Download Chapter 14 Capacitance (E)

Physics beyond 2000
Chapter 14
Capacitance
Capacitor
• A capacitor is a device which can store charges
and thus electrical energy.
• The most common capacitor consists of two
conducting plates separated by a non-conducting
material called dielectric (such as air, paper,
plastic film etc.)
• There are two terminals, one connected to high
potential and the other to low potential.
1. conducting plates
2. dielectric
-
3. terminals
-
Capacitor
• Electrolytic capacitor: note the connection
of the terminals. The - terminal must be at a
higher potential than the - terminal.
Other kinds of capacitors
Charge storage of
a simple capacitor
• Charging a capacitor: store charges in a
capacitor.
battery
+
2. Connect the +
terminal of the
capacitor to the
+ terminal of the
power supply
+
1. Prepare a d.c. power
supply e.g. battery
-
-
capacitor
3. Connect the terminal of the
capacitor to the
- terminal of the
power supply
Charge storage of
a simple capacitor
• Charging a capacitor: store charges in a
capacitor.
battery
1. Positive charges
flow into the capacitor
through its - terminal
+
+
-
2. Negative charges
flow into the capacitor
through its - terminal
-
capacitor
Charge storage of
a simple capacitor
• Charging a capacitor: store charges in a
capacitor.
battery
+
-
I=0
+Q -Q
+
capacitor
When the capacitor is fully charged,
its potential difference is equal to
that of the power supply. No more
charges flow now.
Charge storage of
a simple capacitor
• Charging a capacitor: store charges in a
capacitor.
3. The charged
capacitor can act
like a d.c. battery.
It can supply
electric current.
battery
+
-
+Q -Q
+
capacitor
1. Disconnect the capacitor and the
power supply. The capacitor now
stores some amount of charges,
positive on one plate and negative
on the other.
2. Though there are equal amount
of positive and negative charges in
the capacitor, we still say that the
amount of charge is Q.
Charge storage of
a simple capacitor
• http://maxwell.ucdavis.edu/~electro/dc_circ
uits2/chrg_capacitor.html
Discharging a capacitor
• Discharging a capacitor: let the charges in a
capacitor to flow out through an external
circuit and use up the electrical energy.
3. Connect the
capacitor to
the circuit
+Q
-Q
+
-
1. A charged capacitor
2. A circuit with a resistor R
R
Discharging a capacitor
• Discharging a capacitor: let the charges in a
capacitor flow out through an external
circuit and use up the electrical energy.
1. Positive charges
flow out of the
capacitor from its
+ terminal. This is
the direction of
current I
4. Electrical energy
is consumed in the
resistor
+Q
-Q
+
-
R
2. Negative charges
flow out of the
capacitor from its
- terminal.
3. The amount of
charges in the
capacitor decreases.
Discharging a capacitor
• Discharging a capacitor: let the charges in a
capacitor to flow out through an external
circuit and use up the electrical energy.
Q=0
1. The current stops when
when there is not any
charge in the capacitor.
I=0
R
Charging and discharging
a capacitor
• http://www.fed.cuhk.edu.hk/sci_lab/ntnujav
a/rc/rc.html
Charging and discharging
a capacitor
1. The switch S can connect
either X or Y.
X
Y
2. A center-zero milliammeter
to indicate the direction of current
S
+
-
G
C
4. Circuit for charging
the capacitor when S is
connected to X.
R
3. Circuit for discharging
the capacitor when S is
connected to Y.
Charging and discharging
a capacitor
X
Y
S
_
-
current I
0
G
R
C
charging the capacitor
time t
discharging the capacitor
Charging and discharging
a capacitor
current I
time t
0
charging the capacitor
+Q
+
-Q
-
• Charges are flowing into the capacitor.
• The current decreases exponentially.
• The current stops when the capacitor is fully charged.
current I
Charging and discharging
a capacitor
time t
0
+Q
Discharging the capacitor
+
-Q
-
• Negative current indicates that charges are flowing
out of the capacitor.
• The current decreases exponentially.
• The current stops when the capacitor is fully discharged
(empty).
Charge storage
and potential difference
•
Charging a capacitor
1. Charges accumulate on the plates of the
capacitor.
2. The potential difference (p.d.) V increases.
3. More electrical energy is stored in the
capacitor.
+Q
-Q
V
Charge storage
and potential difference
+Q
d
-Q
V
• Consider a parallel-plate capacitor with area A and
charge Q.
• The p.d. across the plates is V.
• The separation between the plates is d.
Charge storage
and potential difference
+Q
d
-Q
V
• Find the electric field strength E between the plates.
 Q
E 
 A
Charge storage
and potential difference
+Q
d
-Q
V
• The voltage between the plates depends on the separation d.
V  E.d
