Contents
ELECTRICITY AND MAGNETISM
18.1 Capacitance
18.2 Parallel plate capacitors
18.3 Dielectrics
18.4 Capacitors in series and in parallel
18.5 Energy stored in a charged capacitor
18.6 Charging and discharging of a Capacitor
13 Capacitors
1
2
Objectives
Objectives
a) define capacitance (C = Q/V)
b) describe the mechanism of charging a
parallel plate capacitor
c) use the formula C = Q/V to derive C =
0A/d for the capacitance of a parallel
plate capacitor
d) define relative permittivity r (dielectric
constant)
e) describe the effect of a dielectric in a
parallel plate capacitor
f) use the formula C = r 0A/d
g) derive and use the formulae for
effective capacitance of capacitors in
series and in parallel
h) use the formulae U = ½ QV, U = ½
Q2/C, U = ½CV2 (Derivations are not
required.)
i) describe the charging and discharging
process of a capacitor through a
resistor
3
4
Objectives
Objectives
j)
define the time constant, and use the
formula = RC
k) derive and use the formulae
l)
derive and use the formulae
,
for discharging a capacitor through a
resistor
m) solve problems involving charging and
discharging of a capacitor through a
resistor.
and
for charging a
capacitor through a resistor;
5
6
Defining Capacitors
Short-term Charge Stores
13.1 Capacitance
7
8
Capacitors
Capacitors
Device for storing electrical
energy which can then be
released in a controlled
manner
A capacitor consists of 2
conductors of any shape
placed near each other
without touching
The region between the 2
conductors is usually filled
with an electrically insulating
material called a dielectric
Symbol in circuits is
It takes work, which is then stored as
potential energy in the electric field that
is set up between the two plates, to
place charges on the conducting plates
of the capacitor
Since there is an electric field between
the plates there is also a potential
difference between the plates
9
10
Capacitors
Capacitors
When a capacitor
each of its 2 plates carries the
same magnitude, q, of
charge
the potential of the +q plate
exceeds that of the q plate by an
amount V
The charge q and potential
difference V are related by
Q = C V where C is the capacitance
The SI unit for capacitance is farad
(F), where 1 farad = 1 coulomb/volt
We usually talk
about capacitors in
terms of parallel
conducting plates
They in fact can be
any two conducting
objects
11
Capacitance of Parallel-Plate
Capacitor
Dielectric Constant
If a dielectric is inserted
between the capacitor plates,
the electric field E inside the
capacitor is weaker than the
field E0 inside the empty
capacitor, assuming the charge
on the plates is unchanged
This reduction of the field is
described by the dielectric
constant , defined as k = E/E0
( or Er= E/E0)
The capacitance of a parallelplate capacitor without a
dielectric is
A
C
But if a dielectric
is placed between
the plates,
C
0
d
0
A
d
Thus air or vacuum has
=1
13
Capacitance of Capacitor
Capacitance C depends
on capacitor
on dielectric type
NOT on capacitor
potential difference V
12
14
Electric Energy Storage
Electric energy can be stored in
capacitors
q or
Since > 1, capacitance C
increases when a dielectric is present
16
17
Key Ideas - Capacitors
Comparison to a Battery
Capacitors are short-term charge stores.
They act as electrical springs.
There are a number of electrical parameters
(measurements of properties) that are of
interest to the electronics engineer.
These include working voltage, leakage
current, and temperature coefficient.
These are written in data tables
Property
Battery
Capacitor
Charging time
Long (hours)
Short (fraction of a
second)
Number of charge and 1000 1500 at
discharge cycles
best
Charge held
Many coulombs
How energy is stored
analogy
Billions of times
Micro coulombs
Chemical
reaction
Fuel tank
Electric field
Electrical spring
18
Example
19
Capacitor Sandwich Construction
Two metal plates
Separated by
insulating material
20
Swiss Roll Construction
21
Metal Plates + Dielectric
Capacitors consist
of two metal plates
separated by a
layer of insulating
material called a
dielectric.
22
Types of Caps
23
Types of Caps
There are two types of capacitor,
electrolytic and non-electrolytic.
Electrolytic capacitors hold much more
charge
Electrolytic capacitors have to be
connected with the correct polarity,
otherwise they can explode.
24
25
Schematic Symbol
Comparison of Types
The symbol for a capacitor is shown
below:
Advantages:
High capacitance
Can have high working
voltages.
