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Chapter 21 Magnetic fields

21.1 Magnetic field B

Magnetic Field B is a field of force that exist around the magnetic body or current carrying conductor[1].

Magnetic field B, also known as magnetic flux density or magnetic field Intensity[2].

clip_image002

we can summarize the result with the following equation:

(Direction) clip_image004

(Magnitude) clip_image006

21.2 Force on a moving charge

clip_image007

If the charge q moving in a magnetic field B with a velocity v. The angle between B and v is q the charge will experience magnetic force as clip_image006[1].

Unit bagi magnetic flux density adalah Tesla,T atau Wbm-2

To find out the direction of the force also can be shown using Fleming’s left-hand rules.

Contoh 1

Satu eletron bergerak dalam medan magnet. Pada satu ketika laju elektron ialah 3.0 ´ 106 ms-1. Magnitud daya magnet yang bertindak pada elektron ialah 5.0 ´ 1013 N. Sudut di antara arah halaju elektron dan arah daya magnet ialah 30°. Kira ketumpatan fluks magnet pada kedudukan elektron medan itu.

[cas elektron = -1.6 ´ 10-19 C]

21.3 Force on a current-carrying conductor

clip_image010

Above, a current-carrying wire is at right-angles to a uniform magnetic field. The field exerts a force on the wire. The direction of the force is given by Fleming’s left-hand rule . The size of the force depends on the current I, the length l in the field, and the strength of the field. This effect can be used to define the magnetic field strength, known as the magnetic flux density, B:

FB = BIl (1)

B is a vector. The SI unit of B is the testa (T). For example, if the magnetic flux density is 2 T, then the force on 2 m of wire carrying a current of 3 A is 2 x 2 x 3 = 12 N.

If a wire is not at right angles to the field, then the above equation becomes

FB=BIl sinq

where q is the angle between the field and the wire. As q becomes less, the force becomes less. When the wire is parallel to the field, sin q = 0, so the force is zero.

Contoh 2

clip_image012

Satu konduktor lurus panjang 2m membawa arus 1.5A dan di letakkan dalam medan magnet. Medan itu mempunyai ketumpatan fluks magnet bermagnitud 3.0 ´ 10-3 T. Sudut antara arah medan dan konduktor ialah 30°. Kirakan magnitud daya magnet yang bertindak pada konduktor dan nyatakan arah daya itu.

21.4 Magnetic fields due to currents

clip_image014

The Biot-Savart law,

On the top, a short length dl of thin wire, carrying a current I, causes a magnetic flux density dB at P. According to the Biot-Savart law

clip_image016

With a suitable constant, this can be turned into an equation:

clip_image018

For a vacuum (and effectively for air), the value of k is 10-7 T m A-1. However, in practice, another constant, m0, is used, and the above equation is rewritten as follows:

clip_image020 (2)

m0clip_image022 is called the permeability of free space. Its value is 4p x 10-7 T m A-1. This is not found by experiment. It is a defined value, linked with the definition of the ampere .

Calculating magnetic flux density Using equation clip_image020[1] and calculus, it

is possible to derive equations for B near wires and inside coils carrying a current I.

clip_image024

B near an infinitely long, thin, straight wire At a distance a from such a wire (as on the top)

clip_image026

Contoh 3

Arus sebanyak 10A mengalir melalui seutas dawai mencancang lurus yang panjang. Kirakan komponen mengufuk medan magnet Bumi. Jika terdapat satu titik neutral 8.0cm dari dawai itu. [m0= 4p ´ 10-7 Hm-1]

clip_image028

B at the centre of a thin coil Or the axis of such a coil, of N turns and radius r (as on the top),

clip_image030

Contoh 4

Arus yang sama magnitud mengalir secara bergilir-gilir melalui seutas dawai tegak lurus yang panjang dan satu gegelung membulat berjejari 10cm dimana satahnya tegak dan selari dengan komponen mengufuk medan magnet pada titik berjarak 10cm di utara atau di selatan dari dawai itu terhadap paduan medan magnet di pusat gegelung itu.

Contoh 5

Satu gegelung membulat diletakkan, dimana satahnya adalah mencancang dan paksinya membuat sudut kecil q dengan meridian magnet itu. Arus dilalukan mengelilingi gegelung supaya medan yang dihasilkan berkecenderungan untuk menentang medan Bumi. Satu magnet mengufuk yang kecil di letakkan di pusat gegelung itu.

