Physics 2102Physics 2102Exam 2: Review SessionExam 2: Review Session

CH 24CH 24––2828

Physics 2102

Jonathan Dowling

Some links on exam stress:http://appl003.lsu.edu/slas/cas.nsf/$Content/Stress+Management+Tip+1http://wso.williams.edu/orgs/peerh/stress/exams.htmlhttp://www.thecalmzone.net/Home/ExamStress.phphttp://www.staithes.demon.co.uk/exams.html

Exam 2Exam 2• (Ch24) Sec.11 (Electric Potential Energy of a

System of Point Charges); Sec.12 (Potential ofCharged Isolated Conductor)

• (Ch 25) Capacitors: capacitance and capacitors;caps in parallel and in series, dielectrics; energy,field and potential in capacitors.

• (Ch 26) Current and ResistanceCurrent and Resistance: current, currentdensity and drift velocity; resistance andresistivity; Ohm’s law.

• (Ch 27) Circuits: emf devices, loop and junctionrules; resistances in series and parallel; DC singleand multiloop circuits, power; RC circuits.

• (Ch 28) Magnetic Fields: F=vxB, Right Hand Rule,Circular Motion, Force on Wire, Magnetic Dipole.

Potential Energy of a SystemPotential Energy of a Systemof Chargesof Charges

Potential Energy of A System of ChargesPotential Energy of A System of Charges

• 4 point charges (each +Q) areconnected by strings, forming asquare of side L

• If all four strings suddenly snap,what is the kinetic energy of eachcharge when they are very farapart?

• Use conservation of energy:– Final kinetic energy of all four charges

= initial potential energy stored =energy required to assemble the systemof charges

+Q +Q

+Q +Q

Do this from scratch!Don’t memorize theformula in the book!We will change thenumbers!!!

Potential Energy of A System ofPotential Energy of A System ofCharges: SolutionCharges: Solution

• No energy needed to bring infirst charge: U1=0

2

2 1

kQU QV

L= =

• Energy needed to bring in2nd charge:

2 2

3 1 2( )2

kQ kQU QV Q V V

L L= = + = +

• Energy needed to bring in3rd charge =

• Energy needed to bring in4th charge =

+Q +Q

+Q +QTotal potential energy is sum ofall the individual terms shownon left hand side =

2 2

4 1 2 3

2( )

2

kQ kQU QV Q V V V

L L= = + + = +

( )24

2

+L

kQ

So, final kinetic energy of eachcharge = ( )24

4

2

+L

kQ

Electric Field and Potential

in and around a Charged

Conductor : A Summary

in

out

0

All the charges reside on the conductor surface.

The electric field inside the conductor is zero: 0.

The electric field just outside the conductor is: .

The electric fi

E

E!

"

=

=

1.

2.

3.

4. eld just outside the conductor is perpendicular

to the conductor surface.

All the points on the surface and inside the conductor have the same potential.

The conductor is an equipotential

5.

surface.

in0E =

!

out

0

ˆE n!

"=

!

E

!

E

!

n̂

n̂

(24-19)

CapacitorsCapacitorsE = σ/ε0 = q/Aε0E = V dq = C V

C = κ ε0A/d

C = ε0A/d

C=ε0ab/(b-a)

Current and resistanceCurrent and resistancei = dq/dt

V = i RE = J ρ ρ = ρ0(1+α(T−T0))

R = ρL/AJunction rule

DC CircuitsDC Circuits

Single loop Multiloop

V = iRP = iV

Loop rule

Resistors and CapacitorsResistors and Capacitors

Resistors Capacitors

Key formula: V=iR Q=CV

In series: same current same charge Req=∑Rj 1/Ceq= ∑1/Cj

In parallel: same voltage same voltage 1/Req= ∑1/Rj Ceq=∑Cj

Capacitors and ResistorsCapacitors and Resistorsin Series and in Parallelin Series and in Parallel

• What’s the equivalent resistance (capacitance)?• What’s the current (charge) in each resistor (capacitor)?• What’s the potential across each resistor (capacitor)?• What’s the current (charge) delivered by the battery?

RC CircuitsRC Circuits Time constant: RC( )/

/

0

Charging: ( ) 1

Discharging: ( )

t RC

t RC

q t CE e

q t q e

!

!