Disadvantages:
Polarity important
High leakage current
Not stable above 10 kHz
Can be damaged by AC
Advantages:
Do not lose charge
Polarity does not matter
Stable up to 106 Hz (or
more)
Disadvantages:
Low capacitance
26
Numerical Parameters
27
Leakage
A working voltage is given. If the capacitor
exceeds this voltage, the insulating layer will break
down and the component shorts out. The working
voltage can be as low as 16 volts, or as high as
1000 V.
Leakage current means that the capacitor does
not hold its charge indefinitely. A certain amount of
current leaks across the dielectric. This is most
marked in electrolytic capacitors.
Temperature coefficient. The value of a capacitor
can change quite markedly with different
temperatures.
In an electrolytic capacitor there has to be a
current to maintain the aluminium oxide
layer. This is about 1 mA.
Over a period of time the charge leaks
away. This is called the leakage current.
Also it is important that the polarity of the
capacitor is correct, otherwise the aluminium
oxide layer is not made and the component
will conduct. The resulting heating effect can
result in the capacitor exploding.
28
Working Voltage
29
Working Voltage
All capacitors have a maximum
working voltage.
All insulators have a maximum voltage
at which they will retain their insulating
properties.
The breakdown voltage is quoted in
units of volts per metre, so it is actually
an electric field.
The breakdown voltage of air is 3000
V/mm, so a 5 mm gap will insulate up to
15 000 V.
The actual voltage at which the
breakdown occurs depends on the
thickness of the material.
30
Working Voltage
31
Temperature Coefficient
Capacitors, especially electrolytic, can
lose their capacitance, i.e. hold less
charge, when they get hot. The
decrease in capacitance can change
the characteristics of the circuit so much
that it will not work properly. Therefore
it is essential that the temperature in
which the circuit is going to operate at is
taken into consideration when designing
a circuit and choosing the components.
The thinner the material, the lower the
voltage that is needed before sparking
will occur.
If sparking occurs over a dielectric, then
a hole will be burned in the dielectric
and that is the end of the useful life for
the capacitor.
32
33
Temperature Coefficient
Temperature Coefficient
Can be tested
From this graph we can see that:
Mica capacitors are very stable with temperature.
Ceramic bead capacitors have a linear relationship.
Other types of capacitor have a temperature at which
their capacitance is at a maximum. It falls away
either side of the optimum.
34
What happens next?
How Capacitors Work
As the electrons (the charge) build
up on the plate, 2 things happen:
1. The plate becomes more
negative and so becomes less
attractive to the electrons, so the
flow of electrons gradually reduces
which means the current gradually
reduces.
2. The electrons in the other plate
are repelled by the build up of
electrons in the first plate. So they
move off the second plate.
The electrons leaving the second
plate complete the circuit.
When connected to a
potential difference (e.g. a
battery), the battery tries to
push electrons through the
wire away from its negative
terminal.
complete circuit, you get a
flow of electrons to the plate
i.e. you get a current without
a complete circuit, but only
for a short period of time.
35
36
37
If you plot a graph of the potential difference across the
plates against charge stored on the plate you find:
V
Q
38
How it Works
Basic Configuration
Capacitor
Bulb
Battery
40
39
Capacitance
As charge builds up, so does the pd across
the plates
Charged stored is directly proportional to
the potential difference across the plates.
Also, if then, V
Q, Where Q = Charge in
coulombs
Q / V = a constant.
Therefore C = Q / V
We call the constant which relates the two, C,
the capacitance because it is
- the
capacity of the plates to store charge.
Capacitance is measured in farad, F.
41
Capacitance
Farads
C=Q/V
Q - charge in coulombs
C capacitance in farads
V - potential difference in volts
1F = 1 C V-1 (A capacitance of 1 farad will
mean a charge of 1 coulomb can be stored
for each volt across the plates).
Capacitance is measured in units called
farads (F). A farad is a very big unit, and we
are much more likely to use microfarads ( F)
or nanofarads (nF). Or even picofarads (pF)
1 F = 1 × 10-6 F
1 nF = 1 × 10-9 F
1 pF = 1 × 10-12 F
42
Fixed Capacitors
43
Fixed Capacitors
PAPER
CAPACITOR
MICA
CAPACITOR
44
Fixed Capacitors
45
Fixed Capacitors
CERAMIC CAPACITOR
ELECTROLYTIC
CAPACITOR
46
Variable Capacitors
Variable Capacitors
VARIABLE CAPACITOR
A typical variable capacitor
(adjustable capacitor) is the
rotor-stator type. It
consists of two sets of
metal plates arranged so
that the rotor plates move
between the stator plates.
Air is the dielectric. As the
position of the rotor is
changed, the capacitance
value is likewise changed.
