Physics 202 Review Lectures fileTopics With More Weights In this final exam, the following topics...
Transcript of Physics 202 Review Lectures fileTopics With More Weights In this final exam, the following topics...
![Page 1: Physics 202 Review Lectures fileTopics With More Weights In this final exam, the following topics will carry more weights among Exam 1&2 topics: Electric field and electric force.](https://reader033.fdocuments.in/reader033/viewer/2022041423/5e20cd6745bbae1f063556f7/html5/thumbnails/1.jpg)
Physics 202 Review LecturesExam 1&2 materials: todayOptics: Reviewed Dec 11, 2008. (available on Web)Exam 3 materials: Reviewed on Nov. 21/22/23 (available on web).
Also:Also: Exam 1 and Exam 2 were reviewed by Prof. Mellado
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Chapters Covered In Exam 1&2Exam 1 Coverage
Chapter 21: Electric FieldsCh t 22 G ’ LChapter 22: Gauss’s LawChapter 23: Electric Potential Chapter 24: CapacitanceChapter 24: Capacitance
Exam 2 CoveragegChapter 25: Electric CurrentsChapter 26: DC CircuitChapter 27: Magnetism Chapter 28: Source of Magnetic Field
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Exam Topics (1)Basic Quantities:
Electric Charge Electric ForceElectric Field, Field Lines, Electric FluxElectric PotentialElectric Potential EnergyCapacitanceCapacitanceElectrical CurrentResistanceResistance Magnetic ForceMagnetic Field, Magnetic Field Lines, Magnetic Flux
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Exam Topics(2)Electric charge
Two typesT t l h i dTotal charge is conserved.
Electric forceCan be attractive/repulsiveCan be attractive/repulsiveCoulomb’s Law
Electric fieldE. field is a form of matter, it carries energy. E. field is independent of test charge.E. field is a vector quantity.Three ways to calculate E. field
di t t G ’ L d i ti f V• direct vector sum, Gauss’s Law, derivative of VF=qE (note: E does not include the one created by q)
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Exam Topics(3)Electric potential energy.
E. force is a conservative forceE. potential energy depends on both the source and test charge.Like all potential energies the E potential energyLike all potential energies, the E. potential energy is relative to a certain reference state. (Usually, an “infinity” state is taken as U=0.) Use of E. potential energy with energy conservation, work-kinetic energy theorem.
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Exam Topics(4)Electric potential
Electric potential depends only on the source.E. Potential and E. Field are closely related.
• From E to V• From V to E
E. potential and E. potential energy are different quantitiesquantities.
higher V does not necessarily mean higher U.E. potential and E. potential energy are related: U=qVp p gy q
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Exam Topics(5)Capacitance
C=Q/ΔVconnection in parallel and in seriesC of typical configurations (parallel plates, ….)
Current and resistance. Ohm’s LawPower consumption on R
DC circuitDC circuitKirchhoff’s RulesSimple 1-loop, 2-loop circuit of R’s and ε’sSimple 1 loop, 2 loop circuit of R s and ε s
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Exam Topics(6)Magnetic Force
Magnetic force has a form of qvxB.• always perpendicular to v and B• always perpendicular to v and B.• never does work• charged particle moves in circular/helix path in uniform
B field (ω=qB/m, r=mv/qB)• On current segment, it has the form ILxB
Uniform B closed loop ΣF=0 Στ=μxB– Uniform B, closed loop ΣF=0, Στ=μxBMagnetic Field:
Field lines, “north” and “south”. B field contains energy, u= ½ B2/μ0
B field never does work.
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Exam Topics(7)Magnetic Fields can be produced by:
moving charge (Biot-Savart law)change of E field (displacement current, conceptual level only)
Ampere’s LawAmpere s LawAmpere’s law simplifies the calculation of B field in some symmetric cases.
• (infinite) straight line, (infinite) current sheet, Solenoid, Toroid
Gauss’s Law in Magnetism no magnetic charge.g g gForces between two currents
Can be attractive/repulsivef fNo force if perpendicular
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Topics With More WeightsIn this final exam, the following topics will carry more weights among Exam 1&2 topics:
Electric field and electric force.Electric potential and potential energyMagnetic field and magnetic forceMagnetic field and magnetic forceCapacitors and energy storage in capacitors
Other Exam 1&2 topics are also important, but will be de-emphasized to make your preparation easier.
The following exercises reflect this special treatment.
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Reminder:Three Ways to Calculate Electrostatic FieldThree Ways to Calculate Electrostatic Field
Superposition with Coulomb's Law (first principle):
rdqkrqkE i ˆˆ ∫∑ ==r
Apply Gauss’s Law:
rr
krr
kE eii
e 22 ∫∑ ==
Apply Gauss s Law: (Practical only for cases with high symmetry)
∑∫q
0ε∑∫ =•=Φ in
E
qdAE
q1q2
From a known potential: qi
VEVEVE ∂−=
∂−=
∂−=
zE
yE
xE zyx ∂
=∂
=∂
= ,,
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Exercise 1: Seven Point ChargesSix point charges are fixed at corners of a hexagon as shown. A seventh point charge q7=2Q is placed t th tat the center.
1. What is the force on q7? 2 What is the minimum energy required to bring q2. What is the minimum energy required to bring q7
from infinity to its current position at the center?
Solutions: (See board) q = 2QSolutions: (See board)First thoughts:
Point charges: Coulomb’s Law q6=-Q
q2=2Qq1=-2Q
q7Point charges: Coulomb s Law
Be aware of symmetry
Energy: ΔU=qΔV
q6
q3=3Qq7
aaEnergy: ΔU=qΔV
q4=-2Qa
q5=2Q
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Exercise 2: Three shells of chargeAs shown below three thin sphere shells have radius R, 2R, 4R, and charges Q, -Q, 2Q, respectively.
