Topic Seven: Capacitance (AS 90523, Physics 3.6)
Electric Field Strength
Objectives
By the end of this section you should be able to:
1 define an electric field qualitatively
2 define the strength of an electric field
3 state what is shown by the direction of an electric field
4 give an example of a situation in which there is a uniform electric field
5 carry out calculations involving the work done in moving a charged particle in a uniform
electric field
6 state and use two formulae for electric field strength, and state the two units for electric field
strength which correspond to these formulae.
A field is a space in which forces act.
An electric field is a space in which an electric charge experiences a force. If a charge q
experiences a force F, then the electric field strength E is defined by:
E= F
q
Units for E are N C–1 or V m–1
This formula is usually remembered as:
F = qE.
Electric field strength E is a vector. Its direction is given by the direction of
the force on a positive charge.
We can show the field by drawing field lines to represent it. The field lines
always show the direction of the force on a positive charge.
The field is stronger where the field lines are closer together.
+
–
Field between large parallel plates
When the dimensions of the plates are much bigger than the
plate separation, then the field between the plates is uniform.
Near the edges (e.g. D) the field is somewhat weaker. A
charge would experience the same force whether at A, B or
C. The force it would experience at D would be less.
–
–
+
+
–
–
–
+
+
+
A
–
–
+
+
B
C
–
–
+
+
D
Another important formula for E
Suppose in the above we move a charge q from the negative plate to the positive plate. Then
the change in potential energy for q is given by Ep = qV where V is the potential difference
between the plates. Now:
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Continuing Physics
Work done = Change in energy
⇒ F × d = qV
⇒ qE × d = qV
⇒E×d=V
where d is the distance between the plates
where E is the field strength between the plates
E=V .
d
Making E the subject, we get
This shows why E may also be measured in units of V m–1.
Example
+
In calculations like this
one, we can almost
always ignore the force of
gravity. This is because
the force produced by the
electric field will be very
much stronger. Gravity is
a relatively weak force.
It only dominates where
huge masses are involved.
A 12 V battery is
connected to two metal
plates 6.0 mm apart.
6 mm
12 V
(a) Find the strength of the electric field between the plates.
12
E=V =
= 2.0 × 103 V m –1
d 6.0 × 10−3
(b) What is the force on an electron placed between the plates?
F = qE = 1.6 × 10–19 × 2.0 × 103 = 3.2 × 10–16 N
(c) An electron leaves the positive plate with negligible kinetic energy. How much kinetic
energy does it have on reaching the positive plate?
As the electron moves from the negative plate to the positive plate, it is losing electrical
potential energy and gaining kinetic energy. Because energy is conserved, we know
that the loss in electrical potential energy will equal the gain in kinetic energy.
Since it started out with no kinetic energy:
final E = change in Ep
= q ×V
= 1.6 × 10–19 × 12
= 1.92 × 10–18 J
The situation above is similar to what happens to a falling body. As a body falls, it
loses gravitational potential energy and gains kinetic energy. If we can ignore friction,
the loss of gravitational potential energy will equal the gain in kinetic energy.
(d) Find the speed with which the electron above arrives at the positive plate. The mass
of an electron is 9.1 × 10–31 kg.
We can work this out by using the kinetic energy calculated above:
Ek = 1.9 × 10–18 J
⇒ 1 mv2 = 1.9 × 10–18
2
1.9 × 10−18
⇒
v2 =
1/ 2 m
1.9 × 10−18
⇒
v2 =
0.5 × 9.1 × 10 –31
2
⇒
v = 4.2 × 1012
⇒
v = 2.1 × 106 m s–1
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Topic Seven: Capacitance
QUESTION
1 An electron gun in a colour TV set uses a potential of 30 kV to accelerate
electrons.
Assuming the electrons start with negligible speed, find the speed given to them
by the electron gun.
The speed you have calculated for the electron is of the same order as the speed
of light (3 × 108 m s–1). In fact the calculation was not precise. To get an accurate
answer where speeds near the speed of light are concerned we need to use Einstein’s
theory of relativity.
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Capacitance
Objectives
By the end of this section you should be able to:
1 say what capacitance is and give the defining equation
2 describe a parallel plate capacitor, say how its capacitance depends on three quantities, and
state the relevant equation
3 carry out calculations involving stored energy, capacitances in series, and capacitances in
parallel.
Capacitors
A capacitor typically consists of two separated metal plates, as its symbol suggests. A charged
capacitor normally has a charge Q on one plate and an equal and opposite charge (–Q) on
the other. In storing charge, energy is also stored.