This type of capacitor is
used for tuning most
radio receivers.
47
TRIMMER CAPACITOR
A screw adjustment is used to vary the distance
between the plates, thereby changing the capacitance.
48
49
NOT IN EXAM
Color Codes
Color Codes
Although the capacitance value may be
printed on the body of a capacitor, it may
also be indicated by a color code. The
color code used to represent
capacitance values is similar to that
used to represent resistance values. The
color codes currently in use are the Joint
Army-Navy (JAN) code and the Radio
Manufacturers' Association (RMA) code
For each of these codes, colored dots or
bands are used to indicate the value of the
capacitor. A mica capacitor, it should be
noted, may be marked with either three
dots or six dots.
Both the three- and the six-dot codes are
similar, but the six-dot code contains more
information about electrical ratings of the
capacitor, such as working voltage and
temperature coefficient.
50
51
Color Codes
Color Codes
The capacitor shown in the above figure
represents either a mica capacitor or a molded
paper capacitor. To determine the type and
value of the capacitor, hold the capacitor so
that the three arrows point left to right (>).
The first dot at the base of the arrow sequence (the
left-most dot) represents the capacitor TYPE. This dot
is either black, white, silver, or the same color as the
capacitor body. Mica is represented by a black or
white dot and paper by a silver dot or dot having the
same color as the body of the capacitor. The two dots
to the immediate right of the type dot indicate the first
and second digits of the capacitance value
52
Color Codes
53
Color Code - Caps
The dot at the bottom right represents the
multiplier to be used. The multiplier represents
picofarads. The dot in the bottom center
indicates the tolerance value of the capacitor
54
Color Code - Example
55
Reading Color Code
Step 01
Hold the capacitor so the
arrows point left to right
Read the First Dot
Read the second digit dot
and apply it to the first digit.
Read the multiplier dot and
multiply the first two digits by
multiplier (Remember that the
multiplier is in picofarads).
56
57
Reading Color Code
Step 02
Ceramic Capacitor Color Code
Lastly, read the tolerance
dot.
According to the left coding,
the capacitor is a mica
capacitor whose capacitance
is 9100 pF with a tolerance
of 6%.
58
Calculate the Values
59
Calculate the Values
60
61
Parallel-plate Capacitors
made of two plates each of area A (the
plates are separated by a distance d.
13.2 Parallel-plate capacitors
62
Parallel-plate Capacitors
63
Parallel-plate Capacitors
The electric field is the sum of the electric
64
65
Parallel-plate Capacitors
Parallel-plate Capacitors
The electric fields outside the plates
cancel out.
66
Parallel-plate Capacitors
67
Parallel-plate Capacitors
The electric fields between the plates add.
The electric fields outside the plates cancel out.
Make the outside fields disappear.
68
Parallel-plate Capacitors
69
Parallel-plate Capacitors
The charges move to the inside of the plates.
Move the + and symbols toward the center.
The electric field inside is uniform.
The electric field outside is small.
70
71
Parallel-Plate Capacitors
Assume electric field uniform between
the plates, we have
E=
Q
oA
=
o
V=Ed=
C=
C=
Q
V
oA
d
13.3 Dielectrics
Qd
oA
=
Q
Qd oA
(As we have argued before)
Pictures from Serway & Beichner
72
73
Dielectrics
Dielectrics
A dielectric is an insulator with polar
molecules that is placed between the plates
of a capacitor.
Polar molecules rotate in the electric field
of the capacitor.
74
75
Dielectrics
Dielectrics
The net charge inside the dielectric is zero.
But there is leftover charge on the
surfaces of the dielectric.
76
77
Dielectrics
Dielectrics
A dielectric is an insulating material in which the individual
molecules polarize in proportion to the the strength of an
external electric field.
This reduces the electric field inside the dielectric by a
factor , called the dielectric constant.
This charge produces an electric field that
opposes the electric field of the plates.
For fixed charge Q on plates
E of plates
E
E0
and V
V0
Capacitance is
increased by .
E of dielectric
C
01/31/2005
C0
Phys112, Walker Chapter 20
79
78
79
Dielectric Properties
of common materials
Dielectrics Strength
Dielectrics are insulators: charges are not
free to move (beyond molecular distances)
Above a critical electric field strength,
however, the electrostatic forces polarizing
the molecules are so strong that electrons
are torn free and charge flows.
