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9702 AL Physics Revision Contents Motion in a Circle ...................................................................................................... 2 Gravitational Fields .................................................................................................... 8 ............................................................ 12Thermodynamics .................................................................................................... 28 Ideal Gases ............................................................................................................. 36 Alternating Currents ................................................................................................ 80 ..................................................... 100 Page 1
MOTIONIN1 Angular Displacement () If: Figure then sand are related to each other by the equation: ………. (i)where is the (magnitude of) ar displacement of the body expressed in radians (rad). For one = 2rr= 2 radians Likewise, when arc length is equal to the radius of the circle (i.e. r),calculated as: rs= 1 radianSo, one radian (of the circle) by an arc equal in length to the SI unit of angle).Angular displacement θ: = Centripetal acceleration:: one radian is the angle subtended at the centre (of the circle) by an arc equal in length tothe radius of the circle.Tf1=T2frc=a rCentripetal force:rFc=rFmPage
MOTIONINACIRCLE2 Angular Speed () Figure 15.2then the linetand its angular speed is given by the expression: =tSo, angular speed may be defined as the angle swept out (by the radius) per unit time. [1 or 2]It can be shown that: Note The angular speed (garded as the angular frequency (of the revolving body). T), Frequency (Period (), frequency () and angular frequency (equations: TTNote f are both frequencies. The only difference between them is that f is measured in cycles per second (i.e. Hz), whereas Page 3 of 132
3 Uniform Circular MotionFigure 15.3A body is said to execute uniform circular motion if it moves in a circle with constantac)In uniform circul[2]ExampleFigure 15 gravitational force gtowards the centre of the orbit. So, must also be directed towards the centre of the orbit. r2and:a rc=where rv is the angular speed of the body in uniform circular motion (see Fig. 15.3). MOTIONINACIRCLEPage of 132
MOTIONACCentripetal Force(.3, the resultantdirected towards the centre of the circle, and it is known as the centripetal force.The magnitude of centripetal force is given by the expressions: rc2and:crmF2=where mis of the body in uniform circular motion (see Fig. 15.3). ExampleIf: Figure 15.5f) between the cube and turntable surface to angular speed () SolutionAs the cube executes uniform circular motion, so for the cube: resultant force = centripetal force f(The weight of the cube and the normal reaction acting on it cancel out; so the resultant force acting f only) Page 5 of 132
4Vertical Circular MotionFigure 15.The motion of a mass on a string in a vertical circle must satisfy the constraints of centripetal force to The force along the string provides the centripetal force:mgθθmg cos =rCB =r°mgcos180° =rTr2=rmv- mgB and D, the angle °θ = 270°, cosθ = 0.TrA mass on a string:A person is on a Ferris wheel which is rotating at a constant speed. At the top and bottom of the and a normal force N. The resultant force It always points towards the At the top of the circular arc,r- mgrmv2MOTIONC6 of
ExampleIf: Figure 15.7at which or L, is the string mSolutionFor the stone resultant force = centripetal force TrTH =rmv–W.......... (i)Now, for the stone at L:resultant force = centripetal force L–rTLrmv2+ W.......... (ii)L> THAs tension in the string at L (TL) is greater than that at H (THMOTIONINCPage 7
GRAVITATIONAL FIELDS 1 Gravitational FieldFigure .1Gravitational field is the region of space where a mass experiences a force. Representation of GravitationalFigure .(a) 16.2 (b) are radial, and appear to come from the centre of the sphere. So,for points outside the sphere, the mass of the sphere may be considered to act as a point mass at[2]At a point, the direction of gravitational field line indicates the direction of force on a small mass.[1] gravitational field strengthA mass mM, the gravitational force provides the required centripetal force.=Fg1 2=Gm mFr22=22rFor mass –GMmTotal energy: Etotal = GPE + KE = –2GMmrKE = mv =2r21Page of