= !

=

i(t)=dq/dt

Capacitors: Checkpoints,Capacitors: Checkpoints,QuestionsQuestions

Problem 25-21When switch S is thrown to the left, the plates of capacitor 1acquire a potential V0. Capacitors 2 and 3 are initiallyuncharged. The switch is now thrown to the right. What are thefinal charges q1, q2, and q3 on the capacitors?

Current and Resistance: Checkpoints, QuestionsCurrent and Resistance: Checkpoints, Questions

Problem 26-56

A cylindrical resistor of radius 5.0mm andlength 2.0 cm is made of a material that has aresistivity of 3.5x10-5 Ωm. What are the (a)current density and (b) the potential differencewhen the energy dissipation rate in the resistoris 1.0W?

Circuits: Checkpoints, QuestionsCircuits: Checkpoints, Questions

Problem: 27.P.018. [406649]Figure 27-33 shows five 5.00 resistors.(Hint: For each pair of points, imagine that a battery is connectedacross the pair.)

Fig. 27-33(a) Find the equivalent resistance between points F and H.

(b) Find the equivalent resistance between points F and G.

Proble: 27.P.046. [406629]

In an RC series circuit, E = 17.0 V, R = 1.50 MΩ, and C =1.80 µF.

(a) Calculate the time constant.

(b) Find the maximum charge that will appear on thecapacitor during charging.

(c) How long does it take for the charge to build up to 10.0µC?

Magnetic Forces and TorquesMagnetic Forces and Torques

F qv B q E= ! +! ! !!

BLdiFd!!!

!=

B

!!!!= µ"

L

vF qB

mvr =

C

C

Top viewSide view

Consider the rectangular loop in fig. with sides of lengths and and that carries

ˆa current . The loop is placed in a magnetic field so that the normal to th

a a b

i n

Magnetic Torque on a Current Loop

1 3

2 4

e loop

forms an angle with . The magnitude of the magnetic force on sides 1 and 3 is

sin 90 . The magnetic force on sides 2 and 4 is

sin(90 ) cos . These forces cancel

B

F F iaB iaB

F F ibB ibB

!

! !

= = ° =

= = " =

!

net

2 4

1 3

in pairs and thus 0.

The torque about the loop center of and is zero because both forces pass

through point . The moment arm for and is equal to ( / 2)sin . The two

torques tend to

F

C F F

C F F b !

=

1 3

rotate the loop in the same (clockwise) direction and thus add up.

The net torque + =( / 2)sin ( / 2)sin sin sin .iabB iabB iabB iAB# # # ! ! ! != + = =

netsiniAB! "=

net0F =

(28-13)

U Bµ= U Bµ= !

The torque of a coil that has loops exerted

by a uniform magnetic field and carries a

current is given by the equation .

We define a new vector associated with the

N

B

i NiAB!

µ

=

Magnetic Dipole Moment

! coil,

which is known as the magnetic dipole moment of

the coil.

The magnitude of the magnetic dipole moment is

Its direction is perpendicular to the plane of the coil.

The sense of is defined by the right-hand rule. We curl the fingers of the right an

i

.

h d

NiAµ

µ

=

!

n the direction of the current. The thumb gives us the sense. The torque can be

expressed in the form sin where is the angle between and .

In vector form:

The potential energy

.

B

B

B! µ " " µ

! µ=

=

#

!

!! !

!

of the coil is:

has a minimum value of for 0 (position of equilibrium).

has a maximum value of for 180 (position of equilib

cos

rium).

For both posit

.

U B

U

B B

B

U

µ "

µ "

µ " µ

$

= %

=

°

$ = $

=

stable

unstable

Note :

!!

ions the net torque is 0.! =

B! µ= "!! !

(28-14)

U Bµ= ! "!!

Ch 28: Checkpoints and QuestionsCh 28: Checkpoints and Questions

Problem: 28.P.024. [566302]In the figure below, a charged particle moves into a region of uniformmagnetic field , goes through half a circle, and then exits that region. Theparticle is either a proton or an electron (you must decide which). Itspends 160 ns in the region.

(a) What is the magnitude of B?

(b) If the particle is sent back through the magnetic field (along the sameinitial path) but with 3.00 times its previous kinetic energy, how muchtime does it spend in the field during this trip?