Capacitance
This is defined by C = Q
V
where V is the potential difference between the plates.
The unit is the farad, F . This is a particularly large unit, and everyday capacitances are
measured in μF, nF and pF.
Area A
(one plate)
d
F0 = 8.84 s 10–12 F m–1
When a battery charges
up a capacitance, it
delivers a charge Q at a
constant voltage V. Thus
it supplies energy QV.
The capacitance receives
energy 1 QV. The rest of
2
the energy, 1 QV, appears
2
as heat in the resistance of
the circuit.
The parallel plate capacitor
Most capacitors are essentially two parallel metal plates of comparatively large area, separated
by a small distance d, usually by some insulating material (the dielectric).
εA
The capacitance of such an arrangement is C =
where ε is the dielectric’s permittivity.
d
This is often expressed as ε = εrεo where εo is the permittivity of a vacuum (free space) and
εr is the relative permittivity.
Energy stored in a capacitor
1
1
This is given by E = 2 CV2 which may also be written E = 2 QV.
The energy is stored in the electric field between the plates. It is released to a circuit when
the charge Q flows from the positive plate through the circuit to the negative plate, thus
discharging the capacitor.
Capacitances in parallel
Clearly, combining capacitances in parallel increases the equivalent capacitance since it
increases the area for storing charge. For the setup shown, the stored charge Q is given by:
Q = Q1 + Q2 + Q3
Q Q1 Q2 Q3
=
+
+
V V
V
V
Q
⇒ = C1 + C2 + C3
V
⇒
But
Q
is the equivalent capacitance C. Thus, C = C1 + C2 + C3 .
V
Q1 C1
Q2 C2
Q3 C3
V
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Topic Seven: Capacitance
Capacitances in series
Effectively only a charge Q is stored, since connecting A to B would result in only a charge
Q flowing through the external circuit. The arrangement thus stores a charge Q at a voltage
Q
V, and the equivalent capacitance is given by C = .
V
Now: V = V1 + V2 + V3
A
C1
Q
C2
Q
C3
Q
V1
V2
V3
B
V
V V V
⇒V = 1+ 2+ 3
Q Q Q Q
⇒ 1 = 1 + 1 + 1
C C1 C2 C3
We see capacitances in series are like resistances in parallel, and that capacitances in parallel
are like resistances in series.
QUESTION
2 Suppose you have three 60 pF capacitances. Explain how to use some or all
of them to produce capacitance values of: 180 pF, 120 pF, 90 pF, 40 pF, 30 pF
and 20 pF.
Uses of capacitors
Capacitors are often used for their energy-storing abilities. Some computers use them as
backup power sources for CMOS RAMs. In the event of a power failure, up to 30 days’
backup power is typically provided by the energy stored in the capacitor.
They are also used to “concentrate” energy. In a camera flash, a battery takes several seconds
to charge up a capacitor. The capacitor then delivers this energy to the flash in about onehundredth of a second. For this situation, the power delivered by the capacitor to the flash is
several hundred times greater than the power the battery could deliver.
Capacitors can also be used as detection and measuring devices.
Large C
reading
Thin paint
layer
Smaller C
reading
Thick paint
layer
The thickness of paint on a car body could be measured by measuring the capacitance formed
by the car body, a metal plate, and the paint between them. A lower capacitance would indicate
greater plate separation and a thicker paint layer. This variation of capacitance with plate
separation can also be used to detect earthquakes, and to monitor the breathing of a baby.
Air humidity can also be measured by measuring the capacitance of two metal plates with
the air between them. Water has a relative permittivity of about 80, so even a small increase
in the amount of moisture in the air increases the capacitance significantly. This variation
of capacitance with the nature of the dielectric can also be used to measure the amount of
petrol in a tank, or the amount of food in a food storage bin.
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Continuing Physics
QUESTIONS
3 A 100 μF capacitance is connected to a 12 V battery. Find:
(a) the charge on each plate
(b) the energy stored by the capacitance
(c) the energy delivered by the battery.
4 A parallel plate capacitor consists of two square metal plates 1 mm apart with air
between them.
What must be the length of a side of each square if the capacitance is to be 1 F?
5 A capacitor which consists of two parallel metal plates is connected to a battery.
In this situation the charge stored is E.
How much energy is stored if:
(a) the battery voltage doubles
(b) the plate separation is doubled while the capacitor is still connected to the
original battery
(c) the capacitor is disconnected from the original battery and then the plate
separation is doubled.
