Transcript Review MTE 2 - UW-Madison Department of Physics

Exam 2 covers Ch. 27-32,
Lecture, Discussion, HW, Lab
Exam 2 is Wed. Mar. 26, 5:30-7 pm,
2103 Ch: Adam(301,310), Eli(302,311), Stephen(303,306),
180 Science Hall: Amanda(305,307), Mike(304,309), Ye(308)
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Chapter 27: Electric flux & Gauss’ law
Chapter 29: Electric potential & work
Chapter 30: Electric potential & field
Chapter 28: Current & Conductivity
Chapter 31: Circuits
Chapter 32: Magnetic fields & forces

(exclude 32.6,32.8,32.10)
1
Electric flux

Suppose surface make angle  surface normal
E  E|| sˆ  E  nˆ
A  Anˆ
nˆ
Component || surface
Component  surface

Only  component
‘goes through’ surface

E = EA cos 
E =0 if E parallel A
 E = EA (max) if E  A
sˆ



Flux SI units are N·m2/C
E  E  A

2
Gauss’ law

net electric flux through closed surface
= charge enclosed / 
E 
 E  dA 
Qenclosed
o

3
Field outside
uniformly-charged sphere

Field direction: radially out from charge

Gaussian surface:





Sphere of radius r
Surface area where E  dA  0 :
2
4

r

Value of E  dA on this area:

 E
Flux thru Gaussian surface:
  E 4  r 2
Charge enclosed:

Q

Q
Gauss’ law: E 4  r  Q /o  E 
2
4 o r
2
1
4
Electric potential energy
Work is Force x distance (taking into account cosθ between 2 vectors!)
W hand  U >0
since they repel! potential energy increases
If opposte charges they attract => W <0 and potential energy decreases
5
Electric Potential
Electric potential energy per unit charge
units of Joules/Coulomb = Volts
Example: charge q interacting with charge Q
Qq
Electric potential energy  UQq  ke
r
UQq
Q
Electric potential of charge Q  VQ r  q  ke r
Q source of the electric
it
potential, q ‘experiences’

6
Example: Electric Potential
Calculate the electric potential at B
q q 
VB  k   0
d d 
Calculate the electric potential at A
q
q 
2q
VA  k 
 k
d1 3d1  3d1
B
y
x
d
-12  C
d2=4 m
-
+12  C
A
+
d1=3 m
3m
3m
Calculate the work YOU must do to move a Q=+5 mC charge
from A to B.

WYou
2qQ
 U  U B  U A  Q(VB  VA )  k
3d1
7
Work and electrostatic potential energy
Question: How much work would it take YOU to
assemble 3 negative charges?
A. W = +19.8 mJ
B. W = -19.8 mJ
C. = 0
Likes repel, so YOU will
still do positive work!
q3
W1  0
W2  k
6

C
6
q1q2
110  2 10
 9 109
 3.6m J
r12
5
qq
qq
W 3  k 1 3  k 2 3  16.2m J
r13
r23
W tot  k
q1q2
qq
qq
 k 1 3  k 2 3  19.8m J
r12
r13
r23
U E  19.8mJ
5m
q1

C
5m
q2
5m
 C
electric potential energy of the system increases
8
Potential from electric field
dV  E  d



Electric field can be
used to find changes V  V
o
in potential
Potential changes
largest in direction of

E-field.

Smallest (zero)
perpendicular to
E-field
d
d
E

V  Vo  E d

d
V=Vo
 V  Vo  E d

9
Electric Potential and Field
• Uniform electric field of E = 4i+3j N/C
• Points A at 2m and B at 5m on the x axis.
• What is the potential difference VA - VB?
A(2,0)
0
E = 4i N/C
A) -12V
B) +12V
C) -24V
D) +24V
dV
E  V  E x  

dx
VA VB  4  3 12V
B(5,0)

B
A
x(m)
dV  E x  dx 
B
A
10
Capacitors
V  Q /C
Conductor: electric potential proportional to charge:
C = capacitance: depends on geometry of conductor(s)
+Q

Example: parallel plate capacitor
V  Q /C
C
-Q
o A
Area A
d
d
Energy
stored in a capacitor:
Q2 1
1
2
U
 CV  QV
2C 2
2
V

Stored energy
Isolated charged capacitor
Plate separation increased
The stored energy
A)
1) Increases
B)
2) Decreases
C)
3) Does not change
q2
U
2C
Cini 
0 A
d
 C fin 
0 A
D
q unchanged because C isolated
q is the same
E is the same = q/(Aε0)
ΔV increases = Ed
C decreases
U increases
 C fin  Cini  U fin  U ini
12
Conductors, charges, electric fields

Electrostatic equilibrium

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


No charges moving
No electric fields inside conductor.
Electric potential is constant everywhere
Charges on surface of conductors.
Not equilibrium



Charges moving (electric current)
Electric fields inside conductors -> forces on charges.
Electric potential decreases around ‘circuit’
13
L
Electric current

