HKAL Physics Essay Writing : Electromagnetism
HKAL Physics Essay writing : Electromagnetism
Chapter 13 Electrostatics
13.1Electric Potential and Electric Field Intensity
13.1.1Definition
(a)Potential difference between two points is the work done (or change in P.E.) per unit positive charge moving from one point to another.
(b)By selecting a reference point (usually the earth or a point at an infinite distance from any electric charges) in the electric field to be the zero potential, the potential of a point charge is taken as the potential difference between that point and the reference point.
(c)The electric field intensity E at any point is the force per unit positive charge whichit exerts at that point, i.e. E = F/Q, where F is the force acting upon a small charge +Q placed at the particular location. It is in thesame direction as F.
(d)The electric potential V at any point is the work done per coulomb of positive chargefrom infinity (or any zero reference) to that point., i..e. V = W/Q,
where W is the work done in moving a small charge +Q from infinity to that location.
13.1.2Derivation of the relation between E and V
Suppose A and B are two neighbouring points on a line of force, so close together that the electric field intensity between them is constant and equal to E.
(a)Electric potentials at A and B are respectively V and V + δV.
VBA= potential difference between A and B
= VA - VB
= -δV
(b)The work done in taking a charge Q from B to A
Force ×distance= charge ×potential difference VBA
EQ×δx= Q (-δV)
E=
or, in the limit E=
13.2Electrostatic Lines of Force, Equipotential Surface
13.2.1General properties
(a)The lines of force are drawn such that the density of the lines represents the strength of the field.
(b)The tangent at a point on a line gives the direction along which an electric force will act on a positive charge if it is placed at that point and the arrow gives the direction of the force (radially outward in this case).
(c)Since electric field strength is equal to the (negative of) potential gradient ( E = ). The separation between equipotential lines are closer together at the region where the field strength is stronger.
13.2.2Example : Around a charged sphere.
(a)At the region nearer to the conductor, the lines are closer together and thusthe field strength is stronger.
(b)In the diagram, the equipotential lines are closer together at the region nearer to the conductor and thus the field strength there is stronger.
13.3Flame Probe Investigation
13.3.1Working principle of flame probe
A flame probe can be used to investigate the potential in an electric field. The probe, in the form of a small gas flame at the point of a needle, is connected to a calibrated electroscope which has the same potential as the probe, therefore its deflection indicates the potential at the point where the probe is situated.
13.3.2Flame probe investigation of the potential around a charged sphere
(a)The conducting sphere is charged to about 1 kV by an EHT power supply. Negative charges are induced on the probe, and positive charges go to the electroscope. The flame ionizes the surrounding air to produce ions to neutralize the charges on the probe. The potential is therefore unaffected by the presence of the probe.
(b)Move the probe round the sphere keeping a fixed distance from the centre of the sphereshows that the potential does not vary. This shows that the equipotential line is circular.
(c)Move the probe outward from a point quite near the surface of the sphere along a radial direction shows that the potential is inversely proportional to the distance from the centre of the sphere ( V 1/r ).
Chapter 14 Capacitance
14.1Capacitance
(a)Definition
The capacitance (C) or charge-storing ability, of an isolated conductor is the charge (Q) required to cause unit charge in the potential (V) of a conductor.Q = CV.
(b)Markings of capacitor
i.e. Markings ‘50V 470μF’ on a capacitor.
(1)Maximum voltage across the capacitor should not exceed 50 V, otherwise(leakage or) breakdown may occur.
(2)Capacitance of 470μF means the capacitor holds 470μC of charge for every1 V of voltage across it.
14.2Charged sphere as a capacitor
14.2.1Derivation of the capacitance of a charged sphere
At P, if sphere is very small (a 0), and small charge +dq is placed there, then
Sphere Potential
Clearly = 4πε0 a = a constant capacitance C
14.2.2The dependence of the capacitance due to an external conductor
A neutral isolated conductor is pulled nearby to a positively charged, isolated conductor.
After conductor AB is brought near, the end A closer to the sphere is negativelycharged while the far end B is positively charged (accept answer shown in diagram).
As a result the electric potential of the sphere is lowered. Thus the capacitance of the sphere increases as it stores the same amount of charges for a smaller potential.