Charge storage
and potential difference
+Q
d
-Q
V
More charges can
be stored for a
capacitor with large
area, small separation,
high permittivity and
high potential difference.
• Combine the above two equations
Q
V  .d
A
 A 
 Q   .V
d 
Capacitance C
• This is to describe how much charges a
capacitor stores.
• Definition: The charges stored per unit
voltage applied to a capacitor.
Q
C
V
Unit: C V-1 or F, farad.
1 F = 1  10-6 F
Example 1
• The capacitance of a capacitor is a constant.
• The higher the voltage applied to the
capacitor, the more charges the capacitor
stores.
• There is a limit of the voltage, break-down
voltage.
Capacitance of parallel-plate
capacitor
Q A
 A 
Q   .V  C  
V
d
d 
Capacitance of parallel-plate
capacitor
Smaller area A
-
C
A
-
d
• A must be the overlapping area.
-
http://micro.magnet.fsu.edu/electromag/java/capacitance/
Bigger area A
-
Example 2
The dielectric material is air. The permittivity
is o = 8.9  10-12 C2 N-1 m-2
-
-
air
Electrometer
Electrometer
• It is used to measure a potential difference
Vin without drawing a large current from
the external circuit. However it can only
measure a voltage of 1V at most.
• It can measure charge and small current.
Electrometer
•
Measure voltage
1. Disconnect all the contact screws by turning
them fully anti-clockwise.
2. Turn the short-read knob to the read position.
3. Note that the voltage should not be higher
than 1V.
Electrometer
•
Measure charge – Use the built-in capacitor C of
10-8 F.
1. Disconnect all the resistor contact screws by
turning them fully anti-clockwise.
2. Connect the capacitor by turning the capacitor
contact screw fully clockwise
3. Turn the short-read knob to the short position
and then the read position.
4. Take the reading of input voltage Vin.
5. Calculate Q = C.Vin
•
Electrometer
Measure current – Use the built-in resistor R of
either 1010  or 1011  .
1. Disconnect the capacitor contact screw and
one of the resistor contact screw by turning
them fully anti-clockwise.
2. Connect one of the resistor by turning its
contact screw fully clockwise
3. Turn the short-read knob to the short position
and then the read position.
4. Take the reading of input voltage Vin.
5. Calculate
Vin
I
R
Spooning charge
• Use an electrometer to measure the charges
• Note that the spoon carries equal amount of
charges each time. The charges stored in the
capacitor increases linearly. The voltage
thus also increases linearly.
Example 3
• This provides a method to measure the
permittivity of a dielectric.
Investigate the capacitance of a
parallel-plate capacitor
• Use the electrometer to measure the charges
stored in the capacitor.
1
• Verify that Q  A and Q 
d
Investigate the capacitance of a
parallel-plate capacitor
• Use the electrometer to measure the charges
stored in the capacitor.
• Insert a dielectric between the plates. Verify
that more charges can be stored without
change in p.d.
Variable capacitor
• The capacitance
can be changed by
changing the
overlapping area
between the plates.
Capacitors with dielectrics
• When we insert a dielectric between the
plates, the capacitance C increases markedly.
dielectric
parallel-plate capacitor
Capacitors with dielectrics
Consider an isolated capacitor with charges Q unchanged.
1. The positive and negative charges of
the molecules are displaced and form
dipoles due to the electric field.
3. Only the
charges on the
surfaces of the
dielectric have
effect.
+
+
+
+
+
+
+
+
+
+
+
+
+Q
- + - + - +
- + - + - +
- + - + - +
- + - + - +
- +
- +
- +
- +
-
-Q
2. The effect of
charges inside the
dielectric cancels.
Capacitors with dielectrics
Consider an isolated capacitor with charges Q unchanged.
1. Inside the dielectric,
there is an electric field
strength Eo due to the
parallel plates
+
+
+
+
+
+
+
+
+
+Q ++
+
- + - + - +
- + - + - +
- + - + - +
- + - + - +
2. There is also an electric field
strength E’ due to the dielectric
dipoles.
- +
- +
- +
- +
-
-Q
3. The resultant electric field strength is Ed = Eo - Ed
Capacitors with dielectrics
Consider an isolated capacitor with charges Q unchanged.
1. The resultant electric
field strength is Ed = Eo - Ed
+
+
+
+
+
+
+
+
+
+
+
+
+Q
- + - + E- d+
- + - E+ - +
o
- + - + - +
- + - + - +
- +
- +
- +
- +
-
-Q
2. For a smaller electric
field strength, V = E.d is
reduced.
3. C = Q/V. So C increases
when V is reduced.
Use of dielectric
•
Advantages:
1. Keep the plates apart
2. Increase the capacitance
3. Reduce the p.d. and thus reduce the chance
of electrical breakdown.
Use of dielectric
• Let Cd be the capacitance with dielectric of
permittivity 
• Let Co be the capacitance with vacuum as
the dielectric of permittivity o