This is called Dielectric Breakdown, and can
disturb the mechanical structure of the
material
01/31/2005
Phys112, Walker Chapter 20
80
Material
Dielectric
Constant:
Dielectric
Strength (V/m)
Vacuum
Air (lightening)
1
1.00059
( -1) Density
2.5·1018
3.0·106
Teflon
Paper
2.1
3.7
60 ·106
Mica
5.4
100 ·106
01/31/2005
80
Phys112, Walker Chapter 20
16 ·106
81
81
Problem Type 1:
Fixed Charge
Problem Type 1:
Fixed Charge
A capacitor is charged with a battery to a
charge Q. The battery is removed and a
dielectric is inserted.
Without the dielectric: With the dielectric:
A capacitor is charged with a battery to a
charge Q. The battery is removed and a
dielectric is inserted.
With the dielectric:
Q0
C0V0
V
( E0
Ed ) d
Q
CV
Q0
C
Q
V
Q0
V0
V0
Q
Q0
V
V0
C
C0
C0
V
( E0
Ed ) d
Q
CV
Q0
C
Q
V
Q0
V0
V0
C0
82
83
Problem Type 1:
Fixed Charge
Problem Type 2:
Fixed Voltage
A capacitor is charged with a battery to a
charge Q. The battery is removed and a
dielectric is inserted.
A capacitor is connected to a battery with
voltage V and remains connected as a
dielectric is inserted.
With the dielectric:
Without the
dielectric:
1
Q
Q0
V
V0
C
C0
The electric field of the
dielectric reduces the
voltage across the
capacitor, causing the
capacitance to rise.
V
Q C0V0
V0
Q CV
C0V0
84
85
Problem Type 2:
Fixed Voltage
Problem Type 2:
Fixed Voltage
A capacitor is connected to a battery with
voltage V and remains connected as a
dielectric is inserted.
With the dielectric:
A capacitor is connected to a battery with
voltage V and remains connected as a
dielectric is inserted.
Q
V
C
Q0
V0
C0
Q
V
V0
Q
CV
V
C0V0
C
The charge on the
dielectric pulls
additional charge from
the battery to the
plates, causing the
capacitance to rise.
Q0
V0
C0
86
Capacitors with dielectrics
Capacitors with dielectrics
A dielectrics is an insulating material (rubber, glass, etc.)
Consider an insolated, charged capacitor
Notice that the potential difference decreases (k =
V0/V)
Since charge stayed the same (Q=Q0)
capacitance increases
Insert a dielectric
Q
Q
Q
87
C
Q
Q0
V
Q0
V0
Q0
V0
C0
dielectric constant: k = C/C 0
V0
Dielectric constant is a material property
V
88
89
Capacitors with dielectrics - notes
Capacitors with dielectrics - notes
Capacitance is multiplied by a factor
k when the dielectric fills the region
between the plates completely
E.g., for a parallel-plate capacitor
The capacitance is limited from
above by the electric discharge that
can occur through the dielectric
material separating the plates
In other words, there exists a
maximum of the electric field,
sometimes called dielectric strength,
that can be produced in the dielectric
before it breaks down
C
0
A
d
90
91
Example
Dielectric constants and dielectric
strengths of various materials at room
temperature
Material
Dielectric
constant, k
1.00
1.00059
80
3.78
Vacuum
Air
Water
Fused quartz
Take a parallel plate capacitor whose plates
have an area of 2.0 m2 and are separated by
a distance of 1mm. The capacitor is charged
to an initial voltage of 3 kV and then
disconnected from the charging source. An
insulating material is placed between the
plates, completely filling the space, resulting
in a decrease in the capacitors voltage to 1
kV. Determine the original and new
capacitance, the charge on the capacitor, and
the dielectric constant of the material.
Dielectric
strength (V/m)
-3 106
-9 106
92
Take a parallel plate capacitor whose plates have an area of 2 m2 and are separated by a
distance of 1mm. The capacitor is charged to an initial voltage of 3 kV and then
disconnected from the charging source. An insulating material is placed between the
plates, completely filling the space, resulting in a decrease in the capacitors voltage to
1 kV. Determine the original and new capacitance, the charge on the capacitor, and the
dielectric constant of the material.
Given:
V1=3,000 V
V2=1,000 V
A = 2.00 m2
d = 0.01 m
Find:
C=?
C0=?
Q=?
k=?
Take a parallel plate capacitor whose plates have an area of 2 m2 and are separated by a
distance of 1mm. The capacitor is charged to an initial voltage of 3 kV and then
disconnected from the charging source. An insulating material is placed between the
plates, completely filling the space, resulting in a decrease in the capacitors voltage to
1 kV. Determine the original and new capacitance, the charge on the capacitor, and the
dielectric constant of the material.