GRAVITATIONALFIELDSGravitationalFigure (where Fis the force actingSo, at a point, gravitational field strength may be defined as: forcegravitational field strength =mass[1]’s Law of GravitationIf: Figure .4then by Newton’s 3rd law of motion:F1= FF2) andF21rCombining the above relationships gives: F122mm1where G is the gravitational constant (6.10–11 N m2– masses is directly proportional to the product of their29
3 Circular Motion of Satellites For a satellite orbiting the Earth as shown below: Figure .5c). So: F2=r2m rrwhere = 2f2=Twhere fis and Tthe satellite.??????2RKE E−??????−??r 21KE mv =2GMmenergy: ETotal2 = –r, so gravitational potential energy: GPE = = –r2rTGMr rThe orbital period T is related to the radius r of the orbit: 3Energy of the sGRAVITATIONALFIELDS
GRAVITATIONALFIELDSFigure 16 in the plane of the Equator 2NoteAs the Earth's rotates from west to east, so when a geostationary satellite is to be launched into space, it is launched in the direction of Earth's rotation. That way, smaller energy is required to put the [2]Geostationary Satellite 11
ODefining equation of s.h.m.:(magnitudeDisplacement of a simple harmonic oscillator changes with time:=0 0=s.h.m.2v= xcos (v x0 0=Velocity-Displacement for s.h.m.::when xxwhen 0Energy in s.h.m. (in a spring-mass system):U== 21U=mxkinetic energy:2k1=2Emv(where 2 2=0vxx )2 2120E = U + E2=)2Page
SCILLATIONS1Free, Figure The motion of a spring-mass system oscillating on a frictionless surface (with the electric oscillator turned off) is an example of free oscillation. no[2]NotesThe frequency at which free oscillation occurs is called natural frequencyDamped Oscillation (with the electric oscillator turned So, a body (or a system) is said to execute damped oscillation if the amplitude of subsequent No[1]damped oscillation, the amplitude of oscillation decreases because energy flows out of thesystem in the form o[1 or 2]As damping increases, the frequency of oscillation decreases. So, for an oscillating system, thefrequency of damped oscillation is always less than the frequency of free oscillation (i.e. naturalfrequency of the system). [1 or 2]Forced Oscillation oscillator turned on, is an example of forced oscillation. a system) is said to execute forced oscillation if (periodic) driving force. [1]NoteForced oscillation always occurs at the frequency of the driving force, known as the (e.g. the frequency of the electric oscillator). of 132
2If: Figure 18nd law to the motion of the mass Fma………. (i) F= kkxa–xmkax………. (ii)ic motion is a type of (free) oscillation in which: 1 acceleration (ax) from equilibrium position; 2 acceleration and displacement are (always) In Eq. (ii), the negative sign ‘–’ indicates that acceleration (a) and displacement (xDefiningIt can be shown that (for all simple harmonic motions) the ‘constant’ part in equal to 2 is the angular frequency of oscillation). So, Eq. (ii) may be written as:Now, for the spring-mass system (shown in Fig. 18.2): mk= 2m= (2fmk=2where f is the frequency and TEqcharacteristic) equation of simple harmonic motion. i), it follows that the magnitude of 2a x0 0where x=.14 132
OSCILLATIONSa-Graph inthe inhas(which is –below:Figure 3 Some Important Terms Related to OscillationsDisplacement (xAmplitude ()TIt is the time taken for one oscillation. [1]It is the number of oscillations (occurring) per unit time. [1]Angular Frequency (and the direction of motion of the oscillator. (Definition of phase is not included in the syllabus) Page
4 Displacement (x) in Simple Harmonic MotionFigure .4) of a simple harmonic oscillator changes (periodically) with time (=0twhere x0 is the amplitude, is the angular frequency, and is the phase constant. Determining the Value of Phase Constant () ) may have any value from 0° to 360° (depending upon the position of the oscillator at = 0Figure .5 = 0° if the cycle is taken to start from A (i.e. at tx).t t = 0, 0 with the oscillator moving in thenegative direction). = 0, x).Worked Example Fig. 18.6 shows the variation with time (tx) of a simple harmonic oscillator. Figure 18.6Use Fig. 18.6 to = 3.4 s. Solutionxx0sin (t) x0 =T2=22= rad s–1and: = 180° (Fig. 18.6 shows that at t = 0, x = 0 with the body moving the –ve direction) (180 3.4 + 180) 4.76 OSCILLATIONS16
