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HKAL Physics Essay writing : Electromagnetism
Chapter 13 Electrostatics
13.1 Electric Potential and Electric Field Intensity
13.1.1
Definition
(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 which it 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 the same direction as F.
(d) The electric potential V at any point is the work done per coulomb of positive charge from
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.
13.1.2
Derivation 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.
E
ds
V
V + dV
(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)
or, in the limit
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E= 
V
x
E= 
dV
dx
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13.2 Electrostatic Lines of Force, Equipotential Surface
13.2.1
General 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 = 
dV
). The
dr
separation between equipotential lines are closer together at the region where the field strength
is stronger.
13.2.2
Example : Around a charged sphere.
Field lines
+
Equipotentials
(a) At the region nearer to the conductor, the lines are closer together and thus the 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.
13.3 Flame Probe Investigation
13.3.1
Working 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.
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13.3.2
Flame 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 sphere
shows 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 ).
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Chapter 14 Capacitance
14.1 Capacitance
(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 every 1 V of
voltage across it.
14.2 Charged sphere as a capacitor
14.2.1
Derivation of the capacitance of a charged sphere
Q
a
ET
r P
At P, if sphere is very small (a  0), and small charge +dq is placed there, then
Er 
Qdq
4 0 r 2 dq

Q
4 0 r 2
a
Q

4 0 r 2
Sphere Potential V   
Clearly
dr 
Q
4 0 a
Q
= 4πε0 a = a constant  capacitance C
V
14.2.2
The dependence of the capacitance due to an external conductor
A neutral isolated conductor is pulled nearby to a positively charged, isolated conductor.
Potential
C
+
A
B
- +
After
Before
Distance
After conductor AB is brought near, the end A closer to the sphere is negatively charged while the
far end B is positively charged (accept answer shown in diagram).
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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.
14.3 Parallel plate capacitor
14.3.1
Structure of a parallel plate capacitor
(a)
25 cm
25 cm
A
Metal
plates
Polythene
spacer d
Sockets for
4-mm plugs
14.3.2
The 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.
14.3.3
The 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
Q
Q
0
A
0
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
1
energy stored CV 2 would decrease accordingly.
2
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14.4 Measuring the capacitance
14.4.1
Voltmeter is not appropriate to use
For the circuit shown below and with the switch closed for a while.
S

C
R
When the voltmeter is connected across the capacitor, the capacitor would discharge 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 ε as they form a potential divider which share
the voltage equally. As a result, the voltmeter will alter the circuit.
14.4.2
Using a reed switch
Reed switch
(2)
(1)
coil
f ~ 400 Hz from low
impedance output
of signal generator
12 V
R=
100 k 
V
Diode (to rectify a.c.)
A
Capacitor
C
(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-kΩ
protective 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 too large otherwise C does not
completely discharge
(2) Finding effect of electric field at edges of plates affects dependence of C on areas A.
(3) Stray capacitances to earth could affect the effective capacitance of C.
(4) leakage of charge from capacitor
(5) rebound of contact in reed switch.
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14.5 Energy stored in a capacitor
14.5.1
Derivation of the stored energy equation
(a) For capacitor, capacitance C = charge Q / p.d. V.
(b) Consider a small increase of charge Q, work W is done moving this from -ve to +ve plate
against voltage. V (assumed not to change), and W = QV.
(c) If initially uncharged, total work done
W

Q0
0
Q
1 Q02
dQ 
C
2 C
= ½Q0V0 , where Q0 and V0 are final charge, voltage.
This is stored energy EC = ½Q0V0
or ½CV2.
14.6 The charging of capacitor
The discussion below will be focused on the following circuit
C
+
_
R
VC
+
VR
K
_
E
14.6.1
Equation of state
On closing key K, E
= VC + VR
=
so we have:
or
Q
dQ
Q
R
+ IR =
C
dt
C
dQ
E
1

Q
=
dt
R CR
dQ
dt
= A - BQ, where A = E/R, B = 1/CR
14.6.2
Charging curve
(a) Mathematical derivation
Integrating,
1
CR

t
0
dt 

Q
0
dQ
,
EC  Q
t
EC  Q
  ln[
]
CR
EC
Q  EC(1  et / CR )
Q
Q = EC

t
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(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 is fully charged with a charge Q = EC (since VC =
E and I = 0)
14.6.3
Energy convertion
(a) Total work done by the battery,
WB 



[
0
dQ
E ]dt 
dt



[
0
E2
(CR) e t /CR
R

E t / CR
e
E ]dt
R

0
so, WB = E2C = QE

Q
0
0
(b) Energy stored by capacitor,  C   VC dQ  
Q
dQ
C
1
2
so, C  Q2 / 2C  Q E
(c) Difference between WB and  C must be the energy dissipated by Joule heating in the
resistance.


