COLLEGE OF ENGINEERING AND TECHNOLOGY
Electrostatics(Capacitance & Di-electric Circuits)
Capacitance (c) is the capacity of the capacitor to store or
accumulate electric charges -is a measure of how well a capacitor
can store electric charges. Farad (F)- standard unit of
capacitance
Farad is a unit of capacitance, when one coulomb of charge is
given to it raises a potential difference of 1 volt, Named after
Michael Faraday who discovered electro- magnetic induction.
Ex.1 A certain capacitor is charge at 48 volts after which the
stored energy is 5.76x10-2 joules. What is the capacitance of the
capacitor? Ex.2. A 20F capacitor is charged by a 12 volts battery
source. What is the stored energy in the capacitor?
Capacitor is a device in which electric charges can be stored as
to possess electrical potential. It consist of two conducting
plates called electrodes separated by layer of insulating medium
called di-electric aka condenser.
TYPES OF CAPACITORS 1. Air 4. Electrolytic (oil) 2. Ceramic 5.
Paper 3. mica Ampere the rate of electric charge is always equal to
its current.
Capacitance of Parallel Plates CapacitorCASE 1. For two
parallel-plates capacitor with uniform di-electric medium or
insulating material.
d- plate separation (mm) A cross-sectional area of the plate of
the capacitor (cm2, mm2) r relative permittivity of the dielectric
medium (unitless) = 1.0 (for Air or Vacuum) o = permittivity of
free- space (Air) =8.854 x10-12 Farad/meter =8.854 pF/m = 8.854 F/m
= permittivity of any di-electric medium
Energy Stored in a Charged Capacitor Di-electric is an insulator
that would separate the plates or electrodes of a capacitor.
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Ex.1. A capacitor with two parallel plates each of which is 100
cm 2 and 2 mm apart. What charging current would cause the
potential drop across the capacitor to raise a uniform rate of 8
volts/sec?
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LECTURE
COLLEGE OF ENGINEERING AND TECHNOLOGY CASE III. Capacitance of
two parallel plates capacitor with
Ex.2. A capacitor consist of two metal plates each having an
area of 100 cm x 100 cm, spaced 1mm apart. The space between the
plates is filled with di-electric substance having relative
permittivity of 4.0. A potential of 1 kV is magnetized between the
plates. Calculate the force of attraction. CASE II. Capacitance of
multiple plates Capacitor Note: 1. the plate separation or distance
between plates must be the same or identical 2. the plate area of
each plate must be the same 3. the medium or di-electric material
must be the same let: n= number of plates or electrodes c= total
capacitance of the capacitor
Ex.1 A parallel capacitor with 5 plates each 1 sq. meter of
area, and placed at a distance of 1.5 mm and a di-electric of
relative permittivity of 3.5. If the capacitor is charged to 100 V.
How much energy will be stored in it?
CASE IV. Capacitance of multi plates capacitor of composite or
different mediums. medium or di-electric that contains partly air.
d = distance between plates (plate separation) + = thickness of the
di-electric material of the other medium A = area of the plate C =
capacitance of the capacitor
Note: A= area of each plate (assumed to be uniform) C = total
capacitance of the capacitor
Ex.2. A parallel plate capacitor is made of 350 plates,
separately by parafined paper of 0.0010 cm thick (t = 2.5). The
effective size of the plate is 15 cm x 30 cm. What is the
capacitance of the capacitor?
Ex.1. A capacitor has two parallel metal plates 45 cm square
each. The distance between the plates is 10 mm in free air. If a
dielectric material whose t = 4.0 and is 5mm thick and is placed in
the upper plate, leaving an air between the bottom plate and the
di-electric. What is the capacitance of the capacitor?
Ex.1. A capacitor is composed of two plates separated by a sheet
of insulation 3 mm thick and t = 4.0. The distance between the
plates increased by insertion of a second sheet of 5 mm thick and
unknown t. If the capacitance of the capacitor so formed is of the
original capacitance, find the
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Subject: BES 117 / 5:30-8:30 W 5:30-8:30 S E24Page 2
LECTURE
COLLEGE OF ENGINEERING AND TECHNOLOGY value of the permittivity
t. unknown Cable consist of both insulator and a conductor.
Assuming a di-electric of 4.3 for rubber. Calculate the capacitance
per mile of length? A. The total capacitance is equal to the
reciprocal of the sum of the reciprocals of the individual
capacitance of each capacitor.
CASE V. Capacitance of a coaxial or Lead Sheath Cable Lead
Sheath Cable this is a type of cable used for high voltage
underground transmission lines. Submarine cable Coaxial Cable used
for communication system (RG 59 or RG 11) Overhead- transmission
Lines ACSR (aluminum conductor, steel reinforced) - un-insulated,
bare conductor
r=radius of the conductor (inner radius) =d d= diameter of the
conductor R= radius of the cable (outer radius) += Thickness of the
insulating material D= outer diameter = 2R r = permittivity of the
insulating material GMD= distance between conductors = geometric
mean distance
DI-ELECTRIC CIRCUITS I. Capacitors in Series - A number of
capacitors are said to be connected in series if the negative
plates of one capacitor is connected to the positive plate of
another capacitor and so on.
