Physics 1B schedule Winter 2009. Instructor: D.N. Basov dbasov@ucsd.edu
Week
Mon
Wed
Friday
Lecture: the Electric field
1:Jan 5
Lecture: Intro, 15.1-15.3
Lecture: The Coulomb law,
Lecture: Gauss
law/examples
2: Jan 12
Lecture: Electric Flux &
Gauss’s law
3: Jan 19
University Holiday
Lecture: Potential
4: Jan 26
Lecture: capacitance
Lecture: Electric current,
Ohm’s law
Lecture: capacitor
Quiz 2: chapter 16
combinations
Lecture: Resistivity, Electric Lecture: Resistors, series
power
parallel
6: Feb 9
Lecture: Kirchhoff’s rules
Lecture: RC circuits
7: Feb 16
University Holiday
Lecture: current loop,
solenoid
Lecture: Magnetism
Quiz 3: chapter 17
Lecture: torque on current
loop, Ampere’s law
Lecture: Induced EMF
Quiz 4: chapter 18-19
Lecture: Faraday’s law,
Lenz’s law
Lecture: Energy of the
magnetic field
Lecture Inductance,
Inductors
Lecture: AC circuits
OH This week only:
Fri 4-5 pm, 6-7 pm
Quiz 1: chapter 15
5: Feb 2
8: Feb 23
9: March 2
10: March 9
Quiz 5: chapter 19-20
Lecture: discussion of the
final exam
HW: Ch15: 1,10,11,13,15,17,20,24,27,28,30,32,36,38,43,46,48
T.A.:A: Zhoushen Huang
Ch16: 1,3,5,8,12,15,19,22,23,25,29,31,33,35,43,45,47,49,60
zhohuang@physics.ucsd.edu
Ch17: 1,3,8,9,11,13,16,19,20,23,31,33,39,45,52,60
Ch18: 1,3,5,7,13,17,21,26,31,33,35,
B: Andreas Stergiou,
Ch19: 1,3,8,9,11,15,19,22,24,27,29,34,37,38,41,44,47,49,57,61
stergiou@physics.ucsd.edu
Ch20: 1,5,8,11,13,16,18,23,25,27,29,31,34,37,39,
Quizzes:
Final exam: all material in ch 15-21
HW problems, problems in class, more…
no make up final for any reason
4 best out of 5
Last update: Jan 15, 2009
No make-up quizzes for any reason
Potential difference and electric potential [16.1]
Recall physics 1A:
∆PE AB = −WAB
Potential energy
difference
= − F∆x
Work done by
a conservative force
A path
from A to B
Potential difference and electric potential [16.1]
Recall physics 1A:
∆PE AB = −WAB
= − F∆x
A path
Work done by
from A to B
a conservative force
due to moving charged objects in an electric field
Potential energy
difference
E
Physics 1B: ∆PE AB
A
B
+
d
∆PE AB = −WAB = − Fd = −(− qEd ) = qEd
Because force is opposite to the
direction of charge motion
Potential difference and electric potential [16.1]
∆PE AB = −WAB
Recall physics 1A:
= − F∆x
A path
Work done by
from A to B
a conservative force
due to moving charged objects in an electric field
Potential energy
difference
E
Physics 1B: ∆PE AB
A
B
+
∆PE AB = −WAB = − Fd = −(− qEd ) = qEd
d
∆PE
∆V ≡ VB − VA =
q
The potential difference ∆V between
final point B and initial point A: VB-VA
is defined in terms of change of PE
divided by the magnitude of the
charge.
Scalar quantity. Units 1V=1J/1C
Because force is opposite to the
direction of charge motion
sign of charge q is important!
∆PE
= VB − VA = − Ed
q
Potential difference and electric potential [16.1]
Potential: a property of the space due to charges
E
++++++++++
x V1
∆V
x V2
---------------
The potential energy is due to the
charge interacting with the field.
++++++++++
+q PE1
E
∆PE
+q PE2
---------------
∆V ≡ VB − VA =
∆PE
q
∆PE
= VB − VA = − Ed
q
Potential-Depends only position in the
field! Units (V)
Potential Energy- Depends on the
interaction of the field with a charge.
