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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
ELECTROSTATICS
79
FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
UNIT 3: Electrostatics
•
•
Electrostatics is that branch of science which deals with the phenomena associated with
electricity at rest.
Generally an atom is electrically neutral i.e. in a normal atom the aggregate of positive charge
of protons is exactly equal to the aggregate of negative charge of the electrons.
• If, somehow, some electrons are removed from the atoms of a body, then it is left with a
Electrostatics
preponderance of positive charge. It is then said to be positively-charged. If, on the other hand,
4.1.are
Laws
of Electrostatics
some Lesson
electrons
added
to it, negative charge out-balances the positive charge and the body is
said to be negatively charged.
•
In brief, we can say that positive electrification of a body results from a deficiency of the
electrons whereas negative electrification results from an excess of electrons.
•
The total deficiency or excess of electrons in a body is known as its charge.
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
Laws of Electrostatics
 First Law. Like charges of electricity repel each other, whereas unlike charges attract each
other.
 Second Law. The force of attraction or repulsion between two electrically charged bodies is
proportional to the magnitude of their charges and inversely proportional to the square of the
distance separating them, i.e.
Electrostatics
Force F4.1.
∝ Laws of, Electrostatics
F=k
Lesson
, Where k is constant = 9 x 109
This is also known a Coulomb’s Law.
Hence, one coulomb of charge may be defined as that charge (or quantity of electricity) which
when placed in air (strictly vacuum) a distance of 1 m and similar charge repels it with a force
of 9 × 109 N.
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
The value of k depends on the system of units employed. In S.I. system, as well as M.K.S.A.
system k = 1/4πε. Hence, the above equation becomes.
F=
Ɛ
∗
Where Ɛ is the absolute permittivity of the material of the medium within which the charges
exists. For free space or vacuum, i.e.
Ɛo = 8.854 x 10-12 F/m so that k = 9 × 109 N.
The permittivities of all other media are expressed in values relative to that of vacuum, hence
the name Relative Permittivity Ɛr
So that
Ɛ = ƐoƐr
Absolute Permittivity = Permittivity of free space x Relative Permittivity
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
EXAMPLE 1.
Calculate the electrostatic force of repulsion between two α-particles when at a distance of 10−13 m
from each other. Charge of an α-particles is 3.2 × 10−12 C. If mass of each particle is 6.68 × 10−27
N-m2/kg2.
EXERCISE 1.
Determine the resultant force on a 4 µC charge due to a -6 µC and 12 nC charges placed on the
vertices of an equilateral triangle of sides 55 cm.
EXERCISE 2
Two small conducting spheres have charges 4 nC and -0.8 nC. Determine the force between them
when (a) they are placed 5 cm apart. (b) they are brought together and then separate by 4 cm.
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
EXERCISE 3.
Three identical charges, each X C, are placed
in the vertices of an equilateral triangle of
sides 12 cm. Calculate the forces on each
charge.
Electrostatics
Lesson 4.1. Laws of Electrostatics
EXERCISE 4.
Two charges Q C each are placed at two corners
of a square. What additional charge q placed at
the other corners will reduce the resultant electric
force on each charge to zero?
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
ELECTRIC FIELD
 Consider a point charge as shown in Fig. (a) below.
Electrostatics
Lesson 4.1. Laws of Electrostatics
 Figure (a) shows a typical field pattern for an isolated point charge, and figure (b) shows the
field pattern for adjacent charges of opposite polarity.
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
 The figure above represents two parallel metal plates, A and B, charged to different potentials.
Electrostatics
If an electron that has a negative charge is placed between the plates, a force will act on the
Lesson
4.1. Laws
of Electrostatics
electron
tending
to push
it away from the negative plate B towards the positive plate, A.
 Similarly, a positive charge would be acted on by a force tending to move it toward the negative
plate.
 Any region such as that shown between the plates in which an electric charge experiences a
force, is called an electrostatic field.
 The direction of the field is defined as that of the force acting on a positive charge placed in the
field.
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Electric Field Strength, E.
FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
d
V
 The figure above shows two parallel conducting plates separated from each other by air. They
are connected to opposite terminals of a battery of voltage V volts.
 If the plates are close together, the electric lines of force will be straight and parallel and
equally spaced, except near the edge where fringing will occur
𝑽
Electric field strength, E = volts/metre
𝒅
where d is the distance between the plates. Electric field strength is also called potential gradient
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ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
Some Of The Terms Used in Electrostatics.
 Devices specially constructed to store electric charges are called capacitors (or condensers, as
they used to be called).
 In its simplest form a capacitor consists of two plates which are separated by an insulating
material known as a dielectric.