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RC Circuits
Objectives
By the end of this section you should be able to:
1 describe in words and through graphs how the current and voltage values vary with time as
the capacitance charges up
2 find initial and final values for current and the voltages
3 repeat 1 and 2 for the case of capacitance discharge
4 discuss the energy transformations for the above
5 calculate the time constant for an RC circuit
6 use the value of the time constant to make quantitative statements about how the current and
voltage values in an RC circuit vary with time.
Charging a capacitor
Consider the circuit shown with the capacitor initially discharged. At the instant the switch
is closed, the capacitor will still be uncharged and so the voltage across it will be zero.
VC = 0.
Also, V = VR + VC ⇒ VR = V.
The initial current in the resistor will be given by Ohm’s Law: I = V/R. This is also the initial
current in the circuit, I0 = V/R.
As the capacitor charges up, the voltage across it increases. Thus the voltage across the
resistor decreases with time. Hence the current in the circuit decreases with time. All these
relations are exponential in nature. Finally, the capacitor accepts no further charge and so
there is no current in the circuit.
Voltage
Thus VR = IR = 0 × R = 0 and VC = V.
V
1
The battery has provided energy QV, of which the capacitor stores 2 QV,
1
while the resistor has dissipated heat energy 2 QV.
R
C
V
I
I0
t
V = VR+ VC
VC
VR
t
QUESTION
10 k7
6
2.0 V
60NF
In the circuit shown, the capacitance is initially uncharged. Then the switch is
closed and it starts to charge up. Find:
(a) the initial current
(b) the current when the voltage across the capacitance is 0.5 V
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Continuing Physics
(c) the final current
(d) the final stored charge.
(e) On a single set of axes, sketch three current–time graphs for:
(i) the above circuit
(ii) the resistance above tripled
(iii) the capacitance above tripled.
Discharging a capacitor
At all times after the switch has been closed, we have VC = VR.
The current in the resistor, which is also the current in the
circuit, is given by
V
V
I= R = C.
R
R
Q
–Q
C
R
As the capacitor discharges, the voltage across it becomes less and
so the current in the circuit becomes less. Finally, the capacitor is
discharged completely and so VC = 0.
Thus VR = 0 and I = 0.
VR
VC
t
I
t
t
The capacitor has lost its stored energy, and this has been turned to heat in the resistance.
No energy remains in the circuit.
Time constant
Symbol: τ
Units: s
The time constant τ gives a measure of how long it takes a capacitance in an RC circuit to
charge up or discharge. The larger the value of τ, the slower the process. It is defined by
τ = RC.
This equation makes sense: the larger the value of the resistance in the circuit, the smaller the
current—and so the longer it will take to charge or discharge the capacitor. Also, the larger
the capacitance in the circuit, the larger the charge that must be moved to or from it in order
to change the voltage by a certain amount. Thus the process will take longer.
The time constant cannot measure the time taken for the charging or discharging process to
be completed because, in theory, this time is undefined (infinite). Instead, the time constant
gives the time taken for the voltage across the capacitance to change by 63% of the
potential difference in the circuit causing the charging or discharging.
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Topic Seven: Capacitance
Example
An uncharged capacitance of 100 μF is connected in series with a resistance
of 50 kΩ to a 100 V power supply which charges it up.
The time constant τ for the circuit is given by:
τ = RC = 50 × 103 × 100 × 10–6 = 5.0 s
At the start of the first 5 s the potential difference in the circuit causing the
charging is 100 V. Thus, after 5.0 s the capacitance’s voltage will have changed by
100 × 0.63 = 63 V. Since the voltage was zero, the voltage after 5.0 s will
be 63 V.
At the start of the second 5 s the potential difference in the circuit causing
the charging is 100 – 63 = 37 V. Thus, after a further 5.0 s the capacitance’s
voltage will have changed by 37 × 0.63 = 23 V. Since the voltage was 63 V,
the voltage after 10.0 s will be 63 + 23 = 86 V.
V (V)
100
86
63
63%
rise
5
86%
rise
10
t (s)
U
2U
Example
A capacitance of 100 μF is charged to 100 V and then allowed to discharge
through a resistance of 50 kΩ.
As before, the time constant τ for the circuit is given by:
τ = RC = 50 × 103 × 100 × 10–6 = 5.0 s.
At the start of the first 5 s the potential difference in the circuit causing the
discharging is 100 V. Thus, after 5.0 s the capacitance’s voltage will have changed
by 100 × 0.63 = 63 V. Since the voltage was initially 100 V, the voltage after
5.0 s will be 100 – 63 = 37 V.