• The relative permittivity r =
o
Relative permittivity and
breakdown field strength
Material
Relative permittivity εr
Breakdown field
strength E/Vm-1
Vacuum
1.0000
-
Air
1.0006
106
Paper
2.7
107
Mica
5
2 × 108
Glass
7
3 × 108
Water
80
-
Relative permittivity and
breakdown field strength
• What should be a good dielectric?
1. The relative permittivity is large.
2. The breakdown field strength is high.
Electrolytic capacitor
• This is a practical capacitor.
• There may be leakage current.
• Should not exceed its breakdown voltage.
Or it will explode.
Reed Switch
1. Two
magnetic
contacts M
and N
N
M
3. An a.c. supply
2. One magnetic
reed r
r
5. A coil to
magnetize the
reed r.
4. A diode to allow
current to flow in
one direction only.
Reed Switch
N
Currrent through the diode
M
T/2
current
r
time
1. Current (a.c.) flows through
the coil and magnetize the reed
r. The reed touches contact M.
Reed Switch
N
Currrent through the diode
M
T
current
r
time
1. Current is cut by the diode.
The reed r touches the contact N.
Reed Switch
N
N
M
r
• The
M
current
current
r
reed r contacts either N or M at a frequency
f, equal to the frequency of the a.c. supply
Measure C with a reed switch
Measure C with a reed switch
• Assume that the capacitor is fully charged
and discharged in each cycle.
The maximum charge stored in C is
Q = C.V where V is the e.m.f of
the battery
Measure C with a reed switch
• When the capacitor is discharged, the
ammeter shows a current I, which is the
average rate of flow of charge Q.
I = Q.f where f is the frequency
of the a.c. supply.
Measure C with a reed switch
• Q = C.V and
• I = Q.f
I
• C=
Vf
In the experiment, read I from the milliammeter.
V is the e.m.f. of the battery.
f is the frequency of the a.c. power supply.
Measure C with a reed switch
• In the above calculation, the capacitor is
assumed to be fully charged and discharged
in each cycle.
• In this case, the current I is independent on
the resistor R in the circuit.
• The resistor R is to protect the battery and
the milliammeter.
• In experiment, adjust R so that reducing R
would not produce noticeable increase in
current reading.
Measure relative permittivity εr
1. Measure the capacitance Cd of the capacitor
with the dielectric between the plates. Use reed
switch for the measurement.
2. Measure the capacitance Co of the capacitor
without the dielectrics.
3.
Cd
r 
Co
Stray capacitance
• The environment reduces the potential
difference of a capacitor and increases the
capacitance of a parallel-plate capacitor.
• The measured capacitance C of the
capacitor > expected capacitance Co.
• The increase in capacitance is called the
stray capacitance Cs.
• C = Co + Cs
Parallel-plate capacitor
C
C = Co + Cs
Cs
0
1
d