Perihalkan dan terangkan kesan ke atas magnet itu apabila kekuatan arus bertambah. Gegelung itu mempunyai jejari min 10cm,50 lilitan dan arus sebanyak 0.5A mengalir didalamnya. Jika q=12° dan magnet mengarah ke Timur-Barat.Kirakan komponen mengufuk medan magnet Bumi.

clip_image032

B inside an infinitely long solenoid (coil) The field inside such a solenoid (as on above) is uniform. If n is the number of turn per unit length (per m), then

B=µ0nI

This equation is a reasonable approximation for any solenoid which is at least ten times longer than it is wide.

Contoh 6

Berapa banyakkah lilitan mesti dipunyai oleh satu solenoid jika panjang solenoid itu 2.0m dan arus 5.0A mengalir melaluinya supaya arus menghasilkan ketumpatan fluks magnet bermagnitud 2.0 ´ 10-2 T di pusat solenoid ?

B inside a solenoid with a core The value of B is changed by a core. For example, with a pure iron core, B is increased by a factor of about 1000 (depending on the temperature). The previous equation then becomes:

B = µrµ0 nI

µr, is called the relative permeability of the material. So, for pure iron, µr, is about 1000.

An electromagnet is a solenoid with a core of high µr.

21.5 Force between two current-carrying conductors

Force between two current-carrying wires

clip_image034

X and Y above are two infinitely long, straight wires in a vacuum. The current in X produces a magnetic field, whose flux density is B at Y. As a result, there is a force on Y. F is the force acting on length l.

From equation (3) clip_image036

From equation (1) FB = BI2 l

clip_image038

Note:

• The above equation gives the force of X on Y. Working out the force of Y on X gives exactly the same result.

• If the two currents are in the same direction (as above), then the wires attract each other. If the two currents are in opposite directions, then the wires repel.

Contoh 7

clip_image040

Tiga konduktor yang selari, P, Q dan R berada pada satah yang sama.Jarak antara P dan Q adalah 4cm dan antara Q dan R ialah 2 cm. Arus dalam P dan Q adalah 10A dan 5A masing-masing dan mengalir pada arah yang berlawanan.Daya paduan ke atas Q adalah 5 ´ 10-4N per meter dan bertindak ke arah P. Kirakan magnitud dan arah arus dalam R.

21.6 Definition of ampere

Defining the ampere

The SI unit of current is defined as follows:

One ampere is the current which, flowing through two infinitely long, thin, straight wires placed one metre apart in a vacuum, produces a force of 2 x 10-7 newtons on each metre length of wire.

Using the various factors in equation

clip_image038[1] ,

the above definition can be expressed in the following form (for simplicity, units have been omitted):

If I1=I2=1A,a=1m and l=1m,then F= 2×10-7N. Substituting these in equation above gives µ0 = 4p x 10-7. The above definition is not a practical way of fixing a standard ampere. This is done by measuring the force between two current-carrying coils.

Current Balance

- Measure current by using the principle of the force between two paralell wire using mechanical

clip_image042

Prinsip:

clip_image044

m = jisim pemberat W

b = jarak F ke tuas

c = jarak berat ke tuas

L = jarak XY

apabila sistem tuas seimbang kita dapati

clip_image046

Known that

FB = BIL

B=µ0nI

Maka FB = µ0n I . IL

Then clip_image048

clip_image050

21.7 Torque on a coil

clip_image052

Lets consider the force on PQ and RS that involve in producing the torque.

Force that produce on PQ and RS have the same magnitude but different direction.

so force,F

clip_image054

where

L = height of coil(length of the coil in the magnetic field) and N = no. of turns carrying current,I.

Torque

clip_image056

clip_image058

where

q = angle (below the horizontal from the plan view) between magnetic flux density and the coil.

[coil area,A = Lb]

So the equation for torque will be

clip_image060

When

q = 0°

t = BIAN

Contoh 8

21.8 Determination of ratio e/m and q/m

Using an electron gun

Lets consider the electron gun.

clip_image062

Lets consider the movement of electron in an electro gun

1. The kinetic energy gain by electron = Electric potential energy of electron

2. Electron velocity :

clip_image064

clip_image066

3. Electric force, FE = Magnetic Force, FB , when it’s in equilibrium

clip_image068

clip_image070

4. From equation (2) and (3) we can conclude:

clip_image072

clip_image074

Todetermine any positive charge,q/m using spectrometer.