Units (J)
Related by: ∆PE=q∆V
Both PE and V are relative. Only
(∆PE and ∆V) are important!
Example
An electron in the picture tube of an older TV set is accelerated from rest
through a potential difference ∆V = 5000V by a uniform electric field. What is
the change in potential energy of the electron? What is the speed of the
electron as a result of this acceleration (assume it started from rest)?
-e
r
E −?
X
definition of electric potential
Potential due to a point charge [16.2]
∆PE
= VB − VA = − Ed
q
ke q
E= 2
r
ke q
V=
r
keq
V=
r
E=
Fig.16.5
keq
r2
[Eq. 16.4]
V=0 at
r =∞
Æ The electric potential or
work per unit charge
required to move a test
charge from infinity to
distance r from the positive
point charge as the positive
charge moves close to q.
Potential is a scalar
Potential due to a point charge [16.2]
In a crystal of Na+ Cl- the distance
between the ions is 0.24 nm. Find
the potential due to Cl- at the position
of the Na+. Find the electrostatic
energy of the Na+ due to the
interaction with Cl-.
r=0.24nm
Cl-
Na+
−19
Superposition principle:
the total electric potential at
P due to multiple point
charges is the algebraic sum
of the electric potentials due
to the individual charges
ke q 9 x10 ( −1.6 x10 )
V=
=
= −6.0V
−9
r
(0.24 x10 )
PE=qV =1.6x10-19x-6.0 = -9.6x10-19J
ELECTRON VOLT (convenient unit for atomic physics)
1eV=1.6x10-19 J
PE=-6.0 eV
9
Potential due to a point charge [16.2]
ke q
E= 2
r
ke q
V=
r
keq
V=
r
E=
Fig.16.5
keq
r2
[Eq. 16.4]
V=0 at
r =∞
Superposition principle:
the total electric potential at
P due to multiple point
charges is the algebraic sum
of the electric potentials due
to the individual charges
Potential is a scalar
Electric potential: superposition [16.2]
Two charges of +q each are placed at corners of an
equilateral triangle, with sides of 10 cm. If the Electric
field due to each charge is 100 V/m at the A find the
potential at A.
EA = 100 V/m
V at A due to each charge
ke q
E= 2
r
ke q
V=
r
A
d=10 cm
B
C
E 1
=
V r
V = Ed = 10V
Vtotal=VBA +VCA=2V =20 V
ENERGY of a CHARGE DISTRIBUTION [16.2]
q
q
a
How much energy is stored in this square
charge distribution?
W1 = ?
q
-q/2
a
q
q
q2
W2 = k
a
⎛ q2
q2 ⎞
⎟⎟
W3 = k ⎜⎜ +
⎝ a a 2⎠
⎛ q2 q2
q2 ⎞
⎟⎟
−
−
W4 = k ⎜⎜ −
⎝ 2a 2a 2a 2 ⎠
(
)
kq 2 2 2 + 1
W = W2 + W3 + W4 =
2a 2
q
-q/2
ENERGY of a CHARGE DISTRIBUTION [16.2]
Which of the charge distributions is the most stable? (has the lowest PE)
+q
-q
+q
-q
+q
-q
-q
+q
q2
q2
q2
q2
q2
q2
PE = 0 + k
−k
−k
−k
+k
−k
a
a
a
a
a 2
a 2
2q 2
PE = − k
a 2
STABLE
q2
q2
q2
q2
q2
q2
PE = 0 − k
−k
+k
−k
−k
+k
a
a
a
a
a 2
a 2
4q 2
2q 2
PE = − k
+k
a
a 2
Potentials and charge conductors [16.3]
Work on a charge done by electric force: W=-∆PE
Points A and B:
∆PE=q(VB-VA)
W=-q(VB-VA)
No net work is required to move charges between two points
at the same V!
Fig. 16-9
Charged conductor: all points on the surface have the same V.
Why?
At equilibrium:
E is perpendicular to a path between A&B Æ W=0
VA = VB
The electric potential is constant everywhere on the surface of
a charge conductor in equilibrium.
