Chapter 8
Electrical Energy and Capacitance
Work
•Recall: A force is conservative if the work done on
an object when moving from A to B does not depend
on the path followed. Consequently, work was
defined as:
W=PEi-PEf=-PE
• Work done BY gravity decreases PE.
•This was derived for a gravitational force, but as
we saw in the previous chapter, gravitational and
Coulomb forces are very similar:
Fg=Gm1m2/r122 with G=6.67x10-11 Nm2/kg2
Fe=keq1q2/r122 with ke=8.99x109 Nm2/C2
Hence: The Coulomb force is a conservative force.
Work & Potential Energy
•consider a charge +q moving
in an E field from A to B over
a distance d. We can ignore
gravity (why?).
•What is the work done by
the field?
•What is the change in PE?
•If initially at rest, what is
its speed at B?
•WAB=Fdcos ;  = 0, so
•WAB=Fd
•WAB=qEd (since F=qE)
•work done BY the field ON
the charge (W is positive)
• PE=-WAB=-qEd : negative,
so the potential energy has
decreased.
•Conservation of energy:
•PE+KE=0
•KE=1/2m(vf2-vi2)
•1/2mvf2=qEd
•v=(2qEd/m)
work & potential energy II
•WAB=-qEd ; negative, so work
•Consider the same situation
for a charge of –q.
•Can it move from A to B
without an external force
being applied, assuming the
charge is initially (A) and
finally (B) at rest?
must be done by the charge.
This can only happen if an
external force is applied.
•Note: if the charge had an
initial velocity the energy
could come from the kinetic
energy (I.e. it would slow
down)
•If the charge is at rest at A
and B: external work done: qEd
•If the charge has final
velocity v then external work
done:
W=1/2mvf2+|q|Ed
Conclusion
•In the absence of external forces, a positive charge
placed in an electric field will move along the field
lines (from + to -) to reduce the potential energy
•In the absence of external forces, a negative
charge placed in an electric field will move along the
field lines (from - to +) to reduce the potential
energy
+++++++++++++
--------------------
question
P
Q
•a positive charge initially at rest at P moves to Q. Will it
follow the
shortest route (as indicated by the dashed arrow) a) yes b) no
•Answer: b) since the initial force is directed along the field line
•Will the change in potential energy of the charge at Q be different
depending on which path is taken from P to Q? a) yes b) no c) depends on
whether the velocity of the charge at Q is different depending on the path
d) depends on the external forces applied
•Answer: b) no The potential energy depends only on the location in the Efield, not on the path taken.
question
2m
Y
1m
----------------------
X
•a negatively charged (-1 C) mass of 1 g is shot diagonally in an electric
field created by a negatively charge plate (E=100 N/C). It starts at 2 m
distance from the plate and stops 1 m from the plate, before turning back.
What was the initial velocity in the direction along the field lines?
answer
•Note: the direction along the surface of the plate does not
play a role(there is no force in that direction!)
Kinetic energy balance
•Initial kinetic energy: 1/2mv2=0.5*0.001*(vx2+vy2)
•Final (at turning point) kinetic energy: 0.5*0.001*vx2
•Change in kinetic energy: KE=-0.5*0.001*vy2 =-5x10-4vy2
Potential energy balance
•Change in Potential energy: PE=-qEd=1x10-6*100*1=+10-4 J
•Conservation of energy: PE+KE=0
•vy=0.44 m/s
so -5x10-4vy2+10-4=0
Electrical potential
•The change in electrical potential energy of a particle of charge Q in a
field with strength E over a distance d depends on the charge of the
particle: PE=-QEd
•For convenience, it is useful to define the difference in electrical
potential between two points (V), that is independent of the charge that
is moving: V=
PE/Q=-|E|d
•The electrical potential difference has units [J/C] which is usually
referred to as Volt ([V]). It is a scalar
•Since V= -Ed, so E= -V/d the units of E ([N/C] before) can also be given
as [V/m]. They are equivalent, but [V/m] is more often used.
