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8/10/2019 Chapter 3 Electric Field 1
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Introduction
Electrostatics
presence of stationary charges thatcause electric field.
Charge polarity: +ve orve, depends on number of
electrons.
The force of attraction & repulsion acts directly
between 2 adjacent charges.
2 types of materialsconductor & insulator.
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Most electric fields exist between 2 conductors.
The space between 2 conductors needs to be filledwith an insulator (known as dielectric), otherwise the
charges would move towards one another and
therefore be dissipated.
Fundamental electronic charge is -1.6x10-19 C.
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Coulombs Law
The force between 2 point charges is proportional tothe product of the 2 charges, & inversely
proportional to the square of the distance between
the charges.
Newtonr
QQkF
2
21
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kis a constant with a value of 1/4 with =
0r= absolute permittivity
0= free space permittivity = 8.85 10-12F/m
r= relative permittivity (no unit) If vectors are used for directional force,
Newtonrr
kF 12221
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Capacitor & Capacitance
Capacitor
an electronic component consists of 2
conducting surfaces separated by a later of insulating
medium/material (diaelectric)
Capacitance the property of a capacitor to store an
electric charge when its plates ate at different
potentials.
Capacitance Symbol: C Unit: Farad (F)
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1Fof capacitance is when there appears a
potential difference of 1Vbetween the plateswhen it is charged by 1Cof electricity.
Note: In practice, capacitance is usually
expressed in F or pF.
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Example 1
A capacitor having a capacitance of 80F is connected
across a 500V d.c. supply. Calculate the charge.
Since Q = CV
Charge = (8010-6)(500) = 0.04C = 40mC
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Capacitors in Parallel
The charge on C1is Q1coulombs and C2is Q2coulombs
Q1= C1V and Q2= C2V
Suppose 2 capacitors havingcapacitance C1& C2farads
are connected in parallel
across a pd of V volts.
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If we replace C1& C2by a single capacitor ofsuch capacitance C farad that the same total
charge of (Q1+ Q2) coulombs would be
produced by the same p.d., then Q1+ Q2= CV.
Thus, the resultant capacitance, C = C1+ C2
farads
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Capacitors in Series
If V1& V2are corresponding pdacross C1& C2, Q = C1V1+ C2V2
So that
If we replace C1& C2by a singlecapacitance Cfarad such that it
would have the same charge Q
coulombs with the same pd of V
volts, then
2
2
1
1
C
QVand
C
QV
C
QV
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But clearly, V = V1+ V2
Therefore, the resultant capacitance is thesum of reciprocal.
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Example 2If 2 capacitors having capacitances of 6F and 10F
respectively are connected in series across a 200Vsupply, find the potential different across & charge on
each capacitor.
Sol:
a) Let V1& V2be the pd across 6F and 10F capacitors.
From eq. above, therefore
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Solution 2
b) Charge, Q1= (610-6)(125) = 750C
Q2= (1010-6)(75) = 750C
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Example 3
For the circuit below, find the voltage across each capacitor.
First, solve for Ceq, which will be
The total charge is: q = Ceqv = 10 10-330 = 0.3C
This is the charge on the 20-mF and 30-mF capacitors because
they are in series with the source. Therefore:
mFCeq 10
20
1
30
1
60
1
1
VC
q
v 151020
3.0
6
1
1
VC
q
v 101030
3.0
6
2
2
Vv 53
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Gausss Law Here, this law concerns electric field.
It defines that, the electric flux through any closed surface is
proportional to the enclosed electric charge.
The left-side is the surface integral denoting the electric flux E
through a closed surface S, while the right side is the total
charge enclosed by Sdivided by the electric constant.
0
QdsE
S
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Electric Field Strength &
Electric Flux Density
We can investigate an electric field by observing its
effect on a charge.
The magnitude of a force experienced by a unit
charge at any point in a field is termed the electricfield strength at that point.
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Electric field strength is also known as electric stress,
and is measured in newtons per unit charge andrepresented by the symbol E.
1Jof work is necessary to raise the potential of 1Cof
charge through 1V.
Therefore, to move a unit charge through an electric
field so that its potential changes by Vvolts requires
Vjoules of work.
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The most simple arrangement for this is the parallel
charged plates.
Assume that the plates are very large & the distance
between the plates, dis very small & there is free
space between them.
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There is a potential difference of V volts between the
plates, therefore the work in transferring 1C ofcharge between the plates is V joules.
But work is the product of force & distance.
Therefore the force experienced by the charges is theelectric field strength,
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The electric flux density Dis given by:
In electrostatics, the ratio of electric flux density in a vacuum
to the electric field strength is termed thepermittivity of free
space(0).
or
2/ mcoulombs
A
QD
A
Cd
V
d
A
Q
E
D
strengthfieldElectric
densityfluxElectric
0
faradsd
A
C
0
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Permittivity of free space(0) | Unit: farad per
metre (F/m)
From carefully conducted tests,
0= 8.85 10-12F/m
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If the space in between the plates is not vacuum & it
is filled with a certain type of material (glass, rubber,
mica, etc.), the expression becomes
When the vacuum (or air) is replaced with a certain
material as diaelectric, the ratio of the capacitanceincreases greatly.
faradsd
AC r
0
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ris termed as the relative permittivity of that
material and it varies for different type of
materials.
is the absolute permittivity and is equal to
0r
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Example 3
A capacitor is made with metal plates and separated by a
sheet of 0.5mm thick glass with a permittivity of 8. The
area of the plate is 500cm2, what is its capacitance?
Given: d = 0.5mm = 0.0005m ;r= 8;
A = 500cm2= 0.05m2
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Sol 3