The electric potential is constant everywhere inside a
conductor and is equal to the value at the surface.
Equipotential surfaces [16.4]
An equipotential surface is a surface on which all points are the same
potential.
It takes no work to move a particle along an equipotential surface or
line (assume speed is constant).
E
+
The electric field at every point on an equipotential surface is
perpendicular to the surface.
Equipotential surfaces are normally thought of as being imaginary;
but they may correspond to real surfaces (like the surface of a
conductor).
Equipotential surfaces: point charge [16.4]
Point charge: equipotential surfaces are all
spheres centered on the charge.
+
We represent these spheres with
equipotential lines.
Note that the field lines are perpendicular to
the equipotential lines at every crossing.
Equipotential surfaces: point charge [16.4]
Point charge: equipotential surfaces are all
spheres centered on the charge.
+
We represent these spheres with
equipotential lines.
Note that the field lines are perpendicular to
the equipotential lines at every crossing.
Equipotential surfaces: point charge [16.4]
Point charge: equipotential surfaces are all
spheres centered on the charge.
+
We represent these spheres with
equipotential lines.
Note that the field lines are perpendicular to
the equipotential lines at every crossing.
Equipotential surfaces: dipole [16.4]
Fig.16-10
Capacitance [16.6]
Capacitor
a device for storing charge and energy,
can be discharged rapidly to release energy.
Applications
•Camera flash
•Defibrillators
•Electronic devices
•Computer memories (store information)
•Many more!
Parallel plate capacitor
two “infinite” planes of charge area A separated
by distance d where d<< A, carry charge +q, -q
A
+q/A=+σ
++++++++++++
----------------The charges are at the inner
surface of the capacitor
d
-q/A=-σ
Field inside the capacitor plates
By superposition of fields due to sheet of charge
Eout =0
E+
σ
Ein=
εo
A
q/A=σ
++++++++++++
-----------------
EEout =0
σ
σ
=
Ein=E+ +E- = 2
2ε o ε 0
d
E field outside the capacitor using Gauss’s
Law
use a cylinder as the Gaussian surface, ends of the
cylinder parallel to A, sides perpendicular to A
Eout
A
ΦE =
++++++++++++
----------------A
Eout
q −q
εo
= 0 = 2Eout A
Eout = 0
The charge in the
Gaussian surface is
zero.
The E field outside the
capacitor is zero
E field inside the capacitor using Gauss’s
Law
Eout =0
++++++++++++
Ein
-----------------
Φ E = Ein A =
q
εo
q
σ
Ein =
=
Aε o ε o
Capacitance [16.6]
Parallel plate capacitor
-
+
-
-
-
+
+
-
+
+
Capacitance [16.6]
Circuit Diagram:
Wire = conductor
+
Voltage
∆V
source
-
capacitor
+q
C
-q
C
conductor
-
+
-
-
-
+
+
-
+
+
Work is done to separate charges:
Capacitors stores electrical energy
Q
C=
∆V
[
Coulomb ]
=
=1F
[Volt ]
Parallel plate capacitor [16.7]
Circuit Diagram:
Wire = conductor
+
Voltage
∆V
source
conductor
A
capacitor
+q
C
-q
C
Q
C=
∆V
[
Coulomb ]
=
=1F
[Volt ]
+q
∆V
d
-q
Gauss’ law:
rearrange
σ
q
∆V
E field increases with
E=
=
=
charge density
d
ε o Aε o
Aε o
q
Aε o
to increase C:
=
∆V
d
C =
d
Increase A
Decrease d
Parallel plate capacitor [16.7]
Example: A parallel plate capacitor with 2 plates each with area 1.0 m2
separated by a distance of 1.0 mm holds +q,-q, q=10-6C
a) Find the capacitance.