Given:
Since we are dealing with the parallelplate capacitor, the original capacitance
can be found as
C0
V1=3,000 V
V2=1,000 V
A = 2.00 m2
d = 0.01 m
A
0
d
(8.85 10 12 C 2 / N m 2 )
93
2.00m 2
Find:
1.00 10 3 m
C=?
C0=?
Q=?
k=?
18nF
The dielectric constant and the new capacitance
are
C
C0
V1
C0
V2
3 X 18nF
54 nF
The charge on the capacitor can be found to be
Q
C0 V
18 10 9 F
3000V
5.4 10 5 C
94
95
How does an insulating dielectric material reduce electric fields
by producing effective surface charge densities?
Reorientation of polar molecules
13.4 Capacitors in series and
in parallel
Induced polarization of non-polar molecules
Dielectric Breakdown: breaking of molecular bonds/ionization of
molecules.
96
97
Adding Capacitors
Resistors:
Capacitors:
V
Adding Capacitors
Series:
IR
Q Q1 Q2 , V
1
V Q
C
1
R
C
V1 V2 ,
1
C
1
C1
1
C2
Parallel:
V
V1 V2 , Q Q1 Q2 , C
C1 C2
R and 1/C enter the voltage equations in a similar way.
If you replace R with 1/C in series-parallel equations for
resistors, you get the correct result for capacitors!
98
99
Why are the charges the same on
capacitors in series?
As with resistors, the voltages across two
capacitors in series add to get the total
voltage.
As with resistors, the voltages across two
capacitors in parallel are the same.
When we discharge two capacitors in
parallel, the total charge that leaves the
capacitors is the sum of the charges. (Recall
that with resistors the sum of the currents is
the total current in parallel.)
To begin with, there is no charge on either
capacitor.
100
Why are the charges the same on
capacitors in series?
101
Why are the charges the same on
capacitors in series?
q
q
q
q
I
Before we start charging the two capacitors,
the charge within the dashed box is zero.
A the capacitors charge, the charge within
the dashed box remains zero.
102
Why are the charges the same on
capacitors in series?
+Q
103
Why are the charges the same on
capacitors in series?
+Q
Q
When the left plate of the left capacitor
acquires its final charge +Q
charge is Q.
Q
+Q
Q
The charge within the box must remain zero,
so the right capacitor must have the same
charge as the left capacitor.
104
105
Caps in Series - Equation
RT
CT
1
R1
1
1
1
...
R2
Rn
1
C1
1
1
1
...
C2
Cn
2 Caps in Series - Equation
CT
C1 C2
C1 C2
106
107
CT in Series - Example
Determine the total capacitance of a
series circuit containing three capacitors
whose values are 0.01 µF, 0.25 µF, and
50,000 pF,
CT
in Series
Solution
108
Caps in Parallel - Equation
109
CT in Parallel - Example
Determine the total capacitance in a
parallel capacitive circuit containing
three capacitors whose values are 0.03
µF, 2.0 µF, and 0.25 µF, respectively
CT = C1 + C2 + C3
n
110
111
CT in Parallel - Solution
Questions Parallel
What is the single capacitor equivalent of
this circuit below?
What is the charge on each capacitor?
12 V
112
C1
4 F
C2
6 F
113
Questions Series
Answer Parallel
12 V
C1
4 F
C2
What is the single capacitor equivalent of
this circuit below?
6 F
What is the charge on each capacitor?
Ctotal = C1 + C2
=
C1
Charge on C2 , Q2
= 6 10-6 F 12 V
= 7.2 10-5C
= 72 mC
Total charge, Q
= 48 mC + 72 mC
= 120 mC
What are the voltmeter readings?
12 V
C1
4 F
C2
6 F
114
115
Answer Series
13.5 Energy stored in a
charged capacitor
116
Energy Stored in a Battery
117
Energy Stored in a Capacitor
q q+ q
Q
Charge on plates
118
Stored Energy Equation
119
Measuring Stored Energy
Energy stored by the capacitor,
W = 1/2 QV
equation may be written as
W = 1/2 CV2
120
121
Measuring Stored Energy
Stored Energy Question 1
Calculate the charge and energy stored in
a l0 F capacitor charged to a potential
difference of:
a.) 3 V
b.) 6 V
A joulemeter is used to measure the energy transfer from a
charged capacitor to a light bulb when the capacitor discharges. The
capacitor p.d. V is measured and the joulemeter reading recorded
before the discharge starts. When the capacitor has discharged,
the joulemeter reading is recorded again. The difference of the two
joulemeter readings is the energy transferred from the capacitor
during the discharge process. This is the total energy stored in the
capacitor before it discharged. This can be compared with the
calculation of the energy stored using W = ½CV2.