5 Velocity (v) in Simple Harmonic MotionFigure 18vofe harmonic oscillator also changes (periodical) with time (), and is given by the expression: v0cos ( is the angular frequency, xexpression for velocity of simple harmonic oscillator reduces to: From the above equation, it follows that the magnitude of maximum velocity of a simple harmonic oscillator is given by the expression: =Velocity-Displacement Graph forSimple Harmonic MotionIf the displacement () of a simple harmonic oscillator (at a given instant) is known, then the the expression: 22=v x xFrom the above equation (and Fig. 18.7), we can draw the following v vs. Figure OSCILLATIONS17 132
SCILLATIONSWorked 18.9 shows the variation with time (t) of a simple harmonic oscillator. Figure 18On Fig. 18time () of the velocity (v) and acceleration (As the gradient of displacement-time graph is velocity, and the gradient of velocity-time graph is acceleration, so velocity-time and acceleration-time graphs (drawn from follows: 132
O6 Energy in Simple Harmonic MotionFigure 10When the mass is at point P, the elastic PE (21=(where: k)22………. (i) is the spring constant and of= x+ Now, the KE of at 2k1(where: 202 2 2………. (iiE) of the oscillator (also known as the vibrational energy) at point P (or at ) is given by: E= UIt can be shown that: 2=2E m x………. (iii22=2(where Ek0= k2=(where U0)Figure .11Notes As = 200
The displacement (x) and velocity (v) of a simple harmonic oscillator at a point may be positive (+) velocity squared is always positive. and KE of the oscillator at any point, as given by Eqs. (i) and (ii) respectively, are alsoalwaysWorked ExampleIf a simple harmonic oscillator undergoes damped oscillation as indicated by its displacement-time Figure .12hen222E(as: Ex21OSCILLATIONSPage 20 132
OSCILLATIONSSolution Note that, at the point where U = 0 and EE; whereas at the points where = x0x= : UU0 and Ek = 0. So, Et-t graphs (drawn from the x-tWorked Example 2Fig. t) of the displacement () of a simple harmonic oscillator. Figure
OSCILLATIONS7 Damped Oscillation. [2].1Figure 18.14t we can change the magnitude of dissipative forces (and hence the degree of damping) the oscillator in Fig. 18.1. The mass is displaced from the equilibrium position and released from point A (at time t = 0). For different degrees of damping, we obtain different displacement-time graphs as shown in Fig. 184. Figure .15In light damping, the amplitude of subsequent oscillations decreases graduallyoscillation is just sfrequency of the system when it oscillates freely). it in the equilibrium position, takthe pointer to come to rest before he could take the next reading. the amplitude of oscillation decreases because the energy flows out of the[1 or 2]The frequency of oscillation decreases with the increase in damping. So the frequency of dampedoscillation is always less than the frequency of free oscillation (i.e. natural frequency). [1 or 2]Page
8 ResonanceA body (or system) is said to execute forced oscillation if it is made to oscillate by an external(periodic) driving force (e.g. pushes from an electric oscillator)theforced frequencyonoscillator). Let us consider a spring-mass system attached to an electric oscillator, and executing forced oscillation at different frequencies as shown in Fig. 186. Figure 18For the spring-mass system in Fig. 18.16the amplitude vs. forced frequency graph is as shown Figure 18.17So, when (in forced oscillation) the forced frequency, also known as driving frequency (fd), equals the f; this enomenon is called resonance. [2]Effectsof Damping on ResonanceFigure Note that, Useful and Harmful Effects Useful effects of resonance Harmful effects of resonance 1 It is employed in the production of ultrasound. employed in MRI scanning. It may cause (hanging) bridge to collapse. OPage 23