A=1 m2
q
-q
b) Find the E field
C) Find ∆V across the plates
∆V
d=1mm
C=
Aε o
d
E=
q
Aε o
∆V=
Ed
1(8.9 x10 −12 )
=
= 8.9 x10−9 F = 8.9nF
0.001
1x10 −6
5
1.1
x
10
V /m
=
=
−12
(1)(8.9 x10 )
= 1.1x105 (1x10 −3 ) = 1.1x102V
Thin film capacitors
Metal film separated by thin insulators
metal
Aε o
C =
d
insulator
Metal
Making the area large and the insulating gap small
increases C
Energy stored in a charged capacitor [16.9]
A
+q
∆V
Q
C=
∆V
d
-q
Energy
stored
1
1
2
U = Q∆V = C (∆V )
2
2
parallel
plate
C
capacitor
Aε o
=
d
Q2
=
2C
1 Aε 0
2
(Ed ) 1 2
U=
2 d
U = ε 0 E Ad
2
Understanding capacitors
Suppose the capacitor shown here is charged to Q and then the battery is disconnected.
Now suppose I pull the plates further apart so that the final separation is d1.
d
++++
-----
++++
d1
-----
•
How do the quantities Q, U, C, V, E change?
•
•
•
•
•
Q:
U:
C:
V:
E:
1
U = Q∆V
2
Q2
U=
2C
= const: no way for charge to leave.
increases.. add energy to system by separating
decreases.. since energy ↑, but Q remains same
increases.. since C ↓, but Q remains same
remains the same... depends only on charge density
C1 =
d
C
d1
V1 =
d1
V
d
U1 =
Q
∆V =
C
d1
U
d
Understanding capacitors
Suppose the battery (V) is kept attached to the capacitor. Again pull the plates apart
from d to d1.
•
How do the quantities Q, U, C, V, E change?
V
d
•
•
•
•
•
C:
V:
Q:
E:
U:
++++
++++
-----
d1
V
-----
decreases (capacitance depends only on geometry)
must stay the same - the battery forces it to be V
must decrease, Q=CV charge flows off the plate
must decrease ( E =
V
σ
, E= )
D
E0
Q ε0 A
C= =
V
d
must decrease (U = 12 CV 2 )
C1 =
d
C
d1
E1 =
d
E
d1
U
1
=
d
U
d1
Combinations of capacitors [16.8]
Capacitors connected in series and parallel
+
Voltage source
Circuit diagram
resistor
R
V
capacitor
C
Conductor
V
Parallel capacitors [16.8]
Both capacitors are at same potential:
Total charge:
Fig. 16-17
Ceq
∆V = ∆V1 = ∆V2
q = q1 + q2 = C1∆V + C2 ∆V
C1∆V + C2 ∆V
q
=
=
∆V
∆V
Ceq = C1 + C2
Parallel capacitors [16.8]
For N capacitors in parallel:
Ceq = C1 + C2 + ........CN
Both capacitors are at same potential:
Total charge:
Ceq
∆V = ∆V1 = ∆V2
q = q1 + q2 = C1∆V + C2 ∆V
C1∆V + C2 ∆V
q
=
=
∆V
∆V
Ceq = C1 + C2
Capacitors in series [16.8]
∆V
Ceq
q
=
∆V
C1
+q ∆V
1
-q
C2
+q
-q
the charge on both
capacitors in series is q
q=q1=q2
q
q
∆V = ∆V1 + ∆V2 =
+
C1 C2
1
1
1
1
∆V 1 q
q
=
+
=
= ( + )
Ceq
q
q C1 C2 Ceq C1 C2
1
1
1
1
=
+
+ .........
Ceq C1 C2
CN
Combinations of capacitors [16.8]
• What is the equivalent capacitance, Ceq, of the
combination shown?
o
Ceq
C
C
C
o
C
C
1 1 1
= +
C1 C C
C
≡
C
C1 =
2
C1
C
C 3
Ceq = C + = C
2 2
34. Find the equivalent capacitance.
X
Cseries=6µF
X
1
Cseries
Ceq= 4.00+2.00+6.00=12.00 µF
1 1 4
1
=
+ =
=
24 8 24 6
34. Find the charge on each capacitor.
C1
C2
X
X
C3
C4
q = C ∆V
q1=C1∆V=4x10-6(36)=1.44x10-4 C
q2 =C2∆V=2x10-6(36)=0.72x10-4 C
q3=q4=Cseries∆V=6x10-6(36)=2.16x10-4C
Cseries =6µF
Combinations of capacitors [16.8]
• What is the relationship between V0 and V in the
systems shown below?