A
where Co =
d
Parallel-plate capacitor
C
C = Co + Cs
Cs
0
A
where Co =
A
d
Measure the voltage of
a charged capacitor
• What would happen if we use a moving-coil
voltmeter to measure the voltage of a
charged capacitor?
2. Connect a moving-coil
voltmeter to the capacitor
1. A charged
capacitor
+
+
+
+
+
+
+
-
Measure the voltage of
a charged capacitor
• What would happen if we use a moving-coil
voltmeter to measure the voltage of a
charged capacitor?
+
+
+
+
+
+
+
-
3. The charges flow
through the voltmeter
and reduces the amount
of charges in the capacitor
Measure the voltage of
a charged capacitor
• What would happen if we use a moving-coil
voltmeter to measure the voltage of a
charged capacitor?
+
+
+
+
+
-
4. There is
a leakage of charge
Measure the voltage of
a charged capacitor
• The capacitor is discharged through the
voltmeter. The reading on the voltmeter
drops gradually.
+
+
+
+
+
-
4. There is
a leakage of charge
Measure the voltage of
a charged capacitor
• Use a voltage follower (will be studied later)
to measure the voltage of the capacitor.
• The voltage follower draws a very small
amount of charges from the capacitor.
Example 4
• Note that the charge Q on each plate is constant.
• However the area A decreases when a plate is
moved sideway.
Prove that the p.d. between the plates decreases.
Move sideway
Capacitance of
two concentric spheres
1. A small sphere with radius
a and charge +Q at the middle
2. An earthed concentric shell
with radius b.
3. What is the potential of
the shell?
V=0
Capacitance of
two concentric spheres
1. By induction, there is
charge on the inner surface
of the shell.
What is the charge?
V=0
a
+Q
b
-Q
Capacitance of
two concentric spheres
E=?
Electric field strength
between the spheres
V=0
V=?
P.d. between
the spheres
a
+Q
b
-Q
C=?
Capacitance of the
concentric spheres
Capacitance of
two concentric spheres
• Use Gauss’ law to find the electric field
strength Er between the spheres.
V=0
a
+Q
Er = ?
b
r
-Q
1
Q
Er 
. 2
4 r
Capacitance of
two concentric spheres
• Find the p.d. V between the spheres.
• Note that V = 0 at r = b. (Not ∞!)
• Hint: V   Er .dr
V=0
a
+Q
V=?
b
-Q
Q 1 1
V
.(  )
4 b a
Capacitance of
two concentric spheres
• Find the capacitance C.
C=?
V=0
a
+Q
b
-Q
ab
C  4
ba
Capacitance of
an isolated sphere
1. A conducting sphere
with radius a.
2. What is its capacitance C?
3. Put charge +Q on it
V
Q
4a
C  4a
a
+Q
4. Imagine a shell with
radius ∞ enclosing this
sphere
5. Find the p.d. V and
then C.
Example 5
• Find the capacitance of earth.
• Earth is an isolated sphere.
Example 6
• If C = constant, ΔQ = C. ΔV.
Combination of capacitors
• In parallel :
+
-
All positive terminals +
are connected together.
So are the negative
terminals.
+
-
+
-
• In series :
The positive and
negative terminals
are in succession.
+
+
-
-
+
-
+
-
-
Capacitors in parallel
C   Ci
Prove that C = C1 + C2 for two capacitors in
parallel.
C1
+
Hint: Suppose that the p.d. across
each capacitor is V.
-
+
-
+
C2
Capacitors in series
1
1

