Download THE NATURE OF CAPACITANCE

THE NATURE OF CAPACITANCE
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All passive components have three electrical properties
Resistance, capacitance and inductance
Capacitance is a measure of the ability of a component
to store charge
Capacitor – device manufactured to store charge
Capacitance is defined as the ratio of the charge stored
by a capacitor to the voltage across it: C = Q/V
Unit of capacitance is the Farad (F, nF, pF)
+ VC
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A 6v source is required to store 24C of charge on a
certain capacitor. What is the capacitance of the
capacitor? How much charge is stored when a 9v
battery is used as a source? What voltage source is
used when 16C is stored on it?
How much charge is stored by a 220pF capacitor when
a 50v source is connected across it?
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WHAT IS A CAPACITOR?
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A capacitor is similar to a battery
Both store electrical energy, but function differently
Unlike a battery, a capacitor cannot produce new
electrons – it only stores them
The two terminals of a capacitor are connected to two
metal plates separated by an insulator called a
dielectric
Dielectric can be air, paper, plastic – any type of
insulator
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WHAT A CAPACITOR DOES
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Consider the following connection
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Capacitor plate attached to negative battery terminal
accepts electrons produced by battery
Capacitor plate connected to positive battery terminal
loses electrons to battery
Once charged, capacitor has same voltage as battery
Lightning in the sky is essentially charge being released
between two plates – one being the cloud, the other the
ground
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EXAMPLE OF DISCHARGING
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Consider the following connection
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Assuming a large capacitor is being used, bulb would
light up as current flows from battery to capacitor to
charge it up
Bulb would get progressively dimmer and finally go out
once capacitor is fully charged
On replacing the battery with a wire, current will flow
from one plate of capacitor to the other
Bulb will subsequently light, and get dimmer and
dimmer, finally going out once capacitor has completely
discharged
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CAPACITORS IN SERIES
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Every capacitor in a series circuit has the same
charge, regardless of the individual capacitance values
The total charge delivered to series-connected
capacitors equals the charge on each capacitor
Q1
+ C1
=
+
Q2
= QT
C2
V
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Applying KVL:
V  VC1  VC 2
QT
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CT   Q1 C1   Q2 C2 
Yet QT = Q1 = Q2
Cancelling the Q’s gives
1
1
1


CT C1 C2
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EXAMPLE 1
C1=60F
C2=30F
C3=20F
+V1-
+V2-
+V3-
QT
36V
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Find the total equivalent capacitance and the total
charge delivered by the voltage source to the circuit
Find the voltage across each capacitor
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CAPACITORS IN PARALLEL
QT = Q1 + Q2
V
V1
+
C1
-
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V2
+
C2
-
The total charge delivered to parallel connected
capacitors is the sum of the charges on the capacitors
QT = Q1 + Q2 → CTV = C1V1 + C2V2
Since the voltage is the same in each line:
CT = C1 + C2
The total equivalent capacitance of parallel-connected
capacitors is the sum of the capacitances
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EXAMPLE 2
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What is the total
equivalent capacitance
in the circuit on the left?
Does the total charge
delivered by the source
equal the sum of the
charges on the
capacitors?
C1
C2
C3
10V
100pF
Find the voltage across
and charge on each
capacitor
220pF
50pF
C1
12F
C2
10F
C3
5V
8F
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PHYSICAL EXPLANATION OF HOW
A CAPACITOR FUNCTIONS (1)
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A capacitor consists of two conducting surfaces
separated by a non-conducting (dielectric) material
When a source is connected to a capacitor, electrons
will flow from the negative terminal of the battery and
collect on one of the plates
A negative charge develops on the plate collecting
electrons from the negative terminal of the battery
Electrons on this plate repel an equal amount of
electrons from the other plate, forming a positive
charge
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PHYSICAL EXPLANATION OF HOW
A CAPACITOR FUNCTIONS (2)
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An electric field is established in the dielectric between
the plates
This electric field drives electrons off the positively
charged plate
These electrons are attracted to and received by the
positive terminal of the battery
The electric field stores energy
A voltage exists between any two regions containing
opposite types of charge
A voltage is developed across the capacitor as a result
of the charge stored on the plates
As the charge builds up on the plates, the voltage
across the capacitor increases
A capacitor is fully charged when no more electrons
can flow from the battery
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EFFECT OF DIELECTRIC MATERIAL
ON CAPACITANCE (1)
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The permittivity  of dielectric material between
capacitor plates strongly influences the value of its
capacitance
Since the dielectric is an insulator, its electrons are
strongly bound to their parent atoms and are not free to
