03/10/20061 Exam Review I Physics 208. 03/10/20062 Chapters 23-28 Coulomb force between charges Work...
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Transcript of 03/10/20061 Exam Review I Physics 208. 03/10/20062 Chapters 23-28 Coulomb force between charges Work...
03/10/2006 1
Exam Review I
Physics 208
03/10/2006 2
Chapters 23-28• Coulomb force between charges• Work done to move a charge• Electric potential as U/q• Electric field as F/q• Motion of charged particle in electric field• Relation between electric field and electric potential• Electric field lines and equipotentials• Electric fields in and near conductors• Electric flux and Gauss' law• Electric field and potential of a point charge, line charge, sheet charge,
spherical shell• Capacitance of an isolated conductor, and of a pair of conductors• Calculating capacitance, parallel plate, cylinder, sphere• Energy in a capacitor, energy density, energy in E-field• Dielectrics• Current and resistance, Ohms law• Power dissipation in a resistor• Simple circuits (parallel and series) and RC circuits
03/10/2006 3
Charging processes
•Triboelectric:-glass rod rubbed with silk positive-rubber rod rubbed with fur negative
+ + ++ +++Less positively charged rod
+
+++
+
Positively charged metal
+ + + +++++ + + + ++
+++
+- --
-
-
electron flow
•Conduction•Induction
03/10/2006 4
If you rub an inflated balloon against your hair, the two materials attract each other, as shown in this figure. Fill in the blank: the amount of charge present in the system of the balloon and your hair after rubbing is _____ the amount of charge present before rubbing.
(a) less than
(b) the same as
(c) more than
Quiz on Charge Conservation
03/10/2006 5
3 pithballs suspended from thin threads are charged by touching them with a charged object. It is found that pithballs 1 and 2 repel each other and pithballs 2 and 3 repel each other.Which of the following statements are correct?
1) 1 and 3 carry opposite charge
2) All of them carry charges of the same sign
3) We need to do more experiments to determine the sign of the charges
Quiz on charge sign
+ + -1 2 3
- - -1 2 3
+ + +1 2 3
03/10/2006 6
Quiz on charge sign cont.
3 objects are brought close to each other, 2 at a time. When objects A and B are brought together, they attract. When objects B and C are brought together, they repel. From this, we conclude that:
(a) objects A and C possess charges of opposite sign.
(b) all three of the objects possess charges of the same sign.
(c) we need to perform additional experiments to determine information about the charges on the objects.
+ - -A B C
- + +A B C
03/10/2006 7
Answer: (c). In the first experiment, objects A and B may have charges with opposite signs, or one of the objects may be neutral since they can attract one the other even due to induction.
The second experiment shows that B and C have charges with the same signs, so that B must be charged. But we still do not know if A is charged or neutral.
Answer
03/10/2006 8
Insulators and conductorsOne of these isolated charged spheres is copper and the other is rubber. Label which is which and support your answer with an explanation.
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
a) A is rubber and B is copper.b) A is copper and B rubberc) They are both copper
Charged rubber rods are placed near a neutral conducting sphere, causing a redistribution of charge on the spheres. Which of the diagrams below depict the proper distribution of charge on the spheres? List all that apply.
QuickTime™ and aTIFF (Uncompressed) decompressor
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a) A and Db) C and Bc) E
03/10/2006 9
Coulomb’s Law
• Electrical force between two stationary charged particles
• The SI unit of charge is the coulomb (C ), µC = 10-6 C
• 1 C corresponds to 6.24 x 1018 electrons or protons
• ke = Coulomb constant ≈ 9 x 109 N.m2/C2 = 1/(4πo) o permittivity of free space = 8.854 x 10-12 C2 / N.m2
03/10/2006 10
Electric Force
€
Fe = kq1q2
r2
Fg =Gm1m2
r2
03/10/2006 11
Electric Force magnitude
The electric force between 2 like charged objects is F=0.02 N. If the charge of one of the 2 objects is halved and the distance separating the 2 objects is doubled, what is the new force?
€
F = kQ1Q2
r2→ F '= k
Q1Q2
2(2r)2=F
8= 2.5 ⋅10−3N
03/10/2006 12
Electric Force: Newton’s 3rd law
03/10/2006 13
Forces as vectors
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are needed to see this picture.