1. any positivly charge velocity,

clip_image076

2. The charges follow the circular path, the centripetal force of the charge.

clip_image078

3.Magnetic force of charges

clip_image080

clip_image082

Also clip_image084

clip_image086

clip_image088

clip_image090 for any specific charges.

21.9 Hall effect

1. apabila arus mantap I melalui konduktor/semikonduktor

clip_image092

2. kemudian diletakkan kedalam medan magnet yang kuat dimana I berserenjang dengan B.

3. Medan magnet melencongkan cas positif ke bahagian atas dan cas negatif ke bahagian bawah

clip_image094

4. Pemisahan ini menghasilkan medan elektrik E yang berserenjang dengan arah aliran arus I

- Bila cas bertambah

- Medan elektrik juga bertambah

- Daya gerak elektrik juga meningkat

clip_image096

5. Pemisahan akan berhenti bila daya E, FE=FB

Magnetic force = Electric force

clip_image098

clip_image100 where [clip_image102 n ialah halaju hanyut elektron,n bilangan elektron bebas]

clip_image104

Hall voltage clip_image106

clip_image108


[1] Oxford Dictionary

[2] Fajar Bakti STPM V2 m.s 113

chapter 20 Direct current circuits

20.1 Electromotive force

Voltage (PD and EMF)

In the circuit below, several cells have been linked in a line to form a battery. The potential difference (PD) across the battery terminals is 12 volts (V). This means that each coulomb (C) of charge will ‘spend’ 12 joules of energy in moving round the circuit from one terminal to the other.

clip_image002

The PD across the bulb is also 12 V. This means that, for each coulomb pushed through it, 12 J of electrical energy is changed into other forms (heat and light energy).

PD may be measured using a voltmeter as shown above.

PD, energy, and charge are linked by this equation:

energy transformed = charge x PD

For example, if a charge of 2 C moves through a PD of 3 V, the energy transformed is 6 J.

The voltage produced by the chemical reactions inside a battery is called the electromotive force (EMF). When a battery is supplying current, some energy is wasted inside it, which reduces the PD across its terminals. For example, when a torch battery of EMF 3.0 V is supplying current, the PD across its terminals be might be only 2.5 V.

20.2 Internal resistance of sources

Internal resistance

. In reality, when a battery is supplying current, its output PD is less than its EMF. The greater the current, the lower the output PD. This reduced voltage is due to energy dissipation in the battery. In effect, the battery has internal resistance. Mathematically, this can be treated as an additional resistor in the circuit.

clip_image004

The battery above is supplying a current 1 to an external circuit. The battery has a constant internal resistance r. From Kirchhoffs second law E= IR+ lr

But V= IR, so E= V+ lr

So V= E- Ir ……………(1)

clip_image006

The graph above shows how V varies with I. Unlike earlier graphs, V is on the vertical axis.

Note:

• When I is zero, V = E. In other words, when a battery is in open circuit (no external circuit), the PD across its terminals is equal to its EMF

• When R is zero, V is zero. In other words, when the battery is in short circuit (its terminals directly connected), its output PD is zero. In this situation, the battery is delivering the maximum possible current, lmax, which is equal to E/r. Also, the battery’s entire energy output is being wasted internally as heat.

• As Imax = E/r, it follows that r = E/lmax. So the gradient of the graph is numerically equal to the internal resistance of the battery.

If both sides of equation (1) are multiplied by I, the result is Vl = El- I2r. Rearranged, this gives the following:

clip_image008

Contoh 1

Sel kering dengan EMF/dge 1.5 V disambungkan secara sesiri dengan perintang dan bateri. Jika arus 3.0 A mengalir, voltan/PD merentasi sel adalah 0.42 V. Apakah rintangan dalam sel kering?

20.3 Kirchhoff’s law

Kirchhoff’s first law

clip_image010

The currents at junctions X and Y above illustrate a law which applies to all circuits:

TotaI current out of junction = total current into junction

This is known as Kirchhoff’s first law. It arises because, in a complete circuit, charge is never gained or lost. It is conserved. So the total rate of flow of charge is constant.