B. The charges on each capacitor are equal. C. The total voltage
is equal to the sum of the voltages across each capacitor. VOLTAGE
DIVISION THEOREM
Ex.1. A lead sheath cable for underground transmission lines
having a conductor diameter of 0.35 inch surrounded by a 0.2 inches
wall of rubber insulator. Address: San Jose Normal Road Baliwasan
Zamboanga City
E1, E2, E3 potential drops across each capacitor (volts) C1, C2,
C3 constant capacitance of each capacitor respectively (F, F) Q1,
Q2, Q3 electrostatic charged stored in each capacitor (C, C) E
constant emf of the battery source (volts) Qt = total charge in the
circuit (C, C)
Ex.1.Three capacitors are connected in series across a 100 V
source. If the voltages across each are 25, 30, and 45 Volts
respectively, while the total charge taken by the series
combination is 3000 C, determine the value of each capacitance.
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LECTURE
COLLEGE OF ENGINEERING AND TECHNOLOGY charge and to make precise
quantitative measurement of quantity of charge rather than the leaf
microscope. Ex.1. Two capacitors of 2 and 3 F are connected in
parallel. A third capacitor of 6 F is connected in series with the
combination and the whole circuit is connected across a 500 V
supply. Determine the charge accumulated in the 2 F capacitor.
First law of Electrostatic like polarity of charges REPEL, and
unlike polarity of charges ATTRACTS each other. Second Law of
Electrostatics the force of attraction or repulsion between two
charges in space is directly proportional to the product of the
charges and inversely proportional to the square of the distance
between them.
A. The total capacitance is equal to the sum of the capacitance
of each capacitor. B. The total charge accumulated is equal to the
sum of the charges accumulated in each capacitor. C. The voltage is
equal to the voltage across each capacitor. Ex.1. Three capacitors
A, B and C are charged as follows: A=10F, 100V; B=15F, 150 V and
C=25F, 200V. They are connected in parallel with terminals of like
polarity together. What is the voltage across the combination?
Ex.2. The result of capacitance C1=6 F and C2 connected in
series is 3F. Capacitor C2, in F is _____.
II. Parallel Connected Capacitors - A number of capacitors are
said to be parallel if one plate of each capacitor is connected to
the positive terminal of the supply or source while the other plate
of the capacitor is connected to the negative of the supply or
source.
ELECTRIC CHARGES AND FIELDS Electrostatic deals with phenomenon
common to electric charges in motion and at rest. -objects may be
electrified and charged either positively and negatively by the
removal or addition of electrons. Leaf Electroscope is a device
that is used to detect the presence of electric charges.
Electrometers an instrument which uses electronic application in
order to achieve greater sensitivity in measuring the
CHARGE DIVISION THEOREM
F= force of attraction or repulsion between two charged bodies.
S= distance between the two charged bodies Q1 &Q2 = respective
charges K= proportionality k constant Ex.1. Two point charges 10 cm
apart exert a force of 1x10-3 grams on each other. If the charges
are of the same value, what is the charge in statcoulombs?
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Subject: BES 117 / 5:30-8:30 W 5:30-8:30 S E24Page 4
LECTURE
COLLEGE OF ENGINEERING AND TECHNOLOGY 2. Arrangement of charges
must be clearly specified. 3. If the polarity of the charges are
not specified, assumed that all charges carries a + positive
polarity. 4. Units Ex. Three 100 stat-C are arranged in a straight
line. The second charge B is 20 cm to the right of first charged A
and the third charged C is 50 cm to the right of A. What force is
exerted by charges A and B on charge C?
Ex.2. What is the force between a point charge of 100 C and
another of +50 C, when they are 100 cm apart?
V=electrostatic potential (volt) at a distance of d (meter) K=
free space constant in SI units equal to 9x109 Ex. An isolated
positive point charge of 2x10-8 coulombs is in free space. How far
away must the charge be in order to
ELECTRIC FIELD INTENSITY Is the force per unit positive charged
that will act at a point in the field on a very small test charge
placed at the location. A. Electric field intensity near -
Force of Repulsion/ Attraction between 3 or more charges in
space Rules: 1. Established the reference charge which is the basis
of assigning the value or magnitude and direction of forces.
Electrostatic Potential is the electric potential resulting from
the location of charged bodies in the vicinity.
produce an electric potential of 120 volts?
an isolated point charged
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Subject: BES 117 / 5:30-8:30 W 5:30-8:30 S E24Page 5
LECTURE
COLLEGE OF ENGINEERING AND TECHNOLOGY C. Electric field
intensity created by an isolated charged long cylindrical wire in
free space Ex. Two spheres separated from each other by 10 m have
charges of 0.001 coulombs and 0.003 coulombs respectively. In
between the two spheres is a point of zero electric field. What is
the distance from 0.001 coulomb sphere?
E= electric intensity (newton per coulomb) d= distance in meters
of the test charge (+1 C) to the charge (+Q) body r= 1, for free
space K= free space constant in SI units equal to 9x109 B. Electric
field intensity outside an isolated sphere in free space
ELECTRIC FLUX DENSITY - As define is the number of lines of
force per unit area crossing the surface at right angles to the
direction of field.
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Subject: BES 117 / 5:30-8:30 W 5:30-8:30 S E24Page 6
LECTURE