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Exam 2 covers Ch. 28-33,
Lecture, Discussion, HW, Lab
Exam 2 is Tue. Oct. 28, 5:30-7 pm, 2103 Ch
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

Chapter 28: Electric flux & Gauss’ law
Chapter 29: Electric potential & work
Chapter 30: Electric potential & field
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
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Chapter 31: Current & Conductivity
Chapter 32: Circuits


(exclude 30.7)
(exclude 32.8)
Chapter 33: Magnetic fields & forces

(exclude 33.3, 33.6, 32.10, Hall effect)
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
Properties of conductors



E  0 everywhere inside a conductor
Charge in conductor is only on the surface
E  surface of conductor
---
++
+
+
++
4
Gauss’ law example:
Charges on parallel-plate capacitor


Determine fields by
superposition
Apply Gauss’ law:

E=0 inside metal

E=0 to left

 E  0, Qencl  0


-Q
Q


Q
E
Ao
No charge on outer surface
Apply Gauss’ law:

E 0

E 0
E=0 inside metal
E=Q/Ao in middle
Q
E 
Asurf , Qencl  inner Asurf
Ao
 inner  Q/ A


Area A=Length X Width
5
Electric potential: general
U 
F
Coulomb
 ds 
 qE  ds  q  E  ds
Electric potential energy difference U
proportional to charge q that work is done on
U /q  V  Electric potential difference 
 E  ds
Depends only on charges that create E-fields

Electric field usually created
 by some charge
distribution.


V(r) is electric potential of that charge
distribution
V has units of Joules / Coulomb = Volts6
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
 ke
Electric potential of charge Q  VQ r 
q
r
Q source of the electric
it
potential, q ‘experiences’

7
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.

2qQ
WYou  U  UB UA  Q(VB VA )  k
3d1
Work done by electric fields  W E field  U
8
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. W= 0
Likes repel, so YOU will
still do positive work!
q3
W1  0
W2  k
6
 
C
6
q1q2
110  2 10
 9 10 9
 3.6mJ
r12
5
qq
qq
W 3  k 1 3  k 2 3  16.2mJ
r13
r23
W tot  k
q1q2
qq
qq
 k 1 3  k 2 3  19.8mJ
r12
r13
r23
UE 19.8mJ
5m
q1
 
C
5m
q2
5m
  C
electric potential energy of the system increases
9
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

d
V  Vo  E d

V=Vo
 V  Vo  E d

10
Electric Potential and Field
Uniform electric field of E  4 yˆ N /C
What is the electric potential difference VA-VB?
y

A) -12V
B) +12V
C) -24V
D) +24V
5m
A
B
2m
2m
5m
x
11
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
13
Spherical capacitor
Charge Q moved from outer to inner sphere
Gauss’ law says E=kQ/r2
until second sphere
Potential difference V 
b
 E  ds
a
Along path shown
Gaussian surface
to find E
1 1 
kQ
1
V   2  kQ  kQ  
a b 
ra
a r
b

b
1 1 1
Q
C
 k  
a b 
V
Path to find V
+ + +
+
+
+
+ + +
14
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’
15
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:
17
I2
Current conservation
Iin
I1
I3
I1=I2+I3
I1
I3
Iout
Iout = Iin
I2
I1+I2=I3
18
Resistors in Series and parallel



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
R2
I2
=
1
2 resistors in series:
RL
Like summing lengths
 1
1 
  
R1 R2 
L
R
A

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
Quick Quiz
What is the current
through resistor R1?
9V
A. 5 mA
R1=200Ω
R4=100Ω
R1=200Ω
R3=100Ω
B. 10 mA
C. 20 mA
6V
D. 30 mA
E. 60 mA
Req=100Ω
3V
9V
Req=50Ω
21
Capacitors as circuit elements



Voltage difference depends on charge
Q=CV
Current in circuit


Q on capacitor changes with time
Voltage across cap changes with time
22
Capacitors in parallel and series
ΔV1 = ΔV2 = ΔV
Qtotal = Q1 + Q2
Ceq = C1 + C2
Parallel
Q1=Q2 =Q
ΔV = ΔV1+ΔV2
1/Ceq = 1/C1 + 1/C2
Series
23
Example: Equivalent Capacitance
C1 = 30  F
C2 = 15  F
C3 = 15  F
C4 = 30  F
C23  C2  C3 15F 15F  30F
C1, C23, C4 in series
1
1
1
1

 

Ceq C1 C23 C4

1
1
1
1



 Ceq  10F
Ceq 30F 30F 30F
C1
C2
V
C3
C4
Parallel
combination
Ceq=C1||C2
24
Charge
Discharge
R
RC Circuits
Time constant
C

  RC
 C
Start w/uncharged
Close switch at t=0
q(t)  C(1 e
I(t) 

e
t / RC
)
t / RC
R
Vcap t   1 et / RC 
R
C
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
25
Question
What is the current through R1 Immediately
after the switch is closed?
A. 10A
R1=100Ω
B. 1 A
10V
C. 0.1A
D. 0.05A
C=1µF
R2=100Ω
E. 0.01A
26
Question
What is the charge on the capacitor a long
time after the switch is closed?
A. 0.05µC
R1=100Ω
B. 0.1µC
10V
C. 1µC
D. 5µC
C=1µF
R2=100Ω
E. 10µC
27
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
28
FB on a Charge Moving in a
Magnetic Field, Formula
FB = q v x B




FB is the magnetic force
q is the charge
v is the velocity of the
moving charge
B is the magnetic field

SI unit of magnetic field: tesla (T)
N
N
T

C  m /s A  m

CGS unit: gauss (G): 1 T = 104 G (Earth surface 0.5 G)
29
Magnetic Force on a Current
I

Force on each charge qv  B

Force on length ds of wire
Ids  B
N

 Force on straight section

of wire, length L

F  IBL
Current
Magnetic force
S
Magnetic 
field
30
Magnetic field from
long straight wire:
Direction

What direction
is the magnetic
field from an
infinitely-long
straight wire?
o I
B
2 r
y
x
I
r = distance from wire
o  4 107 N / A 2 = permeability of free space
31
Current loops & magnetic dipoles
• Current loop produces magnetic dipole field.
• Magnetic dipole moment: 

  IA
current
Area of
loop
magnitude

direction
In a uniform magnetic field
Magnetic field exerts torque    B,    B sin 
Torque rotates loop to align  with B
32