Let’s consider :

Current Junction:

clip_image012

Arah positif : I1,I2

Arah negatif: -I3,-I4,-I5

Maka menggunakan hukum Kirchoff : SI = I1 + I2 +( -I3) + (-I4) + (-I5) = 0

Resistors in series

If R1 and R2 below have a total resistance of R then R is the single resistance which could replace them.

clip_image014

From Kirchhoff’s first law, all parts of the circuit have the same current /through them.becouse there is only one input and one output

From Kirchhoff’s second law E= IR and E= IR1 + IR2.

So IR = IR1 + IR2

\ R=R1+R2

For example, if R1 = 3 W and R2 = 6 W, then R = 9 W.

Kirchhoff’s second law

clip_image016

The arrangement above is called ‘a circuit’. But, really, there are two complete circuits through the battery. To avoid confusion, these will be called loops.

In the circuit above, charge leaves the battery with electrical potential energy. As the charge flows round a loop, its energy is ‘spent’ – in stages – as heat. The principle that the total energy supplied is equal to the total energy spent is expressed by Kirchhoff’s second law.

Round any closed loop of a circuit, the algebraic sum of the EMFs is equal to the algebraic sum of the PDs (i.e. the algebraic sum of all the IRs).

Note:

•From the law, it follows that if sections of a circuit are in parallel, they have the same PD across them.

•’Algebraic’ implies that the direction of the voltage must be considered. For example, in the circuit below, the EMF of the right-hand battery is taken as negative (-4 V) because it is opposing the current Therefore:

algebraic sum of EMFs = 18 + (-4) = +14V

algebraic sum of IRs = (2 x 3W) + (2 x 4W) = +14 V

clip_image018

Lets consider:

Closed Loops

Paralell

Series

clip_image020

clip_image021

Resistors in parallel

clip_image023

From Kirchhoff’s second law (applied to the various loops):

E = IR (Loop with total Resistor)

and

E = I1 R1 (Loop with Resistor R1)

and

E = I2 R2 (Loop with Resistor R2)

From Kirchhoff’s first law I = I1 + I2.

So

clip_image025

clip_image027

Kegunaan hukum Kirchoff di dalam litar adalah untuk menentukan nilai arus dan arah arus pada satu-satu litar:

clip_image030

Contoh 2

Didalam litar di bawah, dge bateri E1 ialah 21V dan rintangan dalamnya diabaikan dan EMF bateri E2 ialah 84V dan rintangan dalamnya ialah r2 = 4W.

clip_image032

Contoh 3

Calculate the current I, the resistance R and the e.m.f. E for the circuit below.

clip_image034

20.4 Potential divider

Potential divider

A potential divideror potentiometer like the one below passes on a fraction of the PD supplied to it.

clip_image036

In the input loop above, the total resistance = R1 + R2.

So clip_image038

But Vout = IR2,

so clip_image040

Note:

•The above analysis assumes that no external circuit is connected across R2. If such a circuit is connected, then the output PD is reduced.

In electronics, a potential divider can change the signals from a sensor (such as a heat or Iight.detector) into voltage changes which can be processed electrically. For example, if R2 is a thermistor, then a rise in temperature will cause a fall in R2 and therefore a fall in Vout. Similarly, if R2 is a light­dependent resistor (LDR), then a rise in light level will cause a fall in R2, and therefore a fall in Vout.

Potential dividers are not really suitable for high-power applications because of energy dissipation

20.5 Potentiometer

(a) The idea of potentiometer

Function : To measure emf a cell

Key Idea : make sure the galvanometer as a null detector

clip_image042

E = lV

where

V = potential difference per unit length of AB.