Q= 30 C, W = 45 J
Q= 60 C, W = 180 J
122
123
Stored Energy Question 2
Energy in a Capacitor
A 50k F capacitor is charged from a 9 V
battery then discharged through a light
bulb in a flash of light lasting 0.2 s.
Calculate:
a) the charge and energy stored in the
capacitor before discharge,
b) the average power supplied to the light
bulb.
a) = .45 C b) = 10 W
Start with two parallel plates with no charge.
Move one charge from one plate to the other.
There is no electric field and no force, so it
requires no work.
124
Energy in a Capacitor
125
Energy in a Capacitor
After the charge is transferred, the capacitor has
a small charge and a small field.
The field causes a force on the next charge we
move, forcing us to do work.
When the charge on a capacitor is q, the
voltage is q/C and the electric field is
V/d=q/Cd.
The force on a small charge dq is
F
(dq ) E
q
dq
Cd
126
Energy in a Capacitor
127
Energy in a Capacitor
The work done in moving the charge is
1
dW Fd
qdq
C
The work done in charging the capacitor to
its final charge Q is:
Q
1
Q2 1
W
dW
qdq
CV 2 U
C0
2C 2
128
129
Energy in a Capacitor
Energy Density
Energy per unit volume in a an electric field.
U
1
CV 2
2
In a parallel-plate capacitor of volume v=Ad :
U
1
CV 2
2
u
1 0A
Ed
2 d
U
v
1
2
0
2
1
2
0
E 2 Ad
E2
130
131
Energy Density
The density of the energy stored in any
electric field, not just a capacitor, is:
u
1
2
0
13.6 Charging and
discharging of a Capacitor
E2
132
Capacitors in Circuits
133
Capacitors in Circuits
In DC circuits, capacitors just charge or
discharge.
No current flows after a capacitor is fully
charged or discharged.
Describe what happens in this circuit
after the switch is closed.
5
20
1
12 V
134
Capacitors in Circuits
135
Capacitors in Circuits
Initially positive charge on the right plate
of the capacitor pushes charge off the
left plate. It is as if the capacitor were
5
replaced by a wire.
20
When the capacitor starts charging, it
behaves like a battery that opposes the
flow of current.
+
1
5
20
1
12 V
12 V
136
137
Capacitors in Circuits
Capacitors in Circuits
Eventually, the capacitor becomes fully
charged. No more current flows. What is
the final voltage on the capacitor?
5
+
20
First, ignore the branch with the
capacitor.
Rtotal=3 . I = 4 A. V across the 1
5
resistor is IR = 4 V.
+
1
20
1
12 V
12 V
138
Capacitors in Circuits
139
Capacitors in Circuits
V across the 5 resistor is 0. Why?
V across the capacitor is 4 V.
Q on the capacitor = CV = 80 C
5
Summary:
In steady state, no current flows through
the capacitor. Just find the voltage
across the capacitor and you can
determine the charge.
20
+
1
12 V
140
141
RC Discharging
RC Discharging
Charge a capacitor
with a battery to a
voltage V.
Disconnect the
capacitor and attach
it to a resistor.
The initial charge is
Q=CV.
The charge decays
to zero but what is
Q(t)?
I
Q0
Q(t)
t
Look at the voltage
around the circuit. We
I
rule:
I
Q
C
VC
Q
C
VR
IR
0
IR
142
143
RC Discharging
I
VC
Q
C
RC Discharging
I
Q
IR 0
C
I 0, Q 0
VR
IR
dQ
dt
I
VC
The minus sign comes from:
1)I > 0
2)Q is the charge on the capacitor
3)The capacitor is discharging so
dQ
dt
0
144
Q
C
VR
IR
Q
IR 0
C
I 0, Q 0
dQ
I
dt
Q
dQ
R
0
C
dt
dQ
1
Q (t )
dt
RC
145
RC Discharging
dQ
dt
RC Discharging
1
Q(t )
RC
dQ
Q
Q(t )
This is a differential equation, but it is a really easy
one to solve.
dQ
Q
ln
dQ
Q
1
dt
RC
1
dt
RC
ln
Qf
1
dt
RC 0
Q (t )
Q0
1
t,
RC
Q (t )
Qi
t
dQ
Q
Q0
equations.