TEMPERATURE 1 Some Important Terms and Equations Related to Temperature Heat(or )Figure 13.1(or body) of lower temperature. [1]Notes The net rate of heat transfer from A to B decreases with the decrease in temperature differencebetween A and BThermal EquilibriumFigure .2 transfer of heat between [1]Relationship between kelvin Temperature (TKK = + 273.15Specific heat capacity is defined as the amount of heat required to raise the temperature of unitmass of the substance by one degree (or by 1 K).Specific latent heat of fusion is defined as the amount of heat required to change the state of unitmass of the substance from solid to liquid at constant temperature (i.e. melting point).mTc=LvLThe internal energy of a body (or a system) is the sum of a random distribution of kinetic andqwThe work done , on/by the system is given by: w =
TEMPERATUREAbsolute ZeroIt is the temperature at which the KE of molecules becomes zero (i.e. molecular motion ceases). Absolute (or Thermodynamic)Scale of Temperature[1]molecules of a substance increases with its temperature. So, at a particular thermodynamic temperature, the average of molecules is the same for Relationship between kelvin Temperature (T) and Celsius Temperature (T T= + 273.15From the above equation, it follows that: where ∆TK is the change in TKThermometric PropertyA physical quantity that varies with temperature is called thermometric property. [1]inare the length of liquid column and resistance of thermistor respectively. A desirable (or ideal) therFigure Page
2 Thermocouple Thermometer Figure 13.4 assumed that the ) induced in a thermocouple is directly proportional to the Δθ) between its junctions: EDetermining Temperatureiquidwith thethe boiling water (at 100 °C). E100) induced in the thermocouple. 3 Now immerse the hot junction in the liquid of unknown temperature ().θ100×=100EExampleIf: Figure ure 13 (in °C). Δθ21EE=100(cross-multiplication method) 2024=100=20×10024= 183 °CTEMPERATURE
TComparison between Thermocouple and Thermistor ThermometersThermocouple13Its least count is 1 °C°C).It can record rapidly changing temperatures (i.e. it is very responsive to temperature change). Its least count is sensitive than thermocouple.It cannot do so.It cannot do so.Notes substance.When a thermometer is used to measure the unknown temperature of a body, heat transfer takesThe responsiveness of a thermometer is concerned with how long it takes for the thermometrichigher the responsiveness of the thermometer. [2]27 of 132
T1 Specific Heat Capacity (cIf: Figure .1tqΔT(provided and: (provided ΔCombining the above proportional relationships gives: qmΔqcmwhere c is the specific heat capacity of the substance, given by: TmSo, the specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of unit mass of the substance by one degree (or by 1 K). [2] heat capacity of a substance is the amount of heat required to raise the temperature of one moleV=qwhere q is the heat required to raise the temperature of moles of the gas by at constant volume. where R is the molar gas constant. cPage of
HERMODYNAMICS2 Specific Latent Heat (L) 2If: Figure 14then the experimental results show that: =where LffqLSo, the specific latent heat of fusion of a substance is defined as the amount of heat required to change the state of unit mass of the substance from solid to liquid at constant temperature(Lv)Figure .4mq L mvwhere Lvv=qmSo, the specific latent heat of vaporisation of a substance is defined as the amount of heat required to change the state of unit mass of the substance from liquid to vapours at constant temperature 3]Generalof unit mass of substance at constant temperature. 2 or132
3 Determining and Electrical MethodCapacity (a Substance Figure 14.2.m–31 of 4 and ammeter reading I5 Note the final reading T2. write: heat supplied by heater = heat absorbed by substance PtcmΔwhere P= T2T1is the increase in temperature of the substance) 6 Use the above equation (in step ) cExampleIf an electric heater of power Ptat , then write an equP, Liand wAssuming that no heat is lost to the surroundings, we can write: = + q + =TL mf+Ptc m(100)+c m(25w
