physics 2102 exam 2: review session ch 24–28 physics 2102 jonathan dowling some links on exam...
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Physics 2102 Physics 2102
Exam 2: Review SessionExam 2: Review Session
CH 24–28CH 24–28
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 of Charged 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, current density and drift velocity; resistance and resistivity; Ohm’s law.
• (Ch 27) Circuits: emf devices, loop and junction rules; resistances in series and parallel; DC single and 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 System Potential Energy of a System of Chargesof Charges
Potential Energy of A System of ChargesPotential Energy of A System of Charges
• 4 point charges (each +Q) are connected by strings, forming a square of side L
• If all four strings suddenly snap, what is the kinetic energy of each charge when they are very far apart?
• Use conservation of energy:– Final kinetic energy of all four charges
= initial potential energy stored = energy required to assemble the system of charges
+Q +Q
+Q +Q
Do this from scratch! Don’t memorize the formula in the book!We will change the numbers!!!
Potential Energy of A System of Potential Energy of A System of Charges: SolutionCharges: Solution
• No energy needed to bring in first charge: U1=0
2
2 1
kQU QV
L= =
• Energy needed to bring in 2nd charge:
2 2
3 1 2( )2
kQ kQU QV Q V V
L L= = + = +
• Energy needed to bring in 3rd charge =
• Energy needed to bring in 4th charge =
+Q +Q
+Q +Q
Total potential energy is sum of all the individual terms shown on left hand side =
2 2
4 1 2 3
2( )
2
kQ kQU QV Q V V V
L L= = + + = +
( )242
+L
kQ
So, final kinetic energy of each charge = ( )24
4
2
+L
kQ
Electric Field and Potential
in and around a Charged
Conductor : A Summary
in
out0
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
Eσε
=
=
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.
in 0E =r
out0
ˆE nσε
=r
Er
Er
n̂
n̂
(24-19)
CapacitorsCapacitors
E = σ/ε0 = q/Aε0
E = V dq = C V
C = ε0A/d
C = ε0A/d
C=ε0ab/(b-a)
Current and resistanceCurrent and resistancei = dq/dt
V = i RE = J = 0(1+(TT0))
R = L/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 25-21
When switch S is thrown to the left, the plates of capacitor 1 acquire a potential V0. Capacitors 2 and 3 are initially uncharged. The switch is now thrown to the right. What are the final 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 and length 2.0 cm is made of a material that has a resistivity of 3.5x10-5 m. What are the (a) current density and (b) the potential difference when the energy dissipation rate in the resistor is 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 connected across 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 M, and C = 1.80 µF.
(a) Calculate the time constant.
(b) Find the maximum charge that will appear on the capacitor during charging.
(c) How long does it take for the charge to build up to 10.0 µC?
Magnetic Forces and TorquesMagnetic Forces and Torques
F qv B q E= × +r r rr
BLdiFdrrr
×=
Brrr
×=μτ
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
θ
θ θ= = °== = − =
r
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 θ
=
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 iABτ τ τ θ θ θ θ= + = =
net siniABτ θ=
net 0F =
(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 NiABτμ
=
Magnetic Dipole Moment
r 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
=
r
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
Bτ μ θ θ μ
τ μ=
=
×
r
rr r
r
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 θ
μ θ
μ θ μ
−
= ⋅
=
°
− =−
=
stable
unstable
Note:
rr
ions the net torque is 0.τ=
Bτ μ= ×rr r
(28-14)
U Bμ=− ⋅rr
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 uniform magnetic field , goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends 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 same initial path) but with 3.00 times its previous kinetic energy, how much time does it spend in the field during this trip?