travel under the influence of an electric field
Electric field is established in dielectric by charging
capacitor plates – causes dielectric electrons to only
shift position which leaves one side of each atom +ve,
and the other –ve (see next slide)
Each atom in dielectric has a +ve side neutralised by
-ve side of adjacent atom
Atoms at boundaries of capacitor plates atoms have no
adjacent atoms to neutralise their sides
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Consequently this creates an electric field Ed that
opposes field set up by the charged plates
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So net field intensity E – Ed is smaller than it would be
if there were no dielectric present
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Smaller value of E means smaller value of V required
to deposit fixed amount of charge Q on the plates
From C = Q/V, C is increased
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EFFECT OF DIELECTRIC MATERIAL
ON CAPACITANCE (2)
Atoms having +ve and -ve
sides, due to electron shifts
+
- +
- +
- +
- +
-
+vely +
charged
+
plate
- +
- +
- +
- +
-
- +
- +
- +
- +
-vely
charged
plate
+
- +
- +
- +
- +
-
E
Electric field due to charge on
capacitor plates
Neutralised
charges
- +
-
- +
- +
-
+
- +
- +
-
+
- +
- +
-
+
- +
+
Uneutralised
charges
Ed
Electric field due to
boundary atoms
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EFFECT OF DIELECTRIC MATERIAL
ON CAPACITANCE (3)
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A dielectric is effective in increasing capacitance
because it sets up an electric field in opposition to that
created by the charged plates
 is called the permittivity of the dielectric, and is a
measure of how well the material permits the
establishment of an electric field
The greater  of a material, the greater the capacitance
of a capacitor that uses the material as a dielectric
Units of  are Fards/meter (Fm-1)
Permittivity in a vacuum 0 = 8.8410-12 Fm-1
Other insulators have larger values of  as there are no
atoms in a vacuum
Relative permittivity r of a material is defined as the
ratio of its actual permittivity to that of a vacuum
r = /0 = dielectric constant of material
If a material has an r of 5, then its  is 5 times that of
a vacuum, and a capacitor that has this material as a
dielectric will have a capacitance 5 times greater than it
would if the plates were separated by a vacuum
Example When a 24V source is connected across a
capacitor having an air dielectric (r = 1.0006), 1800µC
of charge is deposited on each plate. With the voltage
source fixed, a porcelain dielectric (r = 6) is used in
the space between the plates. How much charge is
then deposited on the plates?
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EFFECT OF CAPACITOR
DIMENSIONS ON CAPACITANCE
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The physical dimensions that effect the capacitance of
a parallel plate capacitor are the area A of its plates,
and the distance d between them
The greater the area of the plates, the greater the
surface on which charge can be stored
Capacitance is directly proportional to the area A of
each parallel plate
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From E = V/d and C = Q/V :
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C = Q/Ed
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From above equation, capacitance is inversely
proportional to the distance d between the parallel
plates
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THE CAPACITANCE EQUATION
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It is possible to express the dependency of capacitance
on the permittivity  of the dielectric, the area A
between each plate and the distance d between the
plates
The capacitance equation can be expressed as
C = A/d = r0A/d
Example Each plate of a parallel plate capacitor
measures 1cm by 2cm. The plates are separated by a
0.2mm thickness of mica ( = 44.2  10-12). Find the
capacitance of the capacitor. Find the capacitance if
each dimension of each plate is doubled. Under these
conditions, find the capacitance if the thickness of the
mica separation is also doubled.
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CURRENT THROUGH AND VOLTAGE
ACROSS A CAPACITOR (1)
V
b
v2
a
c
v1
t1
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t2
t3
t4
t
Current through a capacitor depends on the rate of
change of the voltage across it
Rate of change of voltage deduced from a graph of
voltage v time = slope of different parts of the curve
Slope at a = Δv/Δt = (v2-v1)/(t2-t1)
b = (v2-v2)/(t3-t2)=0; c = (v1-v2)/(t4-t3)
A new rate of change must be calculated over each line
segment that has a different slope
If the voltage is constant over a time interval, the rate of
change is zero
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CURRENT THROUGH AND VOLTAGE
ACROSS A CAPACITOR (2)
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Since current is equal to the rate of change or charge,
i.e. I = dQ/dt and Q = CV
i
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d
dv
CV  C
dt
dt
The above is in differential form, in integral form it is
t
1 2
i dt  v(t )
C t1
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The differential form shows that the faster the voltage
across the capacitor changes (i.e. for small Δt), the
greater the current through it
If there is no voltage change across the capacitor, i.e.
dv/dt = 0, then there is no current flow, so the capacitor
acts as an open circuit.
The energy W stored by capacitance C when it is
storing charge Q is
Q2 1
W
 CV 2
2C 2
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CURRENT THROUGH & VOLTAGE
ACROSS CAPACITOR EXAMPLE
v(t)(V)
i(t) →
100
2.5µF
+ v(t) -
12 14
2
4
t (ms)
-40
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Above is a graph of the voltage across a 2.5µF
capacitor. Plot and draw the current through the
capacitor.
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