+Q
V0
(Area A)
+Q
d
-Q
(Area A)
d/4
V
-Q
d/4
conductor
• The electric field in the conductor = 0.
• The electric field everywhere else is: E = Q/(Aε0)
• To find the potential difference, integrate the electric field:
V0 = Ed
d
d
V = E +0+ E
4
4
1
V = Ed
2
Combinations of capacitors [16.8]
Problem 16-42. Find the equivalent capacitance between a and b.
1
1 1
1
= + =
Cser 7 5 35 /12
X
Cser = 2.9 µ F
X
4.0 µF
a
2.9 µF
6.0 µF
b
Ceq=4.0+2.9+6.0
=12.9 µF
Combinations of capacitors [16.8]
Two identical capacitors. charge one capacitor at 10 V ,
disconnect, connect the charged capacitor to the uncharged
capacitor. What is the voltage drop across the each
capacitor?
One way to do this problem:
q is constant but divided between the two capacitors
∆V0 =10V
q/2
+q
∆V
-q
C
q/2
-q/2
-q/2
C
the charge on each capacitor is reduced by 2 fold
thus the voltage across each capacitor is reduced by 2fold
∆V =
∆Vo
2
Combinations of capacitors [16.8]
Two identical capacitors. charge one capacitor at 10 V ,
disconnect, connect the charged capacitor to the uncharged
capacitor. What is the voltage drop across the each
capacitor?
Another way to do this problem
q is constant but is placed on an equivalent capacitor
∆Vo =10V
Ceq =2C
+q
-q
C
C
The voltage is reduced by 2 fold
∆V =
∆Vo
q
q
=
=
Ceq 2C
2
∆V
Capacitors with dielectrics [16.8]
C = 600 µF
Dielectric material – insulators such as
paper, glass plastic, ceramic.
“Dielectric Strength” - is the electric
field at which conduction occurs
through the material
C = 1F
dielectric material
+q
-q
Capacitors with dielectrics [16.8]
Fig.16-23
∆V =
∆Vo
κ
Potential due to
charge q decreases
by κ
κ= dielectric constant (dimensionless)
Capacitors with dielectrics [16.8]
Originally
+q
∆Vo
Co
-q
Add dielectric
+q
∆V
-q
Capacitance increases
C=
q
κq
=
∆V ∆Vo
C = κ Co
Capacitors with dielectrics [16.8]
Originally
+q
∆Vo
Co
-q
Add dielectric
C = κ Co
+q
∆V ∆Vo
∆V
=
E=
κd
d
-q
Eo
E=
κ
Electric field decreases (when not connected to a battery)
Capacitors with dielectrics [16.8]
Originally
+q
∆Vo
Co
-q
Add dielectric
C = κ Co E =
+q
∆V
-q
Eo
κ
q
q
E=
=
κε o A ε A
ε = κε o
Permittivity is increased (Compared to vacuum)
Capacitors with dielectrics [16.8]
Example: A parallel plate capacitor consists of metal sheets
(A= 1.0m2) separated by a Teflon sheet (κ=2.1) with a thickness
of 0.005 mm. (a) find the capacitance. (b) Find the maximum
voltage. The maximum electric field across Teflon is 60x106
V/m. – this is its dielectric strength.
(a)
Κ=2.1
d=0.005m
A=0.25m2
κε o A
2.1(8.8 x10−12 )(1.0)
=
C=
−3
0.005 x10
d
C = 3.7 x10−6 F
(b)
−3
∆Vmax = Edsd = 60 x10 (0.005 x10 ) = 300V
6
Capacitors with dielectrics summary [16.8]
∆Vo
+qo
Co
-qo
C=
∆Vo
q
-q
κ Co
q= CV = κ CoVo = κ q
∆V ∆Vo
=
= Eo
E=
d
d
1
1
2
C
∆
V
=
κ C ∆Vo 2 = κ PEo
PE=
2
2