Electric potential due to a single
charge
V
+
V=keq/r
r
1C
•the potential at a distance r away from a charge +q is the work done in
bringing a charge of 1 C from infinity (V=0) to the point r: V=keq/r
•If the charge that is creating the potential is negative (-q) then V=-keq/r
•If the field is created by more than one charge, then the superposition
principle can be used to calculate the potential at any point
example
1
1 m
2
-2 C
+1C
r
a) what is the electric field at a distance r?
b) what is the electric potential at a distance r?
-
a) E=E1 E2=ke(Q1/r2)-ke(Q2/[1-r]2)=ke(1/r2)-ke(-2/[1-r]2)=
ke(1/r2+2/[1-r]2) Note the -: E is a vector
b) V=V1+V2=ke(Q1/r)+ke(Q2/[1-r])=ke(1/r-2/[1-r]) Note
the +: V is a scalar
question
•
a)
b)
c)
d)
a proton is moving in the direction of the electric field. During this
process, the potential energy …… and its electric potential ……
increases, decreases
decreases, increases
increases, increases
decreases, decreases
+
+
-
PE=-WAB=-qEd, so the potential energy decreases (proton is
positive)
V= PE/q, so the electric potential that the proton feels
decreases
Note: if the proton were exchanged for an electron moving in the
same direction, the potential energy would increase (electron is
negative), but the electric potential would still decrease since the
latter is independent of the particle that is moving in the field
example
•a particle (q1) with a charge of +4.5C is fixed in space. From
a distance of 3.70 cm, a particle (q2) of mass 6.9 g and charge
–3.10 C is fired with initial velocity of 60 m/s towards to
fixed charge. What is its velocity when it is 1 cm away from
q1?
Use conservation of energy: KE+PE=0
At the start: PE=q2V=q2 x (keq1/r)=-3.39 J (r=0.037 cm)
At collision: PE=q2V=q2 x (keq1/r)=-12.55 J (r=0.01 cm)
The potential energy has decreased (more negative), so
the kinetic energy has increased by (-3.39-(12.55))=9.16 J
KE=0.5mv2 so,
KEfinal-KEinitial=0.5x0.0069vfinal2-0.5x0.0069x602=9.16 J
solve for vfinal = 79.1 m/s
question
+1
A
-2
+1
-1
+1
B
+1
-1
1) the electric potential at A is
a) zero
b) non-zero
2) the electric field at B is
a) zero
b) non-zero
3) a + particle at A would
a) move
b) not move
1) the electric potential at B is
a) zero
b) non-zero
2) the electric field at B is
a) zero
b) non-zero
3) a + particle at A would
a) move
b) not move
equipotential surfaces
compare with a map
A capacitor
+++++++++++++ +Q
d
symbol for capacitor
when used in electric
circuit
-Q
-------------------
•is a device to create a constant electric field. The potential difference
V=Ed
•is a device to store charge (+ and -) in electrical circuits.
•the charge stored Q is proportional to the potential difference V: Q=CV
•C is the capacitance, units C/V or Farad (F)
•very often C is given in terms of F (10-6F), nF (10-9F) pF (10-12F)
•Other shapes exist, but for a parallel plate capacitor: C=0A/d where
0=8.85x10-12 F/m and A the area of the plates
electric circuits: batteries
•The battery does work (e.g. using chemical
energy) to move positive charge from the –
terminal to the + terminal. Chemical energy is
transformed into electrical potential energy.
•Once at the + terminal, the charge can move
through an external circuit to do work
transforming electrical potential energy into other
forms
Symbol used in electric circuits:
+
-
Our first circuit
10nF
12V
•The battery will transport charge from one plate to the other until the voltage
produced by the charge build-up is equal to the battery charge
•example: a 12V battery is connected to a capacitor of 10 nF. How much charge is
stored?
•answer Q=CV=10x10-9 x 12V=120 nC
•if the battery is replaced by a 300 V battery, and the capacitor is 2000F, how
much charge is stored?