 Let the charge be +Q coulombs on one plate and −Q coulombs on the other. The property of
this pair of plates which determines how much charge corresponds to a given p.d. between the
plates is called their capacitance:
Capacitance C =
𝑸
𝑽
,
The unit of capacitance is the farad F.
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ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
TYPES OF CAPACITORS (Write short notes on the following capacitors)
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ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
EXAMPLE
(a) Determine the p.d. across a 4 μF capacitor when charged with 5mC
(b) Find the charge on a 50 pF capacitor when the voltage applied to it is 2 kV.
Ans.
(a) Capacitance C = 4 µF = 4 * 10-6 F, and Charge Q = 5 mC = 5 * 10-3 C.
C=
𝑸
𝑽
,
V=
[ (b) C = 0.1 μF ]
𝑸
,
𝑪
V=
5 ∗ 10−3
= 1250 Volts
4 ∗ 10−6
The charge Q stored in a capacitor is given by:
Q = I x t coulombs
Where I is the current in amperes and t is the time in seconds.
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
Example
A direct current of 4A flows into a previously uncharged 20 μF capacitor for 3 ms. Determine the
p.d. between the plates.
Ans.
I = 4 A, C = 20 μF = 20×10−6 F and, t = 3 ms = 3 * 10−3 s.
Q = Current (A) x time t (s)
It = 4 * 3 * 10−3 C.
V=
CLASSWORK
𝑸
,
𝑪
V=
4 ∗ 3∗ 10−3
= 600 V
20 ∗ 10−6
A 5μF capacitor is charged so that the p.d. between its plates is 800V. Calculate how long the
capacitor can provide an average discharge current of 2 mA.
[Ans. 2 s ]
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
 Unit flux is defined as the amount of lines emanating from a positive charge of 1 coulomb.
Thus electric flux ψ is measured in coulombs, and for a charge of Q coulombs, the flux ψ = Q
coulombs.
 Electric flux density D is the amount of flux passing through a defined area A that is
perpendicular to the direction of the flux:
electric flux density, D =
𝑸
𝑨
coulombs/metre2
 Electric flux density is also called charge density, σ.
 At any point in an electric field, the electric field strength E maintains the electric flux and
produces a particular value of electric flux density D at that point. For a field established in
vacuum (or for practical purposes in air), the ratio D/E is a constant ε0, i.e.
σ
𝑫
, or Ɛ0 =
Ɛ0 =
𝑬
𝑬
where ε0 is called the permittivity of free space or the free space constant.
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
 When an insulating medium, such as mica, paper, plastic or ceramic, is introduced into the
region of an electric field the ratio of D/E is modified:
𝑫
𝑬
material.
= Ɛ0Ɛr .
where εr , the relative permittivity of the insulating
 When an insulating medium, such as mica, paper, plastic or ceramic, is introduced into the region of an
electric field the ratio of D/E is modified:
density in material
ε = flux
𝑫 Relative permittivity,
fluxpermittivity
density in of
vacuum
=Ɛ Ɛ .
where ε , the rrelative
the insulating material.
𝑬
0 r
r
The product ε0εr is called the absolute permittivity, ε, i.e.
ε = ε0εr
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
EXERCISES
(1) Two parallel rectangular plates measuring 20 cm by 40 cm carry an electric charge of 0.2 μC.
Calculate the electric flux density. If the plates are spaced 5mm apart and the voltage between
them is 0.25 kV determine the electric field strength.
[D = 2.5 µC/m2, E = 50 kV/m]
(2) Two parallel plates having a p.d. of 200V between them are spaced 0.8 mm apart. What is the
electric field strength? Find also the electric flux density when the dielectric between the plates
is (a) air, and (b) polythene of relative permittivity 2.3.
[E = 250 kV/m, D1 = 2.213 µC/m2, D2.3 = 5.089 µC/m2]
(3) A charge of 1.5 μC is carried on two parallel rectangular plates each measuring 60mm by 80
mm. Calculate the electric flux density. If the plates are spaced 10mm apart and the voltage
between them is 0.5 kV determine the electric field strength.
[D = 312.5 µC/m2]
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
Constructional Features.
(The parallel plate capacitor)
The figure above shows a parallel-plate capacitor. Experiments show that capacitance C is
proportional to the area A of a plate, inversely proportional to the plate spacing d (i.e. the dielectric
thickness) and depends on the nature of the dielectric:
εεA
Capacitance, C = 0 r farads
d
where ε0 = 8.85×10−12 F/m (constant),
εr = relative permittivity
A = area of one of the plates, in m2, and
d = thickness of dielectric in m
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
Example
A ceramic capacitor has an effective plate area of
4 cm2 separated by 0.1mm of ceramic of relative
permittivity 100. Calculate the capacitance of the
capacitor in pico-farads.