14.3Parallel plate capacitor
14.3.1Structure of a parallel plate capacitor
(a)
14.3.2The dependence of the capacitance to the overlapping surface area A of the plates and separation d between plates.
(a)Capacitance C = εoA/d.
(b)Capacitance C is proportional to A and inversely proportion to d.
14.3.3The dependence of the charges and energy stored in the capacitor due to the changes of A and d
When the capacitor is connected to a battery,
(a)The p.d. and the plate separation are kept constant.The plates can be slid apart to change the area of overlap.Measure Q for different area of overlap A.
Expected result : Q is proportional to A.
(b)The p.d. and plate area are kept constant.Use more spacers to increase the plate separation d.
Measure Q for different plate separations d.
Expected result : Q is proportional to 1 / d
(c)The capacitance of the capacitor would decrease when its plates are pulled apart (C = εoA/d).
As the voltage across it is unchanged, the charge stored would decrease since C = Q/V. The energy storedwould decrease accordingly.
14.4Measuring the capacitance
14.4.1Voltmeter is not appropriate to use
For the circuit shown below and with the switch closed for a while.
When the voltmeter is connected across the capacitor, the capacitor woulddischarge by driving a current through the voltmeter. VC decreases and the battery would then drive a current through the voltmeter and the resistor until the steady state is reached such that the voltage across the voltmeter (or capacitor) and the resistor are both less than εas they form a potential divider which share the voltage equally. As a result, the voltmeter will alter the circuit.
14.4.2Using a reed switch
(a)During the half-cycle when the reed is in contact with (2), the capacitor is charged by the battery. The charge Q = CV stored in the capacitor is then discharged through the 100-kΩprotective resistor and the light beam galvanometer when the reed is in contact with (1) in the following half-cycle.
(b)The number of charge-discharge actions per second equals the frequency f of the a.c. supply to the coil of the reed switch. If this is high enough current pulses follow one another so rapidly that the galvanometer deflection is steady and represents the average current I through it.
I = Qf = CVf, C = I / Vf.
(c)In order to ensure complete discharge, a CRO is connected across the resistor to check whether the discharging current drops to zero within a half-cycle.
(d)Sources of error
(1)R should prevent excessive current pulses but not be toolarge otherwise C does not completely discharge
(2)Finding effect of electric field at edges of plates affectsdependence of C on areas A.
(3)Stray capacitances to earth could affect the effectivecapacitance of C.
(4)leakage of charge from capacitor
(5)rebound of contact in reed switch.
14.5Energy stored in a capacitor
14.5.1Derivation of the stored energy equation
(a)For capacitor, capacitance C = charge Q / p.d. V.
(b)Consider a small increase of charge Q, work Wis done moving this from -ve to +ve plate againstvoltage.V (assumed not to change),and W = QV.
(c)If initially uncharged, total work done
= ½Q0V0 , where Q0 and V0 are final charge, voltage.
This is stored energy EC = ½Q0V0 or ½CV2.
14.6The charging of capacitor
The discussion below will be focused on the following circuit
14.6.1Equation of state
On closing key K, E= VC + VR
= + IR =
so we have: =
or = A - BQ, where A = E/R, B = 1/CR
14.6.2Charging curve
(a)Mathematical derivation
Integrating, ,
(b)Physical interpretation
Initially the capacitor is uncharged (VC = 0) and therefore the current is maximum (ε/R).
Clearly as the charge builds up in C, VC increases, the driving voltage across R (VR =ε- VC) decreases and thus the rate of charging (or current)decreases with time t.
Finally (theoretically t = ) the capacitor isfully charged with a charge Q = EC (since VC = E and I = 0)
14.6.3Energy convertion
(a)Total work done by the battery,
so, WB = E2C = QE
(b)Energy stored by capacitor,
so,
(c)Difference between WB and C must be the energydissipated by Joule heating in the resistance.
This is
i.e.
R = E2C/2 = QE/2, which is correct.
14.7The discharging of capacitor
14.8Applications of capacitor
(a)Flash unit of a camera
(b)Tuning circuit of a radio
(c)Smoothing circuit of a power supply
Chapter 15 Electric Circuits
15.1Drift velocity
15.1.1Free electron gas model for metallic conduction
(a)Metal consists of a large number of free electrons which are in a state of rapid thermal motion, moving randomly within the lattice at high speeds. They make frequent collisions with the lattice ions, changing directions all the time without net displacement.