THERMODYNAMICSDetermining Specific Latent Heat of Fusion (LIceFigure .5Steps1 Build the experimental setup as shown in Fig. a is, equal amounts of ice melt in equal intervals of time. In other words, the reading on the digital Actually, melting occurs at a steady rate when the temperature of the heater becomes stable (i.e. ) andthe[1]3on the digital balance. 4 Calculate mass mtNote1final reading on balance 7 Increase the current in the circuit by decreasing the resistance of the variable resistor. Repeat steps 2to , and obtain another set of readings of I2V and mNote: e is kept the same) heat supplied byheater in time + = tSo, for the two sets of readings obtained above, we can write: 1 s f 1………. (i)P t q L m………. (ii)where P1I1 and P2= I2. P t P t L m L m2 1 f 2 f 1 2 1 f 2 1( ) = ( )………. (iii)132
Determining Specific Latent Heat of Vaporisation (Lv)Figure .6Steps Build the experimental setup as shown in Fig. 143 Start the stopwatch for predetermined time period .tm17 Increase the current in the circuit by decreasing the resistance of the variable resistor. 86, I2Note: theater in time = heat absorbed bywater in time t+ heat lost to theSo, for the two sets of readings obtained above, we can write: P t L m q1 v 1 s………. (i)………. (ii)where I and I2(i) from Eq. (ii) gives: P t P t L m L m ( ) = ( )P P t L m m………. (iii)vTHERMODYNAMICSPage
THERMODYNAMICS4 Internal Energy (U) The internal energy of a body (or a system) is the sum of a random distribution of kinetic and potential energies associated with its molecules; i.e. UKEincrease in temperature.[1]molecules. In other words, heating either changes the temperature, or the state of the substance5 Simple Kinetic Model of Matter Why strong and rigid. hy mel, the soPE of molecules increases); however, the mean, which is related to the temperature, remains the same, and so does the temperature. [3]same substancepushing back the surrounding atmosphere. [3]hy a cooling effect accompanies evaporationDuring evaporation, more energetic molecules escape from the surface of the liquid. As a result, the the temperature. of 132
6 First Law of ThermodynamicsFigure 14.7………. (i)Eq. (i) is actually the standard mathematical form of the first law of thermodynamics. So, the first law of thermodynamics may be stated as: increase in internal energy of a system = heat supplied toon the system [2]The work done, pw p Vwhere ΔV is the change in the volume of the system. Sign Conventions forof ΔUq w + when U of the system when heat flows into the system when volume of the system –when Uwhen heat flows out of the system when volume of the system 0 when U of the system when no heat flows into or out of the system when volume of the system THERMODYNAMICSof
THERMODYNAMICS3 For an ideal gas: Ukits state such that its temperature remains constant (i.e. ΔT = 0), then: Δ = 0 so, the equation of the first law of thermodynamics takes the form: qw 4on the path followed to get the system to that state). So a system, after undergoing a series of changes, returns to its initial state (i.e. completes a completes a cycle (on its V graph) alon.8 (below): Figure 14 = 0 AB BC CANoteThe overall change in the internal energy of an ideal gas in one complete cycle is zero, as the gas returns to its original temperature (i.e. final T = initial T).Very Helpful oncept forFigure 14.9 Applying the First Law of Thermodynamics to Different Situations 1 When a system is completely insulated, OR when it undergoes a sudden change so that no heat can enter or leave it, then: q = 0 Δ2 During melting and freezing processes, the volume of the specimen (i.e. system) remains almost 0 (as ΔV 0 here) So, for melting and freezing processes, the equation of to:Page 35 of
IDEAL GASES 1 Equation of State for an Ideal Gas Figure 12pV=nRT………. (i)where Ris–1A=NNn(where N023mol–is ndA(where J K–)nANpVNkT………. (ii)Eqs. (i) and (ii) are just two different forms of equation of state for an ideal gas. Ideal Gasconstant): = constant pVorTpV= constant So, an ideal gas is the gas that obeys the relationship: at all values of p, Vand T; where ppV, where pVTthermodynamic temperature of gas, mol)pV is number of gas molecules, is the Boltzmann constant (1.38 × 10 J K, where <c2> is the mean-square speed of the gas molecules12=3pV Nm c23= 22c2= 32236 132