Ans.
Area A = 4 cm2 = 4×10−4 m2,
Thickness, d = 0.1mm = 0.1×10−3 m,
ε0 = 8.85 × 10−12 F/m and εr =100
Capacitance, C =
C=
ε0εrA
d
8.85 × 10−12 × 100 × 4 × 10−4
= 3540 pF
0.1 × 10−3
Another method used to increase the
capacitance is to interleave several plates as
shown in (b) above. Ten plates are shown,
forming nine capacitors with a capacitance
nine times that of one pair of plates.
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
If such an arrangement has n plates then capacitance C ∝ (n − 1). Thus capacitance
C=
ε0εrA(n − 1)
Farads
d
EXAMPLES
(1) A parallel plate capacitor has nineteen interleaved plates each 75 mm by 75 mm separated by
mica sheets 0.2 mm thick. Assuming the relative permittivity of the mica is 5, calculate the
capacitance of the capacitor.
[22.4 nF]
(2) The charge on the square plates of a multiplate capacitor is 80 μC when the potential between
them is 5 kV. If the capacitor has twenty-five plates separated by a dielectric of thickness
0.102 mm and relative permittivity 4.8, determine the width of a plate.
[40 mm]
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ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
Capacitors Connected in Series
C1
C2
So that
C3
V=
Q
Q
Q
Q
=
+
+
CT
C1
C2
C3
Note that when capacitors are connected in
series, the charge Q is the same in each
capacitor
Hence
v1
For the circuit above,
v2
v3
V
V = V1 + V 2 + V 3
That is
1
1
1
1
=
+
+
CT
C1
C2
C3
It follows that for n series-connected capacitors:
We already know that the charge Cn
Q = CV
1
1
1
1
1
=
+
+
+…………… +
CT
C1
C2
C3
Cn
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
Capacitors Connected in Parallel
Q1
Q2
QT
Q3
C1
But
QT = Q1 + Q2 + Q3
Hence
C2
CTV = C1V + C2V + C3V
Note that when capacitors are connected in
parallel, the voltage V is the same across
each capacitor
C3
That is
V
For the circuit above, the supply voltage
CT = C1 + C2 + C3
It follows that for n series-connected capacitors:
CT = C1 + C2 + C3 + …………… + Cn
V = V1 = V2 = V3
99
FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
APPLIED ELECTRICITY I (EENG212)
LECTURE NOTES
EXAMPLE 1.
Calculate the equivalent capacitance of two
capacitors of 6 μF and 4 μF connected
(a) in parallel and
(b) in series.
Ans.
(a) When the capacitors are in parallel, the
Calculate the equivalent capacitance
total capacitance
EXERCISE.
What capacitance must be connected in series with a
30 μF capacitor for the equivalent capacitance to be
12 μF?
[ C2 = 20 μF ]
EXERCISE.
Capacitance’s of 3 μF, 6 μF and 12 μF are connected
of two capacitors of 6 μF and 4 μF connected in
(a)series
in across a 350V supply. Calculate
CT = C1 + C2 ,
(a) the equivalent circuit capacitance,
parallel and (b)
in series.
CT = 6 * 10-6 + 4 * 10-6
(b) the charge on each capacitor, and
CT = 10 * 10-6 ,
CT = 10 μF
(c) the p.d. across each capacitor.
(b) When connected in series
1
1
1
=
+
,
CT
C1
C2
CT =
C1C2
,
C1 C2
CT = 2.4 μF
[ (a) 1.714 μF ]
[ (b) QT = 600 μC ]
[ (c) V1 = 200 V, V2 = 100V
V2 = 50 V ]
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FOURAH BAY COLLEGE
ELECTRICAL & ELECTRONIC ENGINEERING DEPARTMENT
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LECTURE NOTES
Lesson 3.7. Energy Stored in a Capacitor
Hence
W=
1 Q 2
( )V joules
2 V
The energy, W, stored by a capacitor is given
1
So that
W = QV joules
by: 2
2
1
W = CV2 joules
2W
2
from which V =
Q
2
∗
1.2
Lesson
3.6. Capacitors
in Parallel
EXAMPLE
1
V=
V = 240V
∗10 3
10
A capacitor is charged with 10 mC. If the energy
stored is 1.2 J find
Q
(b)
C=
(a) the voltage and
V
(b) the capacitance.
10 ∗10 3
Ans.
C=
240
1
Q
(a) W = CV2 joules and
C=
2
V
C = 41.67μF
101
UNIT 3. MAGNETISM & MAGNETIC CIRCUITS