Average current:

Instantaneous value:

SI unit: ampere 1 A = 1 C / s
n = number of electrons/volume
n x AL electrons travel distance L = vd Δt
Iav = Q/ t = neAL vd /L
Current density J= I/A = nqvd
(direction of + charge carriers)

Resistance and resistivity



V = R I (J =  E or E = ρ J
V = EL and E =  J  /A = V/L
R = ρL/A Resistance in ohms ()
Ohm’s Law:
15
I2
Current conservation
Iin
I1
I3
I1=I2+I3
I1
I3
Iout
Iout = Iin
I2
I1+I2=I3
16
Resistors in Series and parallel

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
Series
I1 = I2 = I
Req = R1+R2



Parallel
V1 = V2 = V
Req = (R1-1+R2-1)-1
I1+I2
I
R1
R1+R2
R2
=
I
I1
R1
I
2 resistors in series:
RL
Like summing lengths
R2
I2
=
 1 1 1
  
R1 R2 
L
R
A

17
Quick Quiz
How does brightness of bulb B compare to that of A?
A. B brighter than A
B. B dimmer than A
C. Both the same
Battery maintain constant potential difference
Extra bulb makes extra resistance -> less current
18
Quick Quiz
What happens to the brightness of bulb B
when the switch is closed?
A. Gets dimmer
B. Gets brighter
C. Stays same
D. Something else
Battery is constant voltage,
not constant current
19
Quick Quiz
What happens to the brightness of bulb A
when the switch is closed?
A. Gets dimmer
B. Gets brighter
C. Stays same
D. Something else
20
Capacitors as circuit elements

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
Voltage difference depends on charge
Q=CV
Current in circuit


Q on capacitor changes with time
Voltage across cap changes with time
21
R
RC Circuits
C

C
Start w/uncharged C
Close switch at t=0
q(t)  C(1 e
I(t) 

e
R
 t / RC
)
 t / RC
R
Vcap t  1 et / RC 
Start w/charged C
Close switch at t=0
qt   qoet / RC
qo /C t / RC
It  
e
R
Vcap t   qo /Cet / RC
22
Capacitors in parallel and series
ΔV1 = ΔV2 = ΔV
Qtotal = Q1 + Q2
Ceq = C1 + C2
Q1=Q2 =Q
• ΔV = ΔV1+ΔV2
•1/Ceq = 1/C1 + 1/C2
23
Calculate the equivalent Capacitance
C1 = 10  F
C2 = 20  F
C3 = 30  F
C4 = 40  F
V = 50 Volts
1
1
1
1
 

 Ceq  6.9F
Ceq C1 C2  C3 C4
C1
C2
V
V  V1  V23  V4
Q  Q1  Q23  Q4
C3
V23  V2  V3
Q23  Q2  Q3
Q
Q
Q
Q
V
 

Ceq C1 C2  C3 C4
C4
parallel
C1, C23, C4 in series
24
RC Circuits
What is the value of the time constant of this circuit?
A) 6 ms
B) 12 ms
C) 25 ms
D) 30 ms
25
Magnetic fields and forces
I
FB  q vBsin
F  qv  B


Magnetic force
on currentcarrying wire

B


FB  Ids  B
Magnetic force on moving
charged particle
I
B

Magnetic torque
on current loop
26
Effect of uniform magnetic field

Effect of uniform B-field on charged particle


If charged particle
is not moving - no effect
If particle is moving:
force exerted perpendicular
to both field and
velocity
F  qv  B
vector ‘cross product’

27
Lorentz force
Electron moves in plane of screen the page.
B- field is in the plane of screen to the right.
Direction of instantaneous magnetic force on electron is
A) toward the top of the page
B
B) into the page
C) toward the right edge of the page
F
v
D) out of the page
electron
12/09/2002
U. Wisconsin, Physics 208, Fall 2006
28
Trajectory in Constant B Field
• Charge enters B field with velocity shown. (vB)
• Force is always  to velocity and to B.
x x x x x x x x x x x x
x x x x x x x x x x x vx B
x x x x x x x x x x x x
q
v
F
F

Path is a circle.

mv
Radius determined by velocity: R 
qB
29
Current loops & magnetic dipoles


Current loop produces magnetic dipole field.
Magnetic dipole moment: 

  IA
current
Area of
loop
magnitude

direction
Effect of uniform magnetic field
Magnetic field exerts torque   B,    B sin 
Torque rotates loop to align  with B
30
Question on torque
Which of these loop orientations has the largest magnitude torque?
(A) a (B) b (C) c
μ τ
a
μ
τ
b
c
Answer: (c). all loops enclose same area and carry same current 
magnitude of μ is the same for all.
(c) μ upwards, μ  B and τ = μB. (a),  = 0 (b)  =  Bsin
12/09/2002
U. Wisconsin, Physics 208, Fall 2006
31