(b)When a p.d. is applied, the electric field would cause the free electrons to accelerate (in a direction opposite to the electric field) and hence gain velocity/K.E. On colliding with the lattice ions, the electrons give up their additional K.E. to the lattice ions. The electrons slow down and then accelerate again. (This process is repeated many times.)
(c)The net effect is that the vast number of free electrons drift slowly, giving a net flow of electrons in one direction, which constitute an electric current macroscopically.
15.1.2Derivation of current flow equation
(a)v = average drift velocity
e = electronic charge
A = area of cross-section of the wire
n = no. of conduction electrons per unit volume
(b)Suppose electrons drift a distance l along a wire in time t, the charge q flows through a cross-sectionof the wire is
q= nlAe
Drift velocity v= l/t
Current in the wire i= q/t
= (nlAe)/(l/v)
= nAve
15.1.3Joule heating
(a)Physical interpretation
(1)When collided by the accelerating electrons, the lattice ions gain vibrational energy (both K.E. and P.E. as they vibrate more vigorously and with greater amplitude).
(2)Macroscopically the internal energy of the metal increases resulting in a temperature rise. Consequently thermal energy is released.
(b)Derivation
(1)For a charge Q the convertedenergy in R is (VA - VB)Q.
(2)Energy rate= (VA - VB) or (VA - VB)I
= I2R, since by Ohm's law R =
(3)If all the energy is converted to heat,rate of heat dissipation = I2R.
15.1.4Maintenance of current by a d.c. voltage supply
(a)The outermost electrons of the metal atomsare loosely bound and under the influence of an electricfield E, drift through the wire towards the +ve terminalof power pack (i.e. current in opposite direction).
(b)Electrons collide with atomsbut steady flow of electrons into and outof the power pack is maintained,power being supplied to maintain current.
(c)If V is increased there is a proportionalincrease in I (Ohm law) but for large currents R may increasesignificantly causing departure from relation IV.
15.1.5Drift velocity and speed of electrical signal
(a)average drift velocity: ~10-4 m/s ~ 10-5 m/s
(b)speed of electrical signal: ~ 108 m/s
(c)In a current-carrying conductor, electrons tend to accelerate along the opposite direction of the electric field inside. Due to collisions between electrons and atoms (ions) of the conductor, electrons move in zig-zag paths. (diagram accept) and drift with small displacement in unit time.
(d)Electric field travels at an extremely high speed in a circuitso electrons at every point of the circuit are influenced by the electric field nearly simultaneously as the switch is closed,electrical signal results at once.
15.2Application of current measurement
Measuring the extension of a wire
(a)Change of resistance, R = , provided A does not change.
Assume the diameter and the resistivity of the steel wire are 0.4 mm and ~ 10-6Ωm respectively. Discuss the practicability of the method above.
(b)A = (0.2)2 10-6 m2 ~ 10-7 m2
R = (10-6 / 10-7) 10-5 say
(c)Since resolution of micrometer gauge = mm = 10-5 m.
hence, R = 10-4. Using a current I = 1A; p.d. change = 100 V - which could be accurately measured using a potentiometer method.
15.3Internal resistance of a battery
15.3.1Definition of e.m.f.
E.m.f., , is the energy imparted by the battery percoulomb of charge passing internally between its plates.
15.3.2Measurement of internal resistance
(a)Connect a high resistance voltmeter (0 – 15 V) across the terminals of a low voltage power supply and adjust the output to 12 V, which is the e.m.f. E of the supply.
(b)The voltmeter reading drops slightly after connecting the ray-box lamp and the ammeter.
(c)Record the voltmeter reading V and the ammeter reading I. (I can also be estimated from the ratings of the lamp 12 V 24 W without using the ammeter)
(d)The internal resistance r of the supply is given by r = where I is the current delivered.
(e)Sources of error : -the resistance at the connecting junctions are included in the calculated value of r.
15.3.3Example relating internal resistance
(a)The typical internal resistance of an E.H.T. is of the order of MΩ(106Ω) so as to limit the current it supplies and safety can be ensured.
(b)When the starting motor is working, the ‘voltage lost’ due to the internal resistance is very large, say, a few volts, therefore the terminal voltage delivered to the headlights by the battery is reduced significantly and thus the headlights will become dim.