L = The length of wire

l = length that galvanometer show zero reading

so PD across l,clip_image044

Potentiometer Applications

(a) Measuring a cell’s internal resistance.

clip_image046

clip_image048

clip_image050

clip_image052 or clip_image054

clip_image056

If a graph is plotted,

Gradient, clip_image058

Internal resistance, clip_image060

Intercept, clip_image062

clip_image064

(b) Comparing resistance

clip_image066

clip_image068

clip_image070

clip_image072

20.6 Wheatstone bridge

6 Wheatstone Bridge

clip_image074

clip_image076

20.7 Shunt and multiplier

9 Shunts and Multipliers

(a) Conversion of milliammeter to ammeter

clip_image078

Since clip_image080

clip_image082

clip_image084

(b) Conversion of milliammeter to voltmeter

clip_image086

clip_image088

clip_image090

clip_image092

clip_image094

Chapter 19 Electric current

19.1 Conduction of electricity

Introduction

Static electricity

if two materials are rubbed together, electrons may be transferred from one to another. As a result, one gains negative charge, while the other is left with an equal positive charge. If the materials are insulators , the transferred charge does not readily flow away. It is sometimes called static electricity.

clip_image002

A charged object will attract an uncharged one. The charged rod has extra electrons. Being uncharged, the foil has equal amounts of – and + charge. The – charges are repelled by the rod and tend to move away, while the + charges are attracted. However, the force of attraction is greater because of the shorter distance.

clip_image004

Charge which collects in one region because of the presence of charge on another object is called induced charge.

Current

conventional direction

clip_image006

In the circuit above, chemical reactions in the cell push electrons out of the negative (-) terminal, round the circuit, to the positive (+) terminal. This flow of electrons is called a current.

An arrow in the circuit indicates the direction from the + terminal round to the -. Called the conventional direction, it is the opposite direction to the actual electron flow.

The SI unit of current is the ampere (A).

A current of 1 A is equivalent to a flow of 6 x 1018 electrons per second. However, the ampere is not defined in this way, but in terms of its magnetic effect .

\ One Ampere is the current which, flowing through two infinitely long, thin, straight wires placed one metre apart in a vacuum, produces a force of 2´10-7 N on each metre length of wire

Current may be measured using an ammeter as above.

Conductors and insulators

Current flows easily through metals and carbon. These materials are good conductors because they have free electrons which can drift between their atoms.

Most non-metals are insulators. They do not conduct because all their electrons are tightly held to atoms and not easily moved. Although liquids and gases are usually insulators, they do conduct if they contain ions.

Semiconductors, such as silicon and germanium, are insulators when cold but conductors when warm.

Charge

Charge can be calculated using this equation:

Charge,Q = current,I x time,t

The SI unit of charge is the coulomb (C).

Example 1,

if a current of 1 A flows for 1 s, the charge passing is 1 C. (This is how the coulomb is defined.) Similarly, if a current of 2 A flows for 3 s, the charge passing is 6 C.

19.2 Energy and electrical power

Energy transfer

clip_image008

Above, charge Q passes through a resistor in time t. Work W is done by the charge, so energy W is transformed – the electrons lose electrical potential energy and the lattice gains internal energy (it heats up).

W, Q, and V are linked by this equation :

W= QV

But Q= It,

so W= VIt (1)

Applying V = IR to the above equation gives

W= I2 Rt and clip_image010 (2)

For example, if a current of 2 A flows through a 3 W resistor for 5 s, W = 22 x 3 x 5 = 60 J. So the energy dissipated is 60 J. Double the current gives four times the energy dissipation.

Note:

• Equation (1) can be used to calculate the total energy transformation whenever electrical potential energy is changed into other forms (e.g. KE and internal energy in an electric motor). Equations (2) are only valid where all the energy is changed into internal energy. Similar comments apply to the power equations which follow.

As power clip_image012, it follows from (1) and (2) that

P= VI P= I2R clip_image014

Contoh 1

Resistor of resistance 1.0W,2.0W dan 3.0W disambungkan secara sesiri dengan bateri 12V. Kirakan.

(a) Rintangan berkesan

(b) Arus melalui litar

(c) PD yang melalui setiap

(d) Kuasa yang dibebaskan dari resistor 2.0 W

(e) Jumlah kuasa yang dibebaskan.

19.3 Drift velocity

Current and drift speed

Most electrons are bound to their atoms. However, in a metal, some are free electrons which can move between atoms. When a PD is applied, and a current flows, the free electrons are the charge carriers.

clip_image016

in the wire above, free electrons (each of charge e) are moving with an average speed v. n is the number density of free electrons: the number per unit volume (per m3).