C
Ci
Prove that
parallel.
1
1
1


C C1 C2
for two capacitors in
Hint: Suppose that the charge of
each capacitor is Q.
+
+Q -Q
+Q -Q
-
Example 7
• You may keep the unit μC in your
calculation.
Doubling voltage rating
• If we have capacitors Co with breakdown
voltage Vo , how to combine the capacitors
so that they can stand 2.Vo?
Co
+ -
Use 4 capacitors.
Co
+ -
+
Prove it.
-
+
Co
+
Co
-
Doubling voltage rating
• Note that the p.d. across each capacitor is
only Vo.
Vo
Vo
Co
+ -
Use 4 capacitors.
Co
+ -
+
Prove it.
-
+
Co
+
Co
-
Combining capacitors
1. A capacitor C1
with charge Q.
C1
+Q -Q
C2
2. Two capacitors
are combined in series.
3. Connect C1
to the combined
capacitors.
C3
4. Suppose that all capacitors
have the same capacitance C.
Combining capacitors
• Charges in capacitor C1 will be shared with
capacitor C2 and capacitor C3.
C1
C2
+Q2
-Q2
+Q1-Q1
-Q3
C3
+Q3
Find the charges in each capacitor
Combining capacitors
• Compare the following two circuits. Find
the relations among the charges. Note that
charges are conserved!
C1
+Q-Q
C2
C1
C3 C2
+Q2
-Q2
+Q1-Q1
-Q3
C3
+Q3
Combining capacitors
• Q1 + Q2 = Q  Q2 = Q – Q1
• Q3 – Q2 = 0  Q3 = Q2
C1
+Q-Q
C2
C1
C3 C2
+Q2
-Q2
+Q1-Q1
-Q3
C3
+Q3
Combining capacitors
1.
2.
Q2 Q  Q1
V2 

C
C
V2
C2
Q1
V1 
C
V1
C1
+Q2
-Q2
+Q1-Q1
3.
Q3 Q2
V3 

 V2
C
C
-Q3
+Q3
C3 V3
Combining capacitors
The potential difference is zero round the circuit.
V1 – V2 - V3 = 0  V1 = 2.V2
0V
V1
C1
V2
C2
+Q2
-Q2
+Q1-Q1
-Q3
+Q3
C3 V3
Combining capacitors
1.
2(Q  Q1 )
2. C 
V1
for C2
V1/2
C2
Q1
C
V1 for C1
V1
3. From 1 and 2,
Q1 =2Q/3 and
Q2 = Q3 = Q/3
C1
+Q2
-Q2
+Q1-Q1
-Q3
+Q3
C3 V1/2
Combining capacitors
What is the capacitance of the combination?
Hint: C2 and C3 are in series. C1 is in parallel with C2 and C3.
V1
C1
V1/2
C2
Q
3
Q

3
2Q
2Q

3
3

Q
3
Q
3
C3 V1/2
Combining capacitors
What is the capacitance of the combination?
3C
C'
2
V1
C1
V1/2
C2
Q
3
Q