Use Gauss’s law, find the electric field distribution.Use Gauss s law, find the electric field distribution.
Solution:
The setting is highly symmetrical
Gauss’ surface will be concentric sphere of radius r.
qErAdE 2 14π ==•∫rr
0<r<R: qenclosed = 0 E=0
enclosedS
qErAdE0
4ε
π ==•∫
R<r<2R: qenclosed = Q E=Q/(4πε0r2)
2R<r<4R: qenclosed =Q-Q = 0 E=0
4R<r: qenclosed =Q-Q+2Q = 2Q E=2Q/(4πε0r2)
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Reminder: A Picture to Remember
+q F=qEE-q ΔU=qΔV
W =ΔUΔV=-∫Eds
higher potential V lower V
∫
Field lines always point towards lower electric potentialField lines and equal-potential lines are always at a normal angle. In an electric field:
a +q is always subject a force in the same direction of field line. (i.e. towards lower V) a -q is always subject a force in the opposite direction of field q y j ppline. (i.e. towards higher V)
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Exercise 3: Potential, Field, and EnergyThe equal potential lines surrounding two conductors, +10V and +15V, are shown below.
Draw on the figure the direction of electric field at point CDraw on the figure the direction of electric field at point CIf a charge of Q=+0.5C is to be moved from point B to C, how much work is required?
+12V
+9V
A+11V
+15V +10V
+13V+12V A
C
+14VB
WB C =UC-UB=Q(VC-VB)=0.5*(12-9) =1.5J
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Exercise 4: Potential and FieldThe electric potential of a field is described by
V= 3x2y+y2+yz.
Find the force on a test charge q=1C at (x,y,z)=(1,1,1)
Solution:
First Thoughts:
F=qE
Ex = -dV/dx = -6x = -6 @ (1,1,1)Ey = -dV/dy = -(3x2+2y+z)= -6 @ (1,1,1)F=qE
Ex=-dV/dx, etc.
yEz = -dV/dz = -y = -1 @ (1,1,1)
(F ,F ,F ) = qE = (-6,-6,-1) N(Fx,Fy,Fz) qE ( 6, 6, 1) N
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Exercise 5: Electric Potential EnergyA charged non-conducting ring of radius R has charge -Q distributed on the left side and 2Q distributed on the right side.
How much energy is required to bright a point chargeHow much energy is required to bright a point charge q=0.5Q from infinity to the center of the ring?What if q=-0.5Q instead?
Solution: See boardSolution: See board
First Thought: -Q 2QqEnergy required = external work required to
bringing in q.Follow up: How to calculate work?
-Q 2Qq
o o up o to ca cu ate ointegral dw=Fds (won’t work, to complicated)
☺ use energy conservation & the idea of electric potential
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Reminder: Capacitors+Q -Q1 +Q
Capacitance: C=Q / V (Q=CV)Two capacitors in series: Q=Q =Q V=V +V 1/C = 1/C + 1/C
+Q1
C1C2
1 +Q2-Q2
Q=Q1=Q2, V=V1+V2 1/C = 1/C1 + 1/C2
V1 V2
1
Two capacitors in parallel: V1C
,Q1
V
Two capacitors in parallel:V=V1=V2 ,Q=Q1 + Q2 C = C1 + C2
C1
C2
Energy storage: U= ½ CV2
V2 ,Q2
V
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Exercise 6: Two CapacitorsTwo capacitors, C1 =1 μF and C2=2μF, are connected in series as shown. An voltage of 30V is applied across the two capacitors.capacitors.
What is the combined capacity?What is the charge on C1 ? What is the voltage on C ?
C2C1
What is the voltage on C1?What is the energy stored on C1? 30V
Solution:1/C = 1/C1 + 1/C2 C = 0.67 μFQ = Q1 = Q2 = CV = 0.67 μF * 30V = 2x10-5 CoulombQ Q1 Q2 C 0 6 μ 30 0 Cou o b
V1= Q1/C1 = 2x10-5 /1x10-6 = 20 VU1 = ½ C1V1
2 = 2x10-4 J
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Reminder:Two Ways to Calculate Magnetic Field
Biot-Savart Law: dB
∫×
= 20 ˆ
4 rdI rsB
πμr
Ampere’s Law:
4 rπdB’
I(Practical only for cases with high symmetry)
ds∫ =•any
IdsB 0μr
pathclosedany
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Exercise 7: Biot-Savart LawUse Biot-Savart to find the magnetic field at the point P.
Solutions: (See board)
1∫
×= 2
0 ˆ4
dI rsB μr
Answer:2
∫ 24 rπ
Answer:
segment 1 contribution:
B 03
B=0
segment 3 contribution: B=0P
IdI ˆsegment 2:
(into page) RI
RdI
8ˆ
40
20 μπ
μ=
×= ∫
rsBr
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Exercise 8: Ampere’s Law An infinite straight thin wire is at the center of two concentric conducting cylinders of radius R and 2R.
The currents are I (into the page), 2I (out ), and I (in) , respectivelyThe currents are I (into the page), 2I (out ), and I (in) , respectively for the center wire and the two cylinders. (as color coded).
Find B as function of r.
Solution:
IrBd enclosedμπ2 0==•∫ sBr
Answers:
rIB enclosed
πμ
20=⇒
Answers:
r<R , B= μ0I/2πr (Clockwise)
R<r<2R B= μ0I/2πr (counter-clockwise)R<r<2R , B= μ0I/2πr (counter-clockwise)
r>2R , B= 0