2 small beads with positive charges +3q
and +q are fixed at the opposite ends of
an horizontal rod. A 3rd small charged bead is free to slide on the rod.a) Does the result depend on the sign of the charge on the bead? b) Find its equilibrium position.a) The result is independent on the
sign of the charge of the beadb) Equations to solve the problem
(q’ charge of the bead):
1 2
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+F1F2
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
-F1 F2
€
F1 = F2 ⇒ k3qq'
d12
= kqq'
d22
d = d1 + d2
⎧ ⎨ ⎪
⎩ ⎪⇒ d1 =
3− 3
2d = 0.63d
d1 d2
03/10/2006 14
€
Electrostatic potential
V (r) = kQ
r (point charge Q)
Electrostatic Potential of 2 like charges
03/10/2006 15
An electron initially travels horizontally in a uniform electric field pointing upwards with 3 possible directions of the initial velocity of the same magnitude. Which of the following relation is correct?a) vb > v > vab) va > v > vbc) vb = va > vd) vb = v = va
Answer: a) the greatest change in potential V occurs for the case with the max vertical deflection from the initial position. This occurs when the electron is projected
at an angle below the horizontal.
E field, potential difference and energy conservation
- - - - - - - - - - - - - - - - - - - - - -
+ + + + + + + + + + + + + + + ++
K = -U=-qV
a
bx
L
€
E = −dV
dx⇒ dV = −E dx
a
b
∫a
b
∫ ⇒ ΔV =Va −Vb = EL
03/10/2006 16
Example E Fields and Potential
What is the electric field at point B?
What is the electric field at point A?
Calculate the force on a Q=-3C charge placed at point A:
x
+-
B
A
d1=3 m 3 m
d2=4 m
3 m
y
-12 C +12 C
d
€
E = 2kq
d2cosθ = 2k
qd1
d3=
=18 ⋅109 ×12 ⋅10−6
(32 + 42)3 / 2× 3 = 5.18 ⋅103N /C
€
E = kq1
d12
−1
(2d1)2
⎛
⎝ ⎜
⎞
⎠ ⎟=
3
4kq
d12
F=QE
03/10/2006 17
Example E Potential
Calculate the electric potential at B
Calculate the work YOU must do to move a Q=+5 mC charge from A to B.
Calculate the electric potential at A
x
+-
B
Ad1=3 m 3 m
d2=4 m
3 m
y
-12 C +12 C
d
€
VB = kq
d−q
d
⎛
⎝ ⎜
⎞
⎠ ⎟= 0
€
VA = kq
d1
−q
2d1
⎛
⎝ ⎜
⎞
⎠ ⎟= k
q
2d1
€
W =UA −UB =Q VA −VB( ) = kqQ
2d1
03/10/2006 18
Gauss’ Law
03/10/2006 19
Electric Flux
03/10/2006 20
The electric charge density for plate 1 is - and for plate 2 is + The magnitude of the electric field associated with plate 1 is: /(20) and the filed lines are shown in the figure.When they are placed parallel one to the other, the Magnitude of the field is:
1. /0 between and ± /(20) outside2. /0 between and 0 outside3. None of the above
Plane charge density
03/10/2006 21
Insulator and Conductor sphere
A solid nonconductive spherical ball of diameter d = 2R = 1.0 cm is uniformly charged such that is has an electric field of 20 kV/m at its surface.