Notes An ideal gas may also be defined as the gas that obeys the equation:nRTat all values of p, and T; p is the number of moles and T or 3]All real gases behave as ideal gases at [1]Exampleshown below: Figure 12.2then, for the gas, we can write: 1 1 2 2=Note that the above equation holds true only when: 1 the amount of is fixed (i.e. and n are unchanged) 2 temperatures 2 are both expressed in kelvin (K). Avogadro Constant (N) of atoms in 12 g of carbon-12.mole containing 6.02 23 in a sample of a substance can also be determined by the expression: M=where m is the mass, in grams, of the sample, and Note and that of oxygen molecules (O) is 32 g). IDEALG37 132
IDEALGASES2 Brownian Motion It is the continuous random motion of microscopic particles suspended in a fluid (i.e. gas or liquid). [1]It is caused by the boit provides evidence for the existence and movement of molecules. Figure 12.3 (a) Figure 12.3 (b) (i) Temperature As the temperature of fluid increases, the average K.E. of its molecules also increases, so the collisions of molecules with the suspended particles become more vigorous. Consequently, the Brownian motion increases (i.e. becomes more haphazard).[1 or 2]ic particles (of the same mass) exhibit less haphazard Brownian motion. This is the number of molecules colliding with them also increases, and as a result, the randomness of collisions tends to average out (just like random [1 or 2]Figure 12.4 The gas molecules are in continuous random motion; that is, they move with a range of speeds in different directions. 2, identical spheres (i.e. they undergo elastic collisions with each other and with the walls of the container). [1]3molecules is negligible compared with the time between two successive collisions). [1] The forces between gas molecules are negligible (except during collisions). [1]38
4 Pressure of an Ideal Gas Qualitative AnalysisFigure 12.5 (a)Figure 12gas molecule collides with the wall of the container, it rebounds and thus undergoes a momentum, which is actually caused by the force exerted on it by the wall (= ).tThe molecule also exerts an equal but opposite force onthe wall (Newrd).constant over time, and gives rise to the pressure of the gas (as pF/ IDEALGASESof 132
GASESQuantitative AnalysisFigure 12.6then the density of the gas is given by: VNm=the mean separation between the gas molecules is given by: 1=and the pressure of the (ideal) gas is given by: 12=3Nmp cVwhere2cis the mean-square speed of the gas molecules (i.e. mean of the squared speeds of gas molecules). Putting=NmVinto the above equation gives: 223pc2=orpcrms=where crms in pressure as long as the temperature of the gas remains constant. Actually, when the pressure (p) increases (at constant temperature), the density (rms remains ) of the gas is doubled (at constant temperature), the V) of the gas becomes half (Boyle’s law: pV
Derivation of the Expression for Pressure of an Ideal GasFigure 12.7then the change in the momentum of the molecule when it undergoes elastic collision with the shaded face of the container is given by: 1= mvx– = 2given by: =L=Now, the mean force (F1ule on the shaded face is calculated as:( )1= =2xxpmvF21=xand) exerted by all Nmolecules on the shaded face is given by: 2< >FLwhere <vx2 gas molecules is the direction. By symmetry:2 21So the total mean force exerted by all 21 < >=Land hence the pressure of the gas is given by: 2A LA(where LAV12Nmp v[6]IDEAL132
5 Average Kinetic Energy of a Gas Molecule pVNkT 2=gives: 123= 2= 3kT………. (i)2= 3In the above equation,212m cis the average translational K.E. of a gas molecule; so we can write: k3=E kT[3 or 4]Notes2krms1From Eq. (i), it follows that, for a given ideal gas:2T2TT………. (ii)ExampleIf an ideal gas undergoes a change in its temperature, as shown below: Figure 12t, 2IDEALG
DEALGASES6 Internal Energy (UThe internal energy of a system is the sum of a random distributionassociated with its molecules. [2]U = KE + PE system respectively. Internal Energy of an Ideal Gas the forces between molecules of an ideal gas are negligibleso the potential energies gas molecules are also negligible. This implies that the internal energy of an ideal gas is solely the kinetic energy of its molecules, i.e. U = KE ………. (i)Now if: Figure .9then the total (translational) kinetic energy of the gas molecules may be expressed as: Nk3×2(where k is the Boltzmann constant) 3=Page of