0
0
This is  R   I 2 Rdt  
i.e.  R 
E2
R

 CR  2t /CR
 2  e



E 2  2t / CR
e
dt
R

0
R = E2C/2 = QE/2, which is correct.
14.7 The discharging of capacitor

R
I/A
t/s
14.8 Applications of capacitor
(a) Flash unit of a camera
(b) Tuning circuit of a radio
(c) Smoothing circuit of a power supply
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Chapter 15 Electric Circuits
15.1 Drift velocity
15.1.1
Free 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.
15.1.2
Derivation 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-section of the wire is
q = nlAe
Drift velocity
v = l/t
Current in the wire i = q/t
= (nlAe)/(l/v)
= nAve
15.1.3
Joule 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
+


E
A
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B
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(1) For a charge Q the converted energy in R is (VA - VB)Q.
(2) Energy rate
= (VA - VB)
dQ
dt
or (VA - VB)I
= I2R, since by Ohm's law R =
VA  VB
I
(3) If all the energy is converted to heat, rate of heat dissipation = I2R.
15.1.4
Maintenance of current by a d.c. voltage supply
+
V
_
I
R
E (electric field)
(a) The outermost electrons of the metal atoms are loosely bound and under the influence of an
electric field E, drift through the wire towards the +ve terminal of power pack (i.e. current in
opposite direction).
(b) Electrons collide with atoms but steady flow of electrons into and out of the power pack is
maintained, power being supplied to maintain current.
(c) If V is increased there is a proportional increase in I (Ohm law) but for large currents R may
increase significantly causing departure from relation I  V.
15.1.5
Drift 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.
electric field direction
with electric field
e-
without electric field
small
displacement
(d) Electric field travels at an extremely high speed in a circuit so 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.
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15.2 Application of current measurement
Measuring the extension of a wire
A
attached to rigid
support
B
1M
Test wire
1M
variable load
(a) Change of resistance, R =
S
l , provided A does not change.
A
Assume the diameter and the resistivity of the steel wire are 0.4 mm and ~ 10-6 Ω m
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 =
1
mm = 10-5 m.
100
hence, R = 10-4 . Using a current I = 1A; p.d. change = 100 V - which could be
accurately measured using a potentiometer method.
15.3 Internal resistance of a battery
15.3.1
Definition of e.m.f.
E.m.f., , is the energy imparted by the battery per coulomb of charge passing internally between its
plates.
15.3.2
Measurement of internal resistance
A
12 V
low voltage
supply
V
24 W
12 V
(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 =
E V
where I is the current delivered.
I
(e) Sources of error : - the resistance at the connecting junctions are included in the calculated
value of r.
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15.3.3 Example relating internal resistance
(a) The typical internal resistance of an E.H.T. is of the order of MΩ (106 Ω) 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.
15.4 Non-ohmic behaviour
I
V
0
(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 value of V.
(b) That is, the resistance (V/I) of the tungsten wire filament increases as the current raises its
temperature and makes it white-hot. It is ohmic only for small V and I.
15.5 Power Transmission and domestic use of energy
The block diagram below shows how electric power is supplied to consumers.
Source of
energy
15.5.1
→
Generator
→
Transformers →
Consumers
General description
(a) The power is transmitted as an a.c. high voltage (132 kV) along transmission lines over long
distances from the power stations to sub-stations where it is transformed down to 6 kV, - to 220
V for domestic consumption.
(b) As it is relatively easy to transform the a.c. voltages up or down using transformers. So we use
a.c. voltage.
15.5.1
Major 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.
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15.5.2
Power loss due to power transmission
(a) Power loss
= I2R
P2R
as P = IV
V2
(b) If output power P and resistance R of the cables are constant, then power loss  1/V2,
where V = output voltage.
(c) Thus we use high voltage in power transmission.
=
15.5.3
Voltage variation at different places in Hong Kong
RL
Vm
I
R
V
(a) There is a voltage line drop due to the resistance of mains cables RL and current I taken by
appliance V = Vm - IRL.
(b) Length (and condition) of cable leads from transformer station will vary RL for any particular
location.
(c) Even at a particular location variation of current taken will vary mains voltage across
appliance.
15.5.4
using 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.
15.5.5 Structure 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.
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15.5.6
The 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.
15.5.7
Major 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%.
15.5.8
High-grade and low-grade energy
High-grade form of energy such as electrical energy can do work more efficiently (i.e. with greater
efficiency).
15.5.9
Energy 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’).
15.5.10
Domestic 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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