1. How much charge is on the ball? Chose gauss surface with r = R
2. What is the magnitude of the electric field halfway to the center of the ball?
3. If the ball were made of a conductive material, how would the answers to 1) and 2) change?
€
ΦE = E ⋅dA =Q
ε0
∫ ⇒ E4πR2 =Q
ε0
⇒
Q =ER2
k=
20 ⋅103 × 0.5 ⋅10−2( )
2
9 ⋅109= 5.55 ⋅10−11C
R
€
E(r) =kQ'
r2 r < R
ρ =Q
43πR
3=
Q'4
3πr3⇒ Q'=Q
r3
R3uniform charge distribution
inside the insulating shell
€
E(r) = kQr
R3
r =R
2⇒ E '=
kQ
2R2=E
2=10kV /m
1. Same (all charge on surface) 2. E = 0 (field inside a conductor is 0)
03/10/2006 22
Capacitors
• C Q/V
• Parallel Plate C = A/d
• Adding– Series
• Veq=V1+V2
• Qeq = Q1=Q2
21
111CCCeq
+
– Parallel• Veq=V1=V2
• Qeq = Q1+Q2
21 CCCeq +
03/10/2006 23
€
U =Q2
2C
Ci =ε0A
d→C f =
ε0A
DD<d ⏐ → ⏐ ⏐ C f >Ci ⇒ U f <U i
The parallel plates of a capacitor with vacuum between them are given equal and opposite charge. If the plates are pushed closer from a distance d to D < d, does the energy stored in the capacitor 1) Increase2) Decrease3) Remain equal
Energy stored in a parallel plate capacitor
03/10/2006 24
Metal slab in a capacitor
In the slab E = 0 and we have 2 equivalent capacitors in series
In the other case the capacitance between the 2 bottom plates is (for the other 2 plates at the same potential 1/C 0)
€
1
Ceq
=1
C+
1
C= 2
d
3ε0A
€
Ceq =3ε0A
d
03/10/2006 25
Capacitor with slab of dielectric
A dielectric slab is introduced between the plates of a capacitor. The capacitor is charged and then the dielectric is removed. Which of the following statements are correct:
1) The stored energy increases2) The capacitance increases3) The potential difference decreases
€
C0 =C
κ V0 = κV κ > 1 U =
Q2
2C
0 = quantities in vacuum
03/10/2006 26
Calculate the equivalent Capacitance C1 = 10 F
C2 = 20 FC3 = 30 FC4 = 40 FV = 50 Volts
Compare V1 with V4
C2V C3
C1
C4
€
1
Ceq
=1
C1
+1
C2 +C3
+1
C4
⇒ Ceq = 6.9μF
€
V =V1 +V23 +V4 V23 =V2 =V3
Q =Q1 =Q23 =Q4 Q23 =Q2 +Q3
V =Q
Ceq
=Q
C1
+Q
C2 +C3
+Q
C4 parallel
C1, C23, C4 in seriesC1<C4 V1>V4
03/10/2006 27
Resistors• R = V/I• R= L/A• P = RI2
• Adding– Series
• Veq=V1+V2
• Ieq = I1=I2
21
111RRReq
+– Parallel• Veq=V1=V2
• Ieq = I1+I2
21 RRReq +
03/10/2006 28
Resistors in Series
03/10/2006 29
Resistors in Parallel
€
1
Req
=1
R∑
03/10/2006 30
Light bulbs in Series
Since the wire has R =0 all current will flow through it and bulb B will fade. Instead A will look brigther since the current in A increaseas from I = V/(RA+RB) to I = V/RA
Consider the resistance of the wire negligible
03/10/2006 31
Which of the following statements arecorrect?1. A looks dimmer than B2. A looks as bright as B3. A looks brighter than B
Light Bulbs of Different Power in Parallel
A 100 W
B 60 W
RA < RB more current flows in A: IA>IB
RAIA2 > RBIB
2
What happens if A and B are connected in series?
03/10/2006 32
Resistors in series and parallel
03/10/2006 33
Kirchhoff’s Rules
• Junction Rule: Iin = Iout• A statement of Conservation of Charge
• Loop Rule:
• A statement of Conservation of Energy
I1 = I2 + I3
€
k = ΔVi
closedloop
∑closedloop
∑
-
03/10/2006 34
Kirchhoff’s laws R1 = 10 R2 = 20 R3 = 30 V1 = 50 VoltsV2 = 10 Volts2 loops
Chose 2 currents and their verse
Calculate total power delivered by the batteries
R2V1R3
R1
+-
- +
€
V1 = R1I1 + R2(I1 − I2) + R3I2
V2 = R3I2 + R2(I2 − I1)⇒
⎧ ⎨ ⎩
I1 =(R2 + R3)I2 −V2
R2
= 2.45A
I2 =R2V1 + (R1 + R2)V2
R1(R2 + R3) + R2R3
=1.18A
V2
I1 I2
€
Ptot =V1I1 +V2I2 =134.5W
PR ,tot = R1I12 + R2(I1 − I2)2 + R3I3
2 =134.5W
03/10/2006 35
RC Circuits: CHARGE OF C R
C
S1
Calculate current immediately after switch is closed.
Calculate current after switch has been closed for 2 time constants
Calculate current after switch has been closed for a very long time
Calculate charge on capacitor after switch has been closed for a long time
R=10, C=30 mF and =20 Volts
I = /R = 2 A
€
Rdq
dt+q
C
€
q(t) =Cε(1− e−t /RC )
I(t) =ε
Re−t /RC
=RC = 0.3 s
€
I(t = 2τ ) =ε
Re−2 = 0.27A
€
I(t = ∞) = 0
€
q(t = ∞) =Cε = 0.6C
03/10/2006 36
RC Circuits
You are given a 5-pack of R=2 resistors and a 5-pack of C=2F capacitors. How do you configure them to produce a circuit of 21 s time constant RC circuit?
Time constant, = RC = 21 s = 7 * 3 s = 7 * 3 F
€
3R +R
2=
7
2R = 3Ω
C
2+C =
3
2C