•answer Q=CV=2000x10-6 x 300V=0.6C
•We will see later that this corresponds to 0.5CV2=90 J of energy, which is the
same as a 1 kg ball moving at a velocity of 13.4 m/s
capacitors in parallel
C1=10nF
A C2=10nF
12V
B
At the points
the potential is fixed to
one value, say 12V at A and 0 V at B
This means that if the capacitances C1
and C2 are equal they must have the
same charge stored and the total charge stored
is Q=Q1+Q2.
•We can replace C1 and C2 with one equivalent capacitor:
Q1=C1V & Q2=C2V is replaced by: Q=CeqV
since Q=Q1+Q2 , C1V+C2V=CeqV so:
•Ceq=C1+C2
•This holds for any combination of parallel placed capacitances
Ceq=C1+C2+C3+…
•The equivalent capacitance is larger than each of the components
capacitors in series
A
B
C1=10nF
12V
C2=10nF
The voltage drop of 12V is over both
capacitors. V=V1+V2
The two plates enclosed in
are
not connected to the battery and must
be neutral on average. Therefore the
charge stored in C1 and C2 are the same
•we can again replace C1 and C2 with one equivalent capacitor but now we
start from:
V=V1+V2 so, V=Q/C1+Q/C2=Q/Ceq and thus: 1/Ceq=1/C1 + 1/C2
•This holds for any combination of in series placed capacitances
1/Ceq=1/C1+1/C2+1/C3+…
•The equivalent capacitor is smaller than each of the components
question
•
Given three capacitors of 1 nF, an capacitor can
be constructed that has minimally a capacitance
of:
a) 1/3 nF
b) 1 nF
c) 1.5 nF
d) 3 nF
The smallest possible is by putting the three in series:
1/Ceq=1/C1+1/C2+1/C3=1+1+1=3 so Ceq=1/3 nF
a more general case: what is the equivalent
C
C4
C3
C5
C6
STRATEGY: replace subgroups of
capacitors, starting at the smallest level
and slowly building up.
C2
C1
12V
•step 1: C4 and C5 and C6 are in parallel. They can be replaced
by once equivalent C456=C4+C5+C6
step II
C3
C456
C1
C2
12V
•C3 and C456 are in series. Replace with equivalent C:
1/C3456=1/C3+1/C456 so C3456=C3C456/(C3+C456)
•C1 and C2 are in series. Replace with equivalent C:
1/C12=1/C1+1/C2 so C12=C1C2/(C1+C2)
step III
C3456
C12
12V
C123456
12V
•C12 and C3456 are in parallel, replace by equivalent C of
C123456=C12+C3456
problem
C3
A
C4
C5
C2
C1
12V
B
C1=10nF
C2=20nF
C3=10nF
C4=10nF
What is Vab?
C5=20nF
•V12=V345=12V
•C45=C4+C5=10nF+20nF=30nF
•C345=C3C45/(C3+C45)=300/40=7.5nF
•Q345=V345C345=12V*7.5nF=90nC
•Q45=Q345
•V45=Q45/C45=90nC/30nF=3V
•check V3=Q3/C3=Q345/C3=90nC/10nF=9V V3+V45=12V okay!
energy stored in a capacitor
+++++++++++++ +Q
V
V
-Q
-------------------
Q
Q
•the work done transferring a small amount Q from – to + takes an
amount of work equal to W=VQ
•At the same time, V is increased, since V=(Q+Q/C)
•The total work done when moving charge Q starting at V=0 equals:
W=1/2QV=1/2(CV)V=1/2CV2
•Therefore, the amount of energy stored in a capacitor equals:
EC=1/2CV2
example
•A parallel-plate capacitor is constructed with plate area of
0.40 m2 and a plate separation of 0.1mm. How much energy is
stored when it is charged to a potential difference of 12V?
answer:
First calculate C=0A/d=8.85x10-12 x 0.40 / 0.0001=3.54x1
Energy stored: E=1/2CV2=0.5x3.54x10-8x122=2.55x10-6 J
Now let’s assume a 2000F capacitor being charged with a
300V battery: E=1/2CV2=90J
This is similar to a ball of 1 kg being fired at 13.4 m/s!!
capacitors II
+++++++++++++ +Q
A
d
-Q
-------------------
material

vacuum 1.00000
air
1.00059
glass
5.6
paper
3.7
water
80
•the charge density of one of the plates is defined as: =Q/A
•The equation C=0A/d assumes the area between the plates is
in vacuum (free space)
•If the space is replaced by an insulating material, the
constant 0 must be replaced by 0 where  (kappa) is the
dielectric constant for that material, relative to vacuum
•Therefore: C=0A/d
why does inserting a plate matter?
