + + +

- - - - - - -

Barrier with hole E (down)

B (into screen)

+q v

MASS SPECTROMETER

• 1st: ionize molecules and fragments • 2nd: use velocity selector to pick out

fragments with sss BEv /=

• 3rd: inject ions with speed sv into uniform Br

with magnitude 0B

o Radius of path is 0Bq

vmr s=

• SO: s

s

EBrB

qm 0=

+ + + + + + +

- - - - - - -

velocity selectorBs, Es

+qv

+qvs

Detector

r

B0

CYCLOTRON – device to accelerate charged particles to high energy • for charge q in uniform B

r

o found Bqvmr =

o period of orbit is: Bqm

vr

vcircumfT ππ 22.

===

• NOTICE: period (i.e. timing of “kick”) does NOT depend on speed

alternating voltage source

• Potential difference across gap changes each time particle reaches gap

o “kick” raises K by q∆V

+q

Bup

MAGNETIC FORCE ON A CURRENT-CARRYING CONDUCTOR • Force on each charge in wire is BvqF dB

rrr×=

• For cross-section area A and n carriers per unit volume,

force on wire segment of length l is ( ) lAnBvqF dB

rrr×=

• Current in wire is AvqnI d= so force on length l of wire is

BlIFB

rrr×=

o Vector l

r points in direction of current; magnitude is length of segment

• Force on segment sd r of arbitrary shaped wire is

BsdIFd B

rrr×=

I

Bin

FB

l

A

ds

Magnetic Dipole Moment • For current I circulating around loop of area A

o AIrr

=µ is the magnetic dipole moment of the loop o Magnitude: IA=µ o Direction:

• vector µr

perpendicular to the plane of the loop • direction by Right-hand rule:

• fingers curl in direction of current • thumb shows direction of µ

r

• Units of magnetic dipole moment: A·m2 • For a coil of n loops, nIA=µ

I A

µ = IA

Can express torque as vector product of magnetic dipole moment and field

Brrr

×= µτ • Magnitude of torque is θτ sinBAI= • Direction of B

rr×µ is into screen/page

• True for any current loop in a magnetic field!

2 4

B

F2

F4

θ

µ = IA

τ

BIOT-SAVART LAW: 20 ˆ

4 rrsdI

Bd×

=rr

πµ

• For drawing, direction of rsd ˆ×

r is out of screen/page

o So Bd

r at P due to sdr points out of screen/page

• Magnitude θsinˆ dsrsd =×

r

o For a given r, contributions Bd

r from sdr are maximum for points on

plane perpendicular to sdr

o Current in sdr makes NO contribution to Bdr

at points along direction sdr

I

ds

r

P

θ

r

MAGNETIC FIELD AROUND A LONG (INFINITE) WIRE • Result of using Biot-Savart Law:

o Magnetic field lines circle wire → no component of Br

parallel to wire

• Magnitude of Br

inversely proportional to perpendicular a distance from wire

aIB

πµ2

0=r

(IMPORTANT RESULT)

• Direction of magnetic field lines:

o Another Right-Hand Rule: thumb along I ; fingers curl in direction of Br

B I

Iout

USE aIB

πµ2

0= TO FIND MAGNETIC FORCE BETWEEN PARALLEL WIRES

• Field at I1 due to I2 is aI

Bπ

µ2

202 =

• Force on I1 per unit length is aII

lF

πµ2

2101 = toward I2 (for currents in same dir.)

• Force on I2 per unit length is aII

lF

πµ2

2102 = toward I1 (for currents in same dir.)

o By Newton’s 3rd law.

a B2

I1

I2F1

F1

I2

I1B2

a

• For parallel conductors, current in same direction:

o Wires ATTRACT with force per unit length aII

lF

πµ2

210=

• For parallel conductors, current in opposite direction:

o Wires REPEL with force per unit length aII

lF

πµ2

210=

Provides definition of AMPERE: • For 2 wires, 1 m apart, A 121 == II → force per unit length is N/m 102 7−×