1
dt
RC
Q0e
RC
t/
146
147
RC Time Constant
RC Discharging
Discharging capacitors with three different
time constants.
= RC.
has units of seconds.
When is big, capacitors charge and
discharge slowly.
If R is large, not much current flows, so
is big.
If C is large, there is a lot of charge that
has to flow, so is big.
The time constant is the time it takes the
charge to drop to 1/e of its original value.
=3 s
=2 s
=1 s
1
e
148
149
RC Charging
RC Charging
I
A capacitor is initially
uncharged.
C
V0
Use a battery with
voltage V0 to charge the
capacitor.
R
We again use
VR
V0
IR
VC
V0
The voltage increases to
V0.
IR
Q
C
0
Q
C
The charge increases to
Q=CV0.
150
RC Charging
I
VR
V0
RC Charging
V0
We again use
V0
IR
VC
Q
C
I
V0
151
IR
Q
C
R
dQ
dt
Q
C
0
This differential equation has the solution:
0
dQ
0
dt
dQ Q
R
dt C
Q (t ) Q f 1 e
Qf
t/
CV0
RC
0
Try plugging the solution into the differential
equation and see if it works!
152
153
Charging and Discharging
RC Charging
1
1
e
=1
s
=2
s
=3
s
Charging Capacitors
What happens to
current as time
passes?
Charging
capacitors with
three different
time constants.
The time
constant is the
time it takes the
charge to rise to
1-1/e of its final
value.
Current falls away as
it becomes less
attractive for
electrons to move to
the plate from the
cell.
Note: The area under
the current-time
graph is equal to the
amount of charge
stored on the plates.
154
What happens to the charge on the
plate?
155
Charging Capacitors
When the capacitor is fully charged,
the pd across the plates will equal the
emf of the cell charging it.
Charge builds up quickly at first (a lot
of electrons arriving
each second) and
then more slowly.
The potential
difference is
proportional to
charge, so the p.d.time graph is exactly
the same shape as
the charge-time
graph.
Look at the diagram.
The cell is trying to push
electrons clockwise (with its
156
capacitor is trying to push
electrons anticlockwise (with
its push of 2V). Neither wins
so no charge flows.
157
Discharging Capacitors
Discharging Capacitors
Initially there is a
large current due to
the large potential
difference across the
plates.
The current drops as
pd drops.
Notice that the electrons are now moving the
opposite way round the circuit so the graph
shows the current as negative to show this.
158
Discharging Capacitors
Charge drops quickly at first (due to the
large current - which is of course, a large
flow of charge).
As the charge and therefore the pd across
the plates drops, so the charge drops
more slowly.
159
Charge/Discharge Cycle
As the potential difference across the plates
is directly proportional to the charge on the
plates, the p.d.-time graph is the same shape
as the charge-time graph as before.
160
161
R-C circuits
R-C circuits
So far we have assumed
that all emfs and resistances
are constant, timeindependent quantities.
This assumption fails when
we consider a circuit with a
capacitor.
Lets assume an ideal
source with emf
connected to a resistor of
resistance R and capacitor
with capacitance C.
162
When the switch is
open i is zero, as is the
charge on the capacitor
q.
When the switch is
initially closed the
current is maximized,
and q is zero. As the
capacitor charges i
decreases as q
increased until q is
maximized and i is zero.
163
Decay
Decay
164
Charging a Capacitor
165
Charging a Capacitor
The capacitor in the figure
is initially uncharged. That
means that vbc is zero at
time t = 0.
According to the loop rule
the voltage across the
resistor, vab, is equal to the
emf of the source E.
The initial current through
the resistor: Io = vab/R =
E/R
As the capacitor charges,
its voltage vbc increases
and the potential difference
across the resistor, vab,
decreases.
The sum of these two
voltages must always be
equal to E.
166
Charging a Capacitor
Charging a Capacitor
When the capacitor is fully charged, vbc = E.
Why does the current decrease as the
capacitor charges?
Let q represent the charge on the capacitor
and i the current in the circuit as some time t
after the switch has been closed. The current
will flow counter clockwise. The instantaneous
voltages for the resistor and capacitor are:
vab
iR
vbc
167
q
C
E iR
0
The potential drops by iR from a to b and
by q/C from b to c. Solving the above
equation for i we get:
i
168
q
C
E
R
q
RC
169
Charging a Capacitor
Charging a Capacitor
When the switch is closed at t = 0 the
capacitor is not yet charged, and the initial
current is, as we have stated before, E/R.
Without the capacitor, this would be the
constant value of the current.