In the wire the number of free electrons = clip_image018

So total charge carried by free electrons = clip_image020

As clip_image022:

\Time,t taken for all the free electrons to pass through clip_image024

As clip_image026

clip_image028

\clip_image030

v is called the drift speed or drift velocity. Typically, it can be less than a millimetre per second for the current in a wire.

Contoh 2

Dawai kuprum mempunyai keratan rentas 1.0 x 105 m2 membawa arus 2.0A. Jika bilangan elekron bebas perunit isipadu dawai kuprum ialah 1029, anggarkan halaju hanyut elektron bebas tersebut.

19.4 Electrical conductivity

Conductance

If a PD V is applied across a conductor, and a current I flows, then V = IR.

However, as V is the cause of the current and 1 is the effect, it is more logical to write this as:

clip_image032

clip_image034 is called the conductance.

Resistivity

clip_image036

The resistance R of a conductor depends on its length l and cross-sectional area A:

clip_image038

This can be changed into an equation by means of a constant, r, known as the resistivity of the material:

clip_image040

With this equation, the resistance of a wire can be calculated if its dimensions and resistivity are known.

And clip_image042is the Electrical conductivity,s

Contoh 3

Dawai sepanjang 50cm mengandungi teras besi, berdiameter 2.0mm disalut dengan tembaga dibahagian luar berdiameter 3.0mm.

clip_image044

Kirakan Rintangan dawai.

[ Kerintangan besi = 1.0 x 10-7 Wm,tembaga=1.7 x 10-8 Wm]

19.5 Current density

The current density is the current per unit cross-sectional area (per m2).

clip_image046 so clip_image048

Note:

• The number density of free electrons is different for different metals. For copper, it is 8 x 1028 m-1.

• When liquids conduct, ions are the charge carriers. The above equations apply, except that e and n must be replaced by the charge and number density of the ions.

Contoh 4

Seutas dawai kuprum mempunyai keratan rentas 2.0mm2 dan arus 3.0 A mengalir melaluinya.Anggapkan bahawa terdapat 1029 elektron bebas permeter padu kuprum itu. Kirakan

(a) Halaju hanyut dalam dawai

(b) Ketumpatan arus

19.6 Dependence of resistance on temperature

Pertimbangkan

1. Current Density

clip_image046[1]

2. Electrical Conductivity

clip_image050

3. Electrical resistivity

clip_image052

Maka clip_image054 tapi (clip_image056)

clip_image058

clip_image060

clip_image062

Maka clip_image064 then

clip_image066

clip_image068

clip_image070constant if temperature constant

E, causes free electron of charges,v, drift opposite the electric field.

clip_image072 tapi clip_image074

clip_image076

clip_image078

Maka electrical Conductivity

clip_image080

Resistance and temperature

A conducting solid is made up of a lattice of atoms. When a current flows, electrons move through this lattice.

clip_image082

Metals When free electrons drift through a metal, they make occasional collisions- with the lattice. These collisions are inelastic and transfer energy to the lattice as internal energy. That is why a metal has resistance. If the temperature of a metal rises, the atoms of the lattice vibrate more vigorously. Free electrons collide with the lattice more frequently, which increases the resistance.

Semiconductors (e.g. silicon) At low temperature, the electrons are tightly bound to their atoms. But as the temperature rises, more and more electrons break free and can take part in conduction. This easily outweighs the effects of more vigorous lattice vibrations, so the resistance decreases. At around 100-150 °C, breakdown occurs. There is a sudden fall in resistance – and a huge increase in current. That is why semiconductor devices are easily damaged if they start to overheat.

The conduction properties of a semiconductor can be changed by doping it with tiny amounts of impurities. For example, a diode can be made by doping a piece of silicon so that a current in one direction increases its resistance while a current in the opposite direction decreases it.

clip_image084

The graphs above are for a typical metal conductor and one type of thermistor. The thermistor contains semiconducting materials

Chapter 18 Capacitor

18.1 Capacitance

Capacitors store small amounts of electric charge.

clip_image002

A capacitor can be charged by connecting a battery across it. The higher the PD V, the greater the charge Q stored. Experiments show that Q a V. Therefore, Q/V is a constant. The capacitance C of a capacitor is defined as follows:

clip_image004

In symbols clip_image006

The higher the capacitance, the more charge is stored for any given PD.