E1 Electric FieldFigure .1 experiences a force. Representation of Electric Field Figure .1.2 (a) Figure 17.1Figure 17.1.2 ()1.2 ()forceElectric =FVCoulomb's Law: the force between two point charges is directly proportional to the product of thecharges, and inversely proportional to the square of their separation.1r, where 00–12 F m–2E4rE=lectric potential at point is defined as the work done per unit positivecharge in bringing a smalltest charge from infinity the point.Electric potential due to a point charge (Q): 01=, where the distance of the point from Figure 17.?ൌെ∆?∆?44
negative charge (or plate).At a point, the tangent to the electric field line (in the direction of the field) indicates the directionsmall positive17.1(d)).[2].2 (e)(i.e. as the are parallel [1]Electric Field Strength(E)If: Figure 17.1then the of) electric Q) is given by: =E(where F)force[1 or 2] The field lines in Fig. 17.1.2considered to act as a pointIn Fig. 17.1.2 , there is no electric field inside the spherical conductor (i.e.Electric Field Strength (Fig. 17.1Figure .1.then the (magnitude of) electric field strength anypoint between =Eandat anyby: F=qVFdELECTRIC132
EFIELDSWorked Example a of mass mcharg+plates (as n Figure net=Fm(F ma=) =a(where F is the electric force) =am( / )=q V dNotesIn the above example, the weight of the particle has been assumed to be negligible (compared tothe electric force The charge on an electron is –1e (where e–19The charge on an alpha particle is +2u (where 1u.Worked Example 2Fi.1.6 of uniform electric field on the motion of a charged particle ht angles to it. Figure .1.6NotesThere is no horizontal force acting on the particle; so, horizontally, velocity (u) is constant.).The motion of the particle inside the fieldsmotion of the particle (inside the field).The initial vertical velocity of the particle (at A) is ‘0’.of 132
Worked Example 17.1.7 shows twd a d. the . Figure 17 mass mmand V, for the velocity of the particle with which it hits the [2 or 3]The system comprising the plates and the particle is an isolated system. So, applying the principle of ∆EkU……… (i)where ∆Umoves from one plate to the other), given by U qVPutting the expressions for ∆EU into Eq. (i) gives: 20 =22vmMethod 2The system comprising the particle alone is a non-isolated system. So, using the work-and WUU = 0 for a single-particle system) 21= 02(where F is the electric force acting on the particle) 2. =2(as Ed) 2ELECTRICF
2 If: Figure .2rd law motion:1–F2The experimental results show that: FqqFq1qa21Combining the above relationships gives:1 2qqr1 224q qFr(where 01= 8.99 109)where 0 is the permittivity of free space (8.10 F mtates that the force between two point charges is directly proportional to the [2 or 3]3 Electric Field Strength (EQIf: Figure 17.2E2)2/Qq r2QE………. (i) of electric field strength at a point in the field of aQ).Eq. (i) can also be used to calculate the magnitude of electric field strength at a point in the fieldof a charged conducting sphere (just remember that there is no electric field inside a chargedconductor, and for any point outside a charged conducting sphere, the charge on the sphere maybe considered to act as a point charge at its centre).44ELECTRICFIELDSPage
4 Electric Potential () due to Point Charge (Example 1 +)If: Figure then electric pot+(where W is work done by F charge in bringing a small [2]It can be shown that electric potential at a point (P) due to a point charge (Q) is given by0Q………. (ii)[1 is the distance of the point (P) from the point charge (Q(Q–Figure then electric potential at point P is given by the expression: 04V(where Qis ‘–ELECTRICFIELDS132
Notes Electric potential (V) at a point in the field of a point charge (Q).Er V………. (iii)Electric Potential Energy (UFigure .8tial energy stored in charge at point P is given by the expression: U qV=(where:0V) 0= .4QU q0=4QqUr………. (iv)[1]) is positive for like charges (i.e. qQELECTRICof 132