3
2Q
2Q

3
3

Q
3
Q
3
C3 V1/2
Energy stored in a capacitor
Suppose that a capacitor of capacitance C
stores charges Qo and the p.d. is Vo.
V voltage
V
Vo
Vo 
0
1
.Q
C
1
.Qo
C
Q charge stored
Qo
Energy stored in a capacitor
If the charges has been increased by ΔQ,
the work done is W  V .Q  Q .Q
C
The work done is represented by the area below
the V-Q graph.
V voltage
V
1
.Q
C
Vo
V
0
ΔQ
Q charge stored
Qo
Energy stored in a capacitor
The energy is
1 Qo2 1
1
U
 CVo2  QoVo
2 C 2
2
V voltage
V
1
.Q
C
Vo
0
Q charge stored
Qo
Energy stored in a capacitor
Another method is to find the sum of ΔW.
Qo
Q
W   W   .Q 
Q 0 C
Qo
Q
1Q
0 C dQ  2 C
V voltage
1
V  .Q
C
Vo
V
0
ΔQ
Q charge stored
Q
2
o
Energy stored in an isolated
conducting sphere
QO
a
1. Find the potential Vo.
2. Calculate the energy stored
1
from U = Vo Qo
2
3.
Qo2
U 
.
8 a
1
Example 8
• Choose a suitable equation for the energy
from
The energy is
1 Qo2 1
1
2
U
 CVo  QoVo
2 C
2
2
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html#c1
Camera flash
• C = 2000 μF and V = 50 V.
• What is the energy stored in the capacitor?
2.5 J
Camera flash
• All the energy stored in the capacitor is
given out in 1 ms.
• What is the power of the flash?
2.5 kW
Charging a capacitor
with a cell
S
C
emf Vo
R
empty capacitor
Charging a capacitor
with a cell
S
+
I
emf Vo
R
-
C
p.d.=V
After closing the switch S, current I starts to flow in
the circuit. Charges are stored in the capacitor.
The current decreases with time when more charges
are stored. The p.d. of the capacitor increases with time.
Charging a capacitor
with a cell
S
Qo ++++
I=0
emf Vo
R
C
p.d. = Vo
-Qo - - - -
When the p.d. of the capacitor equals that of the cell,
current stops. The capacitor is fully charged.
Charging a capacitor
with a cell
S
Qo ++++
I=0
C
p.d. = Vo
-Qo - - - emf Vo
R
1. The electrical potential energy stored in C is
1
1
2
U C  CVo  QoVo
2
2
2. The electrical potential energy provided by the cell is
Uo = QoVo = 2.UC
Charging a capacitor
with a cell
S
Qo ++++
I=0
C
p.d. = Vo
-Qo - - - emf Vo
R
Half of the energy is lost when current flows
through the resistor R.
Example 9
• The efficiency of the circuit is 50%.
Typical examples
• Energy stored in a capacitor is
2
1Q
1
1
2
U
 CV  QV
2 C 2
2
V
Q
C
V
+Q -Q
Joining 2 capacitors with same C
1. A charged capacitor
with charge Q.
Q
+
S
C
C
-
3. Connect both capacitors
by closing switch S.
2. An uncharged
capacitor Q.
S
Joining 2 capacitors with same C
Find the energy stored before the capacitors
are connected.
S
2
1Q
Uo 
2 C
Q
+
C
C
-
S
Joining 2 capacitors with same C
Find the energy stored after the capacitors
are connected.
S
Q
2
+
-
C
Q
2
S
The charge in each capacitor is
+
C
-
Q
2
Joining 2 capacitors with same C
Find the energy stored after the capacitors
are connected.
S
Q
2
+
-
C
Q
2
S
The energy in each capacitor is
+
C
-
Q 2
( )
1 2
1
 Uo
2 C
4
Joining 2 capacitors with same C
Find the energy stored after the capacitors
are connected.
S
Q
2
+
-
C
Q
2
+
C
-
S
1
The total energy is U o. Half of the energy is lost
2
in the wire.
Inserting a dielectric
with constant V
• Fully charge a capacitor. The p.d. of the
capacitor is equal to the emf of the cell, Vo
dielectric
C
Vo
Inserting a dielectric
with constant V
• Insert the dielectric between the plates. Note that
the p.d. of the capacitor is still Vo.
• The capacitance is C’ = εr.C
dielectric
C’
Vo
Inserting a dielectric
with constant V
• As there is an increase in capacitance, the energy
stored and charges stored also increase.
1
U '  C 'Vo2
Q
'
C
'.
V
o
2
dielectric
Q’
C’
-Q’
Vo
Inserting a dielectric
with constant V
• Current flows into the capacitor from the cell
when the dielectric is inserted.