Capacitance is measured in C V-1, known as a farad (F). However, a farad is a very large unit, and the mF (10-6 F) is more commonly used for practical capacitors.

Contoh 1

Cas untuk setiap plat kapasitor selari ialah 20 mC bila beza upaya yang dibekalkan ialah 2 V. Kirakan

(a) Kapasitan Kapasitor

(b) Cas yang tersimpan bila PD ialah 1.5 V?

18.2 Parallel Plate capacitors

Electric field near a charged plate

clip_image008

On the top, a metal sphere has a charge Q uniformly distributed over its surface. The electric field E near the surface is given by equation :

clip_image010

But the surface area of the sphere A = 4pR2. So

clip_image012 (1)

Conclusion:

This equation also applies to a flat, charged metal plate of surface area A.

The simplest form of capacitor is made up of two parallel metal plates, separated by an air gap.

clip_image014

The capacitor above has been connected to a battery so that the PD across its plates is V. As a result, it is storing a charge Q. (This means that charge Q has been transferred, leaving -Q on one plate and +Q on the other.)

From equation (1) above clip_image012[1]

From equation clip_image012[2]

From these, it follows that clip_image016

But capacitance, clip_image006[1]

So clip_image018

From this equation clip_image018[1]

we can conclude that capacitance in a parallel plate is only depend on area A and distance d

clip_image021, so a larger plate area gives a higher C.

clip_image023, so a smaller plate separation gives a higher C.

Contoh 2

The capacitance of paralell-plate capacitor is 8.4 pF. It consist of two paralell plates separated by a distance of 2.0 mm. the capacitor is charged to 6.0 V.

(a) calculate the area of each plate

The charged capacitor is then disconnected from the charging voltage. The separation between the plates is reduced to 0.50 mm.

(b) what is the new capacitance of the capacitor

(c) calculate the PD between the plates.

18.3 Dielectrics

Dielectric

clip_image025

if the gap between the capacitor plates is filled with a material such as polythene, the capacitance is increased. Any insulating material which has this effect is called a dielectric.

In practice, this is achieved by rolling up two long strips of foil with a thin dielectric between them.

clip_image027

The capacitance of the parallel plate capacitor is

clip_image029

The relative permittivity clip_image031, of the dielectric is the factor by which the capacitance is increased.

In electrolytic capacitors, the dielectric is formed by the chemical action of a current. This gives a very thin dielectric, and a very high capacitance. But the capacitor must always be used with the same plate positive, or the chemical action is reversed.

Capacitors have a maximum working voltage above which the dielectric breaks down and starts to conduct.

Contoh 3

Satu kapasitor plat selari mengandungi dua permukaan plat setiap satunya berkeluasan 0.40 m2 dan dipisahkan sejauh 0.20 cm

(a) kirakan kapasitan kapasitor jika ruang di antara plat diisikan

i. vakum

ii. penebat dengan pemalar dielektrik 2.3

(b) Dengan kehadiran penebat (dielektrik) diantara plat, penebatan menurun, medan elektrik menjadi 1.8 x 105 V cm-1. Berapakah PD maksimum yang boleh dibekalkan kepada kapasitor ?

18.4 Capacitor in series and parallel

Capacitors in series

clip_image033

if C1, and C2 have a total capacitance of C, then C is the single capacitance which could replace them.

Two capacitors in series store only the same charge Q as a single capacitor. So , clip_image035 and clip_image037 . But V = V1 + V2

So clip_image039

Or clip_image041

Contoh 4

Tiga kapasitor berkapasitan 6.0 F, 3.0 F dan 2.0 F di sambung secara bersiri dengan bateri 12 V. Tentukan

(a) kapasitan berkesan

(b) Cas pada setiap kapasitan

(c) PD setiap kapasitan

Capacitors in parallel

clip_image043

if C1, and C2 have a total capacitance of C, then C is the single capacitance which could replace them.

Capacitors in parallel each have the same PD across them.

So ,Q = CV

And Q1= C1 V and Q2 = C2V.

Together, the capacitors act like a single capacitor with a larger plate area.

So Q = Q1 + Q2

CV = C1V+C2V

and C = C1 + C2

Contoh 5

clip_image045

Didalam litar diatas kirakan

(a) cas setiap kapasitor

(b) kapasitan berkesan

18.5 Energy stored by a capacitor

Work must be done to charge up a capacitor. Electrical potential energy is stored as a result.