15.4Non-ohmic behaviour
(a)For a tungsten filament lamp, the I-V graph bends over as V and I increase,indicating that a given change of V causes a smaller change in I at larger valueof V.
(b)That is, the resistance (V/I) of the tungsten wire filament increases as thecurrent raises its temperature and makes it white-hot. It is ohmic only for smallV and I.
15.5Power Transmission and domestic use of energy
The block diagram below shows how electric power is supplied to consumers.
→→ →
15.5.1General description
(a)The power is transmitted as an a.c. high voltage (132 kV)along transmission linesover long distances from the powerstations to sub-stationswhere it is transformed down to 6 kV, - to 220 V fordomestic consumption.
(b)As it is relatively easy to transform the a.c. voltages upor down using transformers. So we use a.c. voltage.
15.5.1Major source of energy currently used in electric power stations in Hong Kong
(a)One major source is coal (or oil).
(b)The fuel (coal/oil) is burnt in the furnace for heating water in a boiler. This produces high pressure steam for driving the turbine of the generator (alternator, rotor) to produce electric power.
15.5.2Power loss due to power transmission
(a)Power loss= I2R
= as P = IV
(b)If output power P and resistance R of the cables are constant, then power loss1/V2, where V = output voltage.
(c)Thus we use high voltage in power transmission.
15.5.3Voltage variation at different places in Hong Kong
(a)There is a voltage line drop due to the resistance ofmains cables RL and current Itaken by applianceV = Vm - IRL.
(b)Length (and condition) of cable leads from transformerstation will vary RL forany particular location.
(c)Even at a particular location variation of currenttaken will vary mains voltageacross appliance.
15.5.4using overhead cables
Advantages
(a)Maintence is easy as overhead cables are bare.
(b)Also no insulation is required, so it is many times cheaper.
(c)Less installation compared with underground cables, therefore much cheaper.
(d)Compared with underground cables, overhead cables can use higher voltages for reducing power loss.
Disadvantages
(e)Potential danger of electric shock due to high voltage.
(f)Overhead transmission is suspected to cause environmental pollution (e.g. leukemia – a blood cancer) due to electromagnetic radiation consisting of oscillating electric and magnetic fields.
15.5.5Structure of an overhead power cable
(a)An overhead power cable usually consists of a central core of several steel wires, which is surrounded by many strands of aluminium wire. The reasons are :
(b)Aluminium is lighter, but copper is a better electric conductor.
(c)Copper is easily oxidized when bare but aluminium is not.
(d)Compared with copper, aluminium is less brittle.
(e)Steel has a high breaking stress, so it can increase the strength of the overhead cable.
15.5.6The advantages of using electricity domestically
(a)Electrical energy can be transmitted and distributed efficiently through long distance to remote areas using transformers and cables.
(b)Electrical energy can be easily and efficiently converted into other forms of energy using suitable transducers.
(c)Electrical energy is clean when used, producing no polluting gases or waste materials.
15.5.7Major energy conversions in a coal-fired power station
(a)‘Chemical energy’ of coal in furnace changes to ‘Internal energy’ of steam in boiler, then to ‘Kinetic energy’ of the rotating turbines and finally ‘Electrical energy’ from the generator.
(b)Typical conversion efficiency is 30 – 40%.
15.5.8High-grade and low-grade energy
High-grade form of energy such as electrical energy can do work more efficiently (i.e. with greater efficiency).
15.5.9Energy crisis
(a)ALL different types of energy change into internal energy ultimately.
(b)They turn from high-grade useful form to a low-grade useless form i.e. energy is degraded as time goes on.
(c)This is the reason why there is a need for new sources of high-grade energy (which is known as an energy ‘crisis’).
15.5.10Domestic use of solar energy
Favourable factors
(a)Less pollution.
(b)Reduce the dependence on sources like coal, oil.
(c)Energy is consumed locally, no need for distant transmission.
(d)Practically unlimited supply.
Unfavourable factors
(e)Require large area facing the sun for the installation of solar cells.
(f)The cost for the equipment is high.
(g)Only suitable for tropical/subtropical region where sunlight is available throughout the year.
(h)Supply is not steady due to seasonal changes.
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