• This increases both the energy and the charge in
the capacitor.
dielectric
Q’
C’
-Q’
Vo
Inserting a dielectric
with constant Q
• Charge the capacitor so that it stores charge
Qo.
dielectric
Qo
C
-Qo
Vo
Inserting a dielectric
with constant Q
• Disconnect the capacitor from the cell by opening
the switches.
• Note that the charge is kept in the capacitor.
dielectric
Qo
C
-Qo
Vo
Inserting a dielectric
with constant Q
• Insert the dielectric between the plates. The
capacitance C’ = εr.C.
dielectric
Qo
-Qo
C’
Vo
Inserting a dielectric
with constant Q
• As there is an increase in capacitance, the energy
stored and the p.d. decrease.
1 Qo2
U '
2 C'
dielectric
Qo
-Qo
Qo
V '
C'
C’
Vo
Inserting a dielectric
with constant Q
• There is not any current as Q is a constant.
• Some energy of capacitor is lost to the dielectric.
dielectric
Qo
-Qo
C’
Vo
Moving apart two charged
capacitor with constant V
Charge a parallel-plate capacitor. And then
move the plates apart.
Note that an external force is needed to do
work.
+Q
C
-Q
Vo
Moving apart two charged
capacitor with constant V
As the separation increases, the capacitance
C’ is less.
C’
Vo
Moving apart two charged
capacitor with constant V
Both the energy and charge stored decrease.
1
U '  C 'Vo2
2
Q' C '.Vo
+Q’
C’
-Q’
Vo
Moving apart two charged
capacitor with constant V
There is current flowing from the capacitor to the
cell when we pull the plates apart.
This decreases the energy and the charge of the
capacitor.
+Q’
C’
-Q’
Vo
Moving apart two charged
capacitor with constant Q
Charge a parallel-plate capacitor.
The charge stored in the capacitor is Qo.
+Qo
-Qo
C
Vo
Moving apart two charged
capacitor with constant Q
Off the switch to keep the charges in the
capacitor.
+Qo
-Qo
C
Vo
Moving apart two charged
capacitor with constant Q
Move the plates apart by an external force.
The capacitance C’ is less.
+Qo
-Qo
C’
Vo
Moving apart two charged
capacitor with constant Q
The energy stored and the p.d. increase.
1 Qo2
U '
2 C'
Qo
V '
C'
Q’
C’
-Q’
Vo
Moving apart two charged
capacitor with constant Q
The energy is increased due to the work done
by external force to move the plates apart.
Q’
C’
-Q’
Vo
Charging a capacitor
at constant rate
• Keep the current constant Io while charging the
capacitor.
• Decrease R to keep the current constant. Read the
current from the ammeter.
Io
3. The current can
maintain constant
for a period of time
S
A
1. The p.d. of the capacitor
increases and decreases
the current.
C
Vo
R
2. Decrease R from Ro to zero
to increase the current.
Charging a capacitor
at constant rate
• When the capacitor is full, the current stops
suddenly.
S
A
Qo
C
Vo
-Qo
R
Charging a capacitor
at constant rate
• The charge in the capacitor is Qo and the p.d.
is Vo.
V
Q
I
Vo
Qo
Io
0
to
Vo
Io 
Ro
t
0
to
Qo  I o .to
t
0
to
Qo
Vo 
C
t
Charging a capacitor
at constant rate
• The time required to fully charge a capacitor is to =
C.Ro. It depends on the resistor and the capacitor.
V
Q
I
Vo
Qo
Io
0
to
Vo
Io 
Ro
t
0
to
Qo  I o .to
t
0
to
Qo
Vo 
C
t
Charging a capacitor
at constant rate
• In practice, there are other resistances. The
current does not drop to zero suddenly.
I
Io
0
to
t
Charging a capacitor
via a resistor
• Close the switch to charge an empty capacitor.
Vo
• Initially (at t = 0), the current Io = .
R
Io
S
A
C
Vo
R
Charging a capacitor
via a resistor
• The current I decreases with time as the
charge in the capacitor accumulates.
I
S
A
Q
C
Vo
I
R
VR
-Q
Vc
Charging a capacitor
via a resistor
• There are three p.d. (Vo, Vc and VR)
Vo is constant
I
S
A
Q
C
Vo
dQ
VR  IR  R
dt
I
R
VR
Vc
-Q
Q
VC 
C
Charging a capacitor
via a resistor
• Use Vo = VC + VR, to find the charge Q stored
in the capacitor at time t.
t

