electrostatics bes 117

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. Address: San Jose Normal Road Baliwasan Zamboanga City

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?

Subject: BES 117 / 5:30-8:30 W 5:30-8:30 S E24Page 1

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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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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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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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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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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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electrostatics bes 117

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