At the start of the second 5 s the potential difference in the circuit causing
the discharging is 37 V. Thus, after a further 5.0 s the capacitance’s voltage
will have changed by 37 × 0.63 = 23 V. Since the voltage was initially 37 V,
the voltage after 10.0 s will be 37 – 23 = 14 V.
V (V)
100
63%
fall
86%
fall
37
14
5
10
t (s)
U
After five time constants have passed, we may say that a charging capacitor
is fully charged and that a discharging capacitor is fully discharged.
2U
Example
In the first example above, after five time constants had passed, the voltage
across the capacitor would be over 99 V, within 1% of the fully charged value of
100 V.
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Continuing Physics
Exercise 7:
Capacitors
1 A 1.2 nF parallel plate air gap capacitor is connected to a 10 V battery as shown in the
diagram.
10 V
a Mark, on the diagram, which plate carries a surplus negative charge.
b Show on the same diagram, the electric field that exists between the plates.
c Calculate the size of the charge on each plate.
1.2 nF
d If the plates are 0.0047 m apart, find the strength of the electric field between the
plates.
e Calculate the area of the plates.
2 Sketch the network that would connect three 2.0 μF capacitors to make a total
capacitance of:
a 6.0 μF
b 3.0 μF
c 0.67 μF
3 A 350 nF capacitor is connected, in series as shown in the
diagram, to a 500 kΩ resistor and a 12 V cell.
12 V
a Calculate the time constant of this circuit.
350 nF
500 k7
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Topic Seven: Capacitance
b Sketch the voltage vs time graphs for the first second after
the switch is closed.
i the capacitor
VC
t
ii the resistor
VR
t
iii the cell
VT
t
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Continuing Physics
c Draw a detailed current through the resistor vs time
graph for the first second after the switch is closed.
I (A)
t (s)
d Explain what a current through the capacitor vs time graph would look like.
e What is the total energy, in the first second after the switch is closed,
i stored in the capacitor?
ii supplied by the cell?
iii dissipated by the resistor?
4 (From UB 1996) The distributor in older models of cars makes the spark plug in
the engine create a spark when a switch (called the contact breaker) in the circuit is
opened. One part of the distributor is a capacitor (often called a condenser) which is
connected across the contact breaker. The purpose of the capacitor is to store charge
and hence reduce the possibility of spark occurring across the open contact breaker
while the induced voltage in the circuit is very high.
A typical capacitor in a distributor is 0.20 μF.
a Calculate the charge stored on the capacitor when it is charged up to 12 V.
Once the spark plug has sparked, the capacitor must discharge ready for the next
spark.
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Topic Seven: Capacitance
The greater the capacitance of the capacitor, the greater the amount of charge which
can be stored and so the more effective the capacitor is in reducing the possibility of
an unwanted spark.
The voltage across the capacitor is measured with an oscilloscope. The contact breaker
is closed to discharge the capacitor and the following graph of voltage vs time is
obtained.
V (V)
12
6
0
t (s)
b On the graph above, sketch what the graph would look like if the capacitor had a
capacitance of 0.40 μF.
5 (From UB 1998) One day Janet turned off the 12 V DC electricity supplying her portable
TV and noticed that the screen took some time to finally go dark and the sound some
time to go off. She assumed there must be a large capacitor in the TV power supply
somewhere. She connected an oscilloscope across the 12 V supply at the TV, as in the
diagram below. When she disconnected the battery, the voltage decreased with time,
as in the following graph.
S1
+
RTV
12 V
Oscilloscope
CTV
–
TV
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Continuing Physics
V (V)
12
10
8
6
4
2
0
0
2
4
6
8
10
12
14
16
18
t (s)
a Using the graph, explain why the time constant of this capacitive system is
3.0 s.
b If the resistance of Janet’s TV was 22 Ω, what was its capacitance value?
c Explain, in terms of charge (electrons), why increasing the value of the capacitor
in Janet’s TV would make the time it takes for the TV to go off even longer.
As shown in the left figure below, three 1.0 μF (1.0 × 10–6 F) capacitors are separately
charged to 100, 200 and 300 V, respectively.
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Topic Seven: Capacitance
100 V
+
+
+
–
–
–
A
a
300 V
200 V
B
+
A
–
+
–
B
+
C
–
C
b
1.0
10–6 F each
d Calculate the charge stored in capacitor A.
When the capacitors are then connected in parallel, as shown above on the right, the
total charge stored is 600 μC (6.0 × 10–4 C).
e Calculate the total capacitance of the three capacitors when connected in
parallel.
f Calculate the voltage, Vab, across the combination.
g Calculate the total energy stored by the capacitor combination.
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