Q  Qo (1  e CR )
I
S
A
Q
C
Vo
I
R
VR
-Q
Vc
Charging a capacitor
via a resistor
Q
I
Io
Qo
0
t
Q  Qo (1  e
t

CR
)
0
t
I  I o .e

t
CR
Charging a capacitor
via a resistor
What is the time t1/2 needed
to fill the capacitor by half?
Q
1
Hint: Q = Qo.
2
Qo
1
Qo
2
0
t1/2 = CR.ln2
= 0.693CR
t
t1/2
Q  Qo (1  e

t
CR
)
t1/2 is the half-time.
The product CR is called
the time constant.
Charging a capacitor
via a resistor
• If there is not any resistor in the circuit, a large
current pulse will flow through and the capacitor is
fully charged almost immediately.
I
S
A
Q
Vo
C
I
-Q
Vc
Charging a capacitor
via a resistor
Vc
VR
Vo
Vo
0
t
V  Vo (1  e
t

CR
)
0
t
VR  Vo .e

t
CR
Examples 10 and 11
• Time constant = CR. Longer time is needed
to charge the capacitor fully for a circuit
with large CR.
• Half-time t1/2 = CR.ln2.
Discharging a capacitor
via a resistor
Suppose that the initial p.d. of the capacitor is Vo
and the charge stored is Qo
S
A
Q0
C
Vc = Vo
-Q0
R
Qo = C.Vo
Discharging a capacitor
via a resistor
The initial current Io =
S
A
Vo
Qo

R
CR
Io
Qo
C
Io
R
-Qo
Vc = Vo
Discharging a capacitor
via a resistor
The current I decreases with time. I  I .e
o
S
A
I
Q
C
I
R
VR
-Q
Vc
t

CR
Discharging a capacitor
via a resistor
The charge Q decreases with time. Q  Qo .e
S
A
I
Q
C
I
R
VR
-Q
Vc

t
CR
Discharging a capacitor
via a resistor
Q
I
Qo
Io
0
Q  Qo .e
t

CR
t
0
I  I o .e

t
CR
t
Discharging a capacitor
via a resistor
Vc
VR
Vo
Vo
0
VC  Vo .e
t

CR
t
0
VR  Vo .e

t
CR
t
Discharging a capacitor
via a resistor
VC  Vo .e

t
CR
VR  Vo .e

t
CR
Discharging a capacitor
via a resistor
Find the time needed for the voltage to drop
by half.
Discharging a capacitor
via a resistor
• The time constant = CR.
• The half-time t1/2 = CR.ln2
• Similar to that of charging a capacitor via a
resistor.
Energy lost by resistor during
charging and discharging
• UR = I2.R. t
2

Q
UR =
2
o
I
.
R
.
dt

0
2C
• During charging, half the energy supplied by the
battery is lost by the resistor. And half of the energy
is stored in the capacitor.
• During discharging, all the energy stored in the
capacitor is lost.
Example 12
• For a circuit with large CR, more time is
needed to fully discharge the capacitor.
Investigation with square waves
1.
to CRO2
Use a signal
VR
generator to input
a square wave
(a.c.)
R
2. Use CRO2 to view the
p.d. VR with time
It also shows the current
I with time.
High V
I
C
Vc
Low V
3. Use CRO1 to view the
p.c. VC with time
to CRO1
Investigation with square waves
to CRO2
VR
R
High V
I
C
Vc
to CRO1
Low V
Note that the two CRO cannot be connected
at the same time in practice because of the ground
lead.
Investigation with square waves
Input voltage
0
t
1. Charging time
VC
0
t
2. Discharging time
VR
0
t
Investigation with square waves
Input voltage
0
t
VC
0
t
VR
0
t
If CR is too large,
each period of input
voltage is not
enough for complete
charging and
discharging the
capacitor.
Investigation with square waves
Input voltage
0
t
VC
0
t
VR
0
t
If CR is too small,
complete charging
and discharging
occur before the
end of each cycle
of the input voltage.