• molecules, such as water, can/is be polarized
• when placed in an E-field, the orient themselves along the field
lines; the negative plates attracts the positive side of the
molecules
•near to positive plate, net negative charge is collected; near the
negative plate, net positive charge is collected.
•If no battery is connected, the initial potential difference V
between the plates will drop to V/.
•If a battery was connected, more charge can be added,
increasing the capacitance from C to C
problem
•An amount of 10 J is stored in a parallel plate capacitor with C=10nF. Then
the plates are disconnected from the battery and a plate of material is
inserted between the plates. A voltage drop of 1000 V is recorded. What is
the dielectric constant of the material?
answer:
step 1: Ec=1/2CV2 so 10=0.5x 10x10-9 V2, V=44721 V
step 2: after disconnecting and inserting the plate, the voltage over the
capacitor is equal to Voriginal/ 
So: (44721-1000)=44721/ 
=1.023
problem
•
a)
b)
c)
d)
An ideal parallel plate capacitor is connected to a battery and becomes
fully charged. The capacitor is then disconnected and the separation
between the plates is increased in such a way that no charge leaks off.
The energy stored in the capacitor has
increased
decreased
not changed
become zero
answer: Ec=1/2CV2 with C=0A/d. If d increases, C becomes smaller.
The charge remains the same and Q=CV. So, if C becomes smaller, V
becomes larger by the same factor.
Rewrite: Ec=1/2CV2=1/2QV. Since Q is constant and V goes up,
Ec must increase.
Remember
•Electric force acting on object 1 (or 2):
F=keq1q2/r122
•Electric field due to object 1 at a distance r:
E=keq1/r2
•Electric potential at a distance r away from a
charge q1:
V=keq1/r
Producing a Sound Wave
•Sound waves are longitudinal waves traveling
through a medium.
•A tuning fork can be used as an example of
producing a sound wave.
•A tuning fork will produce a pure musical
note (i.e. one with a Sinusoidal Waveform).
Using a Tuning Fork to Produce a
Sound Wave
•As the tine swings to the
right, it forces the air
molecules near it closer
together
•This produces a high
density area in the air.
•This is an area of
compression
Using a Tuning Fork, cont.
•As the tine moves
toward the left, the air
molecules to the right of
the tine spread out
•This produces an area of
low density
•This area is called a
rarefaction
Using a Tuning Fork, final
•As the tuning fork continues to vibrate, a
succession of compressions and rarefactions spread
out from the fork
•A sinusoidal curve can be used to represent the
longitudinal wave
•Crests correspond to compressions and troughs to
rarefactions
Tuning Fork (DEMO)
Categories of Sound Waves
 The physical quantity that determines
the pitch of sound is the frequency.
•Audible waves
•Normal range of hearing of the human ear.
•Normally between 20 Hz to 20,000 Hz
•Infrasonic waves
•Frequencies are below the audible range.
•Eexample: Earthquakes
•Ultrasonic waves
•Frequencies are above the audible range.
•Example: Dog whistle.
Applications of Ultrasound
•Widely used as a diagnostic and treatment
tool in medicine.
•Ultrasonic Sound Imaging:
•Image small body organs/organs that can’t be
resolved by X-rays (e.g Liver, spleen).
•Observe foetuses in the womb.
•Ultrasonic flow meter to measure blood flow
(Doppler Effect).
•Cavitron Ultrasonic Surgical Aspirator
(CUSA) used to surgically remove brain
tumors.