The current and the
capacitor charge are given
as functions of time to the
right.
As q increases the voltage drop across the
capacitor increases. The drop equals E when
the charge reaches its final value Q f. The
current eventually reaches zero. When this
happens Q f = EC.
Current jumps from 0 to Io
= E/R at t = 0, then
gradually approaches zero.
The charge starts at zero
and gradually approaches
Qf.
170
Charging a Capacitor
Charging a Capacitor
We can derive general
expressions for the charge
and current as functions of
time. Due to our choice of
the direction of current, i
equals the rate at which
positive charge arrives at
the left-hand plate of the
capacitor, so i = dq/dt.
Putting this into equation,
dq
dt
E
R
q
RC
The previous equation rearranged:
dq
q CE
0
CE
0
RC
After a time equal to RC,
the current in the R-C circuit
has decreased to 1/e of its
initial value and the charge
has reached (1 1/e) of its
final value. The product RC
is called the time constant or
relaxation time of the circuit,
= RC. It is a measure of
how quickly the capacitor
charges or discharges.
t
RC
Exponentiating both sides, taking the inverse
logarithm, and solving for q:
t
t
q CE (1 e RC ) Q f (1 e RC )
The instantaneous current i is just the time
derivative:
t
dq E t RC
i
e
I o e RC
R
173
Time Constant
When we carry out the integration we get:
dt
q
172
Charging a Capacitor
q CE
CE
dt
RC
And then integrate both sides. Lets
change the integration variables to and
so we can use q and t as the upper
q
t
limits:
dq
dt
1
q CE
RC
ln
171
174
Time Constant
175
Discharging a Capacitor
When tau is small, the
capacitor charges quickly.
When it is large the
capacitor takes a longer
time to charge. Hence, tau
is the factor RC.
176
Suppose that after the capacitor in figure 26.20 has
acquired a charge Qf we remove the battery and
connect points a and c to an open switch.
When we close the switch, t = 0 and q = Qo, the
capacitor discharges through the resistor and the
charge again returns to zero.
Let i and q represent the time varying current and
charge at some instant after the switch is closed.
We choose our current direction to be the same as
we did before so we can use equation 26.10 for
E = 0. This gives:
dq
q
i
dt
RC
177
Discharging a Capacitor
dq
q
The initial current, when t =0, Io = -Qo/RC. i
dt
RC
To find q as a function of time we take the same steps as
t
charging:
dq
dt
i
q
1 t
dt
RC 0
dq
q
0
ln
q
RC
q
Qo
t
RC
q Qoe RC
i
dq
dt
Qo t RC
e
RC
t
I oe RC
178
179
Summary
By Carsten Denker
180
Capacitance
181
Calculating the Capacitance
q
CV
Parallel-plate
capacitor
A
d
0
C
C
2
C
4
C
4
0
L
ln b / a
Cylindrical
capacitor
ab
b a
Spherical
capacitor
Capacitance
[F = C/V]
A battery maintains a potential difference
between its terminals. It sets up an electric
field E which drives electrons through the wire
towards the positive terminal.
C eq
Cj
j 1
capacitor
in parallel
2
Potential energy
Potential energy
Energy density
1
Ceq
183
Energy Stored in an Electric Field
When a potential difference is
applied across several capacitors
connected in parallel, that
potential difference is applied
across each capacitor. The total
charge stored on the capacitors
is the sum of the charges stored
on all the capacitors.
When a potential difference is
applied across several capacitors
connected in series, the
capacitors have identical charges
. The sum of the potential
differences across all the
capacitors is equal to the applied
potential difference .
Isolated
sphere
0R
182
Capacitors in Parallel & in Series
n
0
n
j
1
1 Cj
capacitors
in series
184
U
U
u
q
2C
1
CV 2
2
1
2
2
0E
The potential energy of a
charged capacitor may be
viewed as being stored in the
electric field between its plates.
Material
Dielectric
Constant
Dielectric
Strength
(kV/mm)
Air
1.00054
3
Polystyrene
2.6
24
Paper
3.5
16
Transformer Oil
4.5
Pyrex
4.7
Ruby Mica
5.4
Porcelain
6.5
Silicon
12
Germanium
16
Ethanol
25
Water (20º C)
80.4
Water (50º C)
78.5
Titania Ceramic
130
Strontium Titanate
310
14
8
185
Dielectric
In a region completely filled by a
dielectric material of dielectric constant
, all electrostatic equations containing
the permittivity constant are to be
modified by replacing with
.
Polar and nonpolar dielectrics
0
E dA
q
dielectric
186