If a charge of 2 C is moved through a steady PD of 10 V, then, using equation

Electrical potential, clip_image047,

\work done W = QV = 2 x 10 = 20 J.

So the stored energy is 20 J. Numerically, this is the area under the graph below.

clip_image049

When a capacitor is being charged, Q and V are related as in the graph below. As before, the energy stored is numerically equal to the area under the graph, which is clip_image051. As clip_image006[2] , this can be expressed in three ways:

Energy stored, clip_image053

clip_image055

Contoh 6

Satu kapassitor 6.0 mF disambungkan secara bersiri kepada kapasitor 10 m.Cas yang tersimpan dalam setiap kapasitor ialah 40 mC. Kirakan jumlah tenaga keupayaan elektrik yang tersimpan dalam kedua-dua kapasitor.

18.6 Charging and discharging

Discharge of a capacitor

clip_image057

The capacitor above is charged from a battery and then discharged through a resistance R.

clip_image059

Graph A shows how, during discharge, the charge Q decreases with time t, according to the following equation:

clip_image061 where e = 2.718

RC is called the time constant. (it equals the time which the charge would take to fall to zero if the initial rate of loss of charge were maintained.)

Increasing R or C gives a higher time constant, and therefore a slower discharge.

The gradient of the graph at any time t is equal to the current at that time.

clip_image063

Graph B shows how the current decreases with time. The area under the graph is numerically equal to the charge lost.

Note:

•Each graph is an exponential decay curve, with the same characteristics as a radioactive decay curve. A half-life can be calculated in the same way.

Graph C shows how voltage decrease with time during discharge

charging a capacitor

The capacitor below is charged through a resistance R. The graph shows how the charge builds up.

clip_image065

The equation for the charging process is clip_image067

The charge reaches a maximum value of Qo which is equal to VC.

The charging current starts at a maximum value of V/R and falls to a lower value in the same way as it does when the capacitor discharges.

Voltage-time graphs

Since the voltage across a capacitor is proportional to the charge on it the variations of voltage with time are the same shape as the charge-time graphs. The equations for calculating the voltage at any time are similar to those for charge, substituting V for Q

Contoh 7

Sebuah kapasitor 2 mF dicaskan dengan bateri dengan PD 1.5 V. Kemudian dinyahcaskan melalui perintang 60 kW.

(a) Berapakah pemalar masa litar nyahcas

(b) Tentukan masa untuk cas berkurang kepada

i.clip_image069 dari asal

ii. clip_image071dari asal

1.13 LATITUD DAN PENCERAP

clip_image002

Bintang hanya kelihatan di sebelah atas horizon pencerap, dan jejak bintang tersebut adalah bergantung kepada latitud pencerap. Bintang yang dapat dicerap adalah berbeza bagi latitud yang berlainan.

Andainya kita dapat mencerap bintang dengan berada di kutub utara dan mencerap pula di kutub selatan nescaya kita akan dapati bintang yang kita akan dapat cerap itu adalah berlainan sama sekali. Perbezaan ini kerana kedudukan latitud kita adalah berbeza.dengan itu bentuk Bumi yang bulat ini telah membuat pandangan kita menjadi hanya separuh dari yang sebenar.

Kita boleh menentukan orientasi sfera langit di tempat sendiri dengan merujuk kepada horizon dan titik zenith di tempat sendiri.Kaedah ini boleh di praktikkan dimana sahaja. Di sebelah hemisfera utara, kutub utara langit berada di atas horizon kutub utara pada ketinggian(sudut) yang sama dengan latitud pencerap (contoh: kota bharu berada pada latitud 6°U, ini bermaksud kutub utara langit berada pada ketinggian 6° dari horizon langit utara). Bintang Kutub (Polaris), ialah bintang yang paling hampir dengan kutub utara langit, kedudukan bintang kutub adalah kurang 1° dari kedudukan kutub utara langit. Bintang kutub boleh di jadikan panduan untuk menentukan kedudukan kutub utara langit. Manakala bagi hemisfera selatan,kutub selatan sfera langit tidak dapat dirujuk kepada mana-mana bintang.

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