Speed of Sound
v
B
v
Y
vair
r
r
Liquids and Gases: B is bulk
modulus, r is density (mass/volume)
Solids: Y is Young’s modulus
m
T
 (331 )
s 273 K
•
•
331 m/s is v at 0° C;
T is the absolute
temperature.
Speed of sound is higher
in solids than in fluids
(coz of atomic/molecular
spacing).
Example 6.1
John Brown hits a steel railroad rail with a hammer.
Betsy Brown, standing one mile (1609 m) down the
track, hears the bang through the cool 32 F air while
her twin sister Boopsie is lying next to her and hears
the bang through the steel by placing her ear on the
track.
DATA: Ysteel=2.0x1011 Pa, rsteel=7850 kg/m3
What is the time difference between the moments
when Betsy and Boopsie hear the bang?
4.54 s
Intensity of Sound Waves
E
P
I

A t A
Power
Area
SI units are W/m2
Intensity is proportional to square
of amplitude (pressure modulation)
Various Intensities of Sound
 The physical quantity that determines
the loudness of sound is Intensity.
•Threshold of hearing
•Faintest sound most humans can hear
•About 1 x 10-12 W/m2
•Threshold of pain
•Loudest sound most humans can tolerate
•About 1 W/m2
•The ear is a very sensitive detector of
sound waves
•It can detect pressure fluctuations as small as
about 3 parts in 1010
Decibel Scale
Sensation is logarithmic
  10 log10
I
Io
I  I 0 10  /10
• I0 is threshold of hearing
(0 dB)
• Threshold of Pain is
therefore 120 dB
Intensity vs. Intensity Level
• INTENSITY is P/A, W/m2
• INTENSITY LEVEL is in decibels (dimensionless)
Example 6.2
A noisy machine in a factory produces a sound with a
level of 80 dB. How many machines can the factory
house without exceeding the 100-dB limit?
100 machines
Spherical Waves
•
•
A spherical wave
propagates radially outward
from the oscillating sphere.
Energy propagates equally
in all directions.
P
I
4 r 2
2
I
r
1

 22
I 2 r1
Exercise
A train sounds its horn as it approaches an intersection.
The horn can just be heard at a level of 50 dB by an
observer 10 km away. Treating the horn as a point
source and neglect any absorption of sound by the air or
ground,
a) What is the average power generated by the horn?
a) 126 W
b) What intensity level of the horn’s sound is observed
by someone waiting at an intersection 50 m from the
train?
b) 96 dB
Doppler Effect
• A change in the frequency is experienced by
an observer due to relative motion between
a source of waves and an observer.
•When the source and the observer are moving
toward each other, the observer hears a higher
frequency.
•When the source and the observer are moving
away from each other, the observer hears a lower
frequency.
Doppler Effect, Moving Observer
(stationary source)
When not moving,
f v 
When observer moving,
f '  (v  vobs ) 
 v  vo 
ƒ'  ƒ 
 v 
Fig 14.8, p. 435
Slide 12
If observer moves away from
source:
 v  vo 
ƒ'  ƒ 
 v 
Fig 14.9, p. 436
Slide 13
Example 14.5
Mary is riding a roller coaster. Her mother who is
standing on the ground behind her yells out to her at a
frequency of 1000 Hz, but it sounds like 920 Hz.
(v=343 m/s)
What is Mary’s speed?
27.4 m/s
Doppler Effect, Case 2 (Source in
Motion)
•As the source moves
toward the observer (A),
the wavelength appears
shorter and the frequency
increases.
•As the source moves away
from the observer (B), the
wavelength appears longer
and the frequency appears
to be lower
Doppler Effect
Source in Motion
 '    vs T

   vs
v
  1  vs v 
f '  v'
v
f' f
v  vs

'
Doppler Effect, Source in Motion
Approaching source:
v
f' f
v  vs
Source leaving:
v
f' f
v  vs
Example 6.4
An train has a brass band playing a song on a flatcar. As
the train approaches the station at 21.4 m/s, a person on
the platform hears a trumpet play a note at 3520 Hz.
DATA: vsound = 343 m/s
a) What is the true frequency of the trumpet?
a) 3300 Hz
b) What is the wavelength of the sound?
b) 9.74 cm
c) If the trumpet plays the same note after passing the
platform, what frequency would the person on the
platform hear?
c) 3106 Hz
Application of Doppler Effect: Speed Radar
v
f' f
v  vs
•By measuring the
frequency shift in the
incident and reflected
wave, the velocity of the
moving object can be
known.
Shock Waves
•A shock wave results
when the source
velocity exceeds the
speed of the wave
itself.
•The circles represent
the wave fronts
emitted by the source
Shock Waves, final
•Shock waves carry
energy concentrated on
the surface of the cone,
with correspondingly
great pressure variations
•A jet produces a shock
wave seen as a fog
Interference of Sound Waves
•Sound waves interfere
•Constructive interference occurs when the path
difference between two waves’ motion is zero or
some integer multiple of wavelengths
•path difference = nλ
•Destructive interference occurs when the path
difference between two waves’ motion is an odd
half wavelength
•path difference = (n + ½)λ
Interference of Sound Waves
Assume sources “a” and “b” are “coherent” (producing
waves of the same wavelength). If observer is located
ra and rb from the two sources,
Source a
ra
Source b
rb
ra  rb  n for maximum
ra  rb  (n  1 2) for minimum
Observer
Exercise 14.12
A pair of speakers separated by 1.75 m are driven by
the same oscillator at a frequency of 686 Hz. An
observer starts at one of the speakers and walks on a
path that is perpendicular to the separation of the
two speakers. (Assume vsound = 343 m/s)
a) What is the position of the last intensity maximum?
a) 2.81 m
b) What is the position of the last intensity minimum?
b) 6.00 m
c) What is the position of the first intensity
maximum?
c) 27 cm
Standing Waves
•When a traveling wave reflects back on
itself, it creates traveling waves in both
directions
•The wave and its reflection interfere
according to the superposition principle
•With exactly the right frequency, the wave
will appear to stand still
•This is called a standing wave
Standing Waves, cont
•A node occurs where the two traveling
waves have the same magnitude of
displacement, but the displacements are in
opposite directions
•Net displacement is zero at that point
•The distance between two nodes is ½λ
•An antinode occurs where the standing wave
vibrates at maximum amplitude
Fundamental, 2nd, 3rd... Harmonics
•The modes of
vibration form a
harmonic series
•ƒ1
is the fundamental
and also the first
harmonic
•ƒ2 is the second
harmonic
Example 14.9
A cello string vibrates in its fundamental mode with a
frequency of 220 vibrations/s. The vibrating segment is
70.0 cm long and has a mass of 1.20 g.
a) Find the tension in the string
a) 163 N
b) Determine the frequency of the string when it
vibrates in three segments.
b) 660 Hz
Forced Vibrations
•A system with a driving force will force a vibration
at its frequency
•When the frequency of the driving force equals the
natural frequency of the system, the system is said
to be in resonance.
•This frequency is called the resonant frequency.
Example 14.12
A pair of speakers separated by 1.75 m are driven by
the same oscillator at a frequency of 686 Hz. An
observer starts at one of the speakers and walks on a
path that is perpendicular to the separation of the
two speakers. (Assume vsound = 343 m/s)
a) What is the position of the last intensity maximum?
a) 2.81 m
b) What is the position of the last intensity minimum?
b) 6.00 m
c) What is the position of the first intensity
maximum?
c) 27 cm
Standing Waves in Air Columns
•If one end of the air column is closed, a node must
exist at this end since the movement of the air is
restricted.
•If the end is open, the elements of the air have
complete freedom of movement and an antinode
exists
Harmonic Series; Tube Closed at One
End
Where n is odd
Harmonic Series; Tube Open at Both
Ends
Where n is even
Example 14.11
An organ pipe (open at one end and closed at the other)
is designed to have a fundamental frequency of 440 Hz.
Assuming the speed of sound is 343 m/s,
a) What is the length of the pipe?
a) 19.5 cm
b) What is the frequency of